Mixing
Find a point $X$, in the plane of regular pentagon $ABCDE$, that minimizes $frac{XA+XB}{XC+XD+XE}$.
$begingroup$
Find such a point $X$, in the plane of the regular pentagon $ABCDE$, that the value of expression $$frac{XA+XB}{XC+XD+XE}$$ is the lowest.
I tried using Ptolemy's theorem but don't know how to make use of inequalities it gives.
I'd be really grateful for any help :)
geometry optimization euclidean-geometry maxima-minima geometric-inequalities
edited Jan 19 at 13:57
the_fox
2,90021537
asked Jan 19 at 8:36
SamHarSamHar
776
$endgroup$
|
show 6 more comments
$begingroup$
Find such a point $X$, in the plane of the regular pentagon $ABCDE$, that the value of expression $$frac{XA+XB}{XC+XD+XE}$$ is the lowest.
I tried using Ptolemy's theorem but don't know how to make use of inequalities it gives.
I'd be really grateful for any help :)
geometry optimization euclidean-geometry maxima-minima geometric-inequalities
edited Jan 19 at 13:57
the_fox
2,90021537
asked Jan 19 at 8:36
SamHarSamHar
776
$endgroup$
$begingroup$
Is X inside of ABCDE? Othwerwise there is no such point...
$endgroup$
– Jakub Andruszkiewicz
Jan 19 at 8:44
$begingroup$
It was not stated in the problem. Why such point does not exist outside $ABCDE$?
$endgroup$
– SamHar
Jan 19 at 8:51
1
$begingroup$
The minimum value is $BA/(EA+DA+CA)$
$endgroup$
– Cesareo
Jan 19 at 11:57
4
$begingroup$
@i. m. soloveichik In fact, numerical simulations show that the whole "little" arc $AB$ is solution, i.e. one can take any $X (cos(a),sin(a))$ with $-frac{pi}{5} leq a leq frac{pi}{5}$.
$endgroup$
– Jean Marie
Jan 20 at 21:21
3
$begingroup$
@SamHar: What is the source of this problem? Knowing that would help readers gauge the intended level of difficulty.
$endgroup$
– Blue
Jan 22 at 5:52
|
show 6 more comments
$begingroup$
Find such a point $X$, in the plane of the regular pentagon $ABCDE$, that the value of expression $$frac{XA+XB}{XC+XD+XE}$$ is the lowest.
I tried using Ptolemy's theorem but don't know how to make use of inequalities it gives.
I'd be really grateful for any help :)
geometry optimization euclidean-geometry maxima-minima geometric-inequalities
edited Jan 19 at 13:57
the_fox
2,90021537
asked Jan 19 at 8:36
SamHarSamHar
776
$endgroup$
Find such a point $X$, in the plane of the regular pentagon $ABCDE$, that the value of expression $$frac{XA+XB}{XC+XD+XE}$$ is the lowest.
I tried using Ptolemy's theorem but don't know how to make use of inequalities it gives.
I'd be really grateful for any help :)
geometry optimization euclidean-geometry maxima-minima geometric-inequalities
geometry optimization euclidean-geometry maxima-minima geometric-inequalities
edited Jan 19 at 13:57
the_fox
2,90021537
asked Jan 19 at 8:36
SamHarSamHar
776
edited Jan 19 at 13:57
the_fox
2,90021537
asked Jan 19 at 8:36
SamHarSamHar
776
edited Jan 19 at 13:57
the_fox
2,90021537
edited Jan 19 at 13:57
the_fox
2,90021537
edited Jan 19 at 13:57
the_fox
2,90021537
2,90021537
asked Jan 19 at 8:36
SamHarSamHar
776
asked Jan 19 at 8:36
SamHarSamHar
776
asked Jan 19 at 8:36
SamHarSamHar
776
776
$begingroup$
Is X inside of ABCDE? Othwerwise there is no such point...
$endgroup$
– Jakub Andruszkiewicz
Jan 19 at 8:44
$begingroup$
It was not stated in the problem. Why such point does not exist outside $ABCDE$?
$endgroup$
– SamHar
Jan 19 at 8:51
1
$begingroup$
The minimum value is $BA/(EA+DA+CA)$
$endgroup$
– Cesareo
Jan 19 at 11:57
4
$begingroup$
@i. m. soloveichik In fact, numerical simulations show that the whole "little" arc $AB$ is solution, i.e. one can take any $X (cos(a),sin(a))$ with $-frac{pi}{5} leq a leq frac{pi}{5}$.
$endgroup$
– Jean Marie
Jan 20 at 21:21
3
$begingroup$
@SamHar: What is the source of this problem? Knowing that would help readers gauge the intended level of difficulty.
$endgroup$
– Blue
Jan 22 at 5:52
|
show 6 more comments
$begingroup$
Is X inside of ABCDE? Othwerwise there is no such point...
$endgroup$
– Jakub Andruszkiewicz
Jan 19 at 8:44
$begingroup$
It was not stated in the problem. Why such point does not exist outside $ABCDE$?
$endgroup$
– SamHar
Jan 19 at 8:51
1
$begingroup$
The minimum value is $BA/(EA+DA+CA)$
$endgroup$
– Cesareo
Jan 19 at 11:57
4
$begingroup$
@i. m. soloveichik In fact, numerical simulations show that the whole "little" arc $AB$ is solution, i.e. one can take any $X (cos(a),sin(a))$ with $-frac{pi}{5} leq a leq frac{pi}{5}$.
$endgroup$
– Jean Marie
Jan 20 at 21:21
3
$begingroup$
@SamHar: What is the source of this problem? Knowing that would help readers gauge the intended level of difficulty.
$endgroup$
– Blue
Jan 22 at 5:52
$begingroup$
Is X inside of ABCDE? Othwerwise there is no such point...
$endgroup$
– Jakub Andruszkiewicz
Jan 19 at 8:44
$begingroup$
Is X inside of ABCDE? Othwerwise there is no such point...
$endgroup$
– Jakub Andruszkiewicz
Jan 19 at 8:44
$begingroup$
It was not stated in the problem. Why such point does not exist outside $ABCDE$?
$endgroup$
– SamHar
Jan 19 at 8:51
$begingroup$
It was not stated in the problem. Why such point does not exist outside $ABCDE$?
$endgroup$
– SamHar
Jan 19 at 8:51
1
1
$begingroup$
The minimum value is $BA/(EA+DA+CA)$
$endgroup$
– Cesareo
Jan 19 at 11:57
$begingroup$
The minimum value is $BA/(EA+DA+CA)$
$endgroup$
– Cesareo
Jan 19 at 11:57
4
4
$begingroup$
@i. m. soloveichik In fact, numerical simulations show that the whole "little" arc $AB$ is solution, i.e. one can take any $X (cos(a),sin(a))$ with $-frac{pi}{5} leq a leq frac{pi}{5}$.
$endgroup$
– Jean Marie
Jan 20 at 21:21
$begingroup$
@i. m. soloveichik In fact, numerical simulations show that the whole "little" arc $AB$ is solution, i.e. one can take any $X (cos(a),sin(a))$ with $-frac{pi}{5} leq a leq frac{pi}{5}$.
$endgroup$
– Jean Marie
Jan 20 at 21:21
3
3
$begingroup$
@SamHar: What is the source of this problem? Knowing that would help readers gauge the intended level of difficulty.
$endgroup$
– Blue
Jan 22 at 5:52
$begingroup$
@SamHar: What is the source of this problem? Knowing that would help readers gauge the intended level of difficulty.
$endgroup$
– Blue
Jan 22 at 5:52
|
show 6 more comments
3 Answers
3
active
oldest
votes
$begingroup$
We use Ptolemy inequality : For an arbitrary quadrilateral $PRQS$ we have: $$PQcdot RS leq PRcdot QS+PScdot RQ$$
Let $AB = s$ and $AC=m$ and $XA =a$, $XB =b$... Now we use this inequality on $3$ different quadrilaterals:
$bullet$ $AXBE$: $$ esleq bs+am$$
$bullet$ $AXBC$: $$ csleq as+bm$$
$bullet$ $AXBD$: $$ dsleq am+bm$$
Adding these three we get $$ s(c+d+e)leq (a+b)(2m+s)$$so we have $$ {sover 2m+s}leq {a+bover c+d+e}$$
So your expression is always $geq {sover 2m+s}$ and this value is achieved if we put $X=A$ or $X=B$ or at any point on arc between $AB$ on circumcircle for $ABCDE$ (Remember that equality is true iff all inequalities become equalities and that is if $AXBC$, $AXBD$ and $AXBE$ are cyclic with that order of points on circle).
edited Feb 1 at 0:08
Jean Marie
30.3k42153
answered Jan 23 at 19:09
greedoidgreedoid
45k1157112
$endgroup$
1
$begingroup$
[+1] Very neat solution for points $A$ and $B$ ! It remains the case of points of arc $AB...$
$endgroup$
– Jean Marie
Jan 23 at 20:48
1
$begingroup$
Well that can be easly done with Ptolomy theorem. :)
$endgroup$
– greedoid
Jan 23 at 20:49
$begingroup$
OK. Clearly, nobody can compete with your solution ! I am going to present modestly an intuitive reasoning that could be helpful for this kind of issues.
$endgroup$
– Jean Marie
Jan 23 at 22:10
1
$begingroup$
Great solution , but aren’t we yet to prove that all the points which satisfy the equality condition lie on arc $AB$ ? We seem to have proved that the points lying on arc $AB$ satisfy the equality condition
$endgroup$
– Sinπ
Jan 24 at 4:01
add a comment |
$begingroup$
What follows is not a solution (an excellent one has been given by @greedoid) but a heuristic method that provides an efficient angle of attack of this locus and loci of the same "flavour". This presentation has a handwaving side, but I think that this kind of intuitive reasoning can have some interest, in particular for students in mathematics in order to convey the spirit of discoverers of infinitesimal calculus.
Let :
$$q(X):=frac{n(X)}{d(X)}, text{with}$$
$$n(X):=XA+XB text{and } $$
$$d(X):=XC+XD+XE$$
Remark :
$lim q(X)=frac23$ when $|overrightarrow{OX}|$ tends to $infty$.
Consider figure $1$ that depicts some contour lines of functions $n$ (in blue) and $d$ (in red) :
Fig. 1. : To each point $X$ in the plane is attached two curves, a red one and a blue one giving the locus of points having resp. the same numerator $n(X)$ and the same denominator $d(X)$. One shouldn't be surprized that the blue curves are confocal ellipses (with foci $A$ and $B$).
What usage can we do of these two families of curves ?
A first result is that a point $X$ can be a minimum of function $q$ (or at least a relative minimum), if the blue and the red contour lines passing through $X$ are tangent ; otherwise, if these curves are transverse in $X$ (transverse = non tangent), there are two directions (see Fig. 2) one can take that are more advantageous.
Fig. 2 : A minimum of function $q$ cannot occur in a point $X$ with transverse intersection of the blue and red contour lines passing through this point : indeed, moving $X$ in one of the given directions decreases the value of $q$.
Caveat : tangency in $X$ of the red and the blue curves passing through $X$ is a necessary condition, not a sufficient one ; let us take the (counter) example of a point $X$ belonging to the interval $I=(-1,cos(pi/5))$ of the $x$ axis; moving $X$ to the right into $X'$ still in $I$, will also give $q(X')<q(X)$ (a particulary favorable case where $n(X)$ decreases while $d(X)$ increases!).
More generally, it is possible to eliminate all points of the $x$-axis (except point $(1,0)$) although blue and red curves are tangent in these points.
What are the remaining candidate points for the minimum of $q(.)$ ? A concrete example is provided by the tangency of blue curve indexed by $1.2$ (the most elongated ellipse) and the red curve indexed by $5.1$ [just at the right of vertical line segment $AB$]. A detailed look show that all solution points $X$ are in this area and constitute a continuum of points ; a more detailed analysis is needed to prove that this "continuum" is the small circular arc $AB$ of the unit circle. I haven't done it.
Besides, and this is a second idea, one can have access to a numerical solution using gradient computations.
Let us first recall that the value of the gradient of distance function defined by $f(M)=AM$, where $A$ is any fixed point of the plane, is the unit norm vector defined in point $M$ as :
$$overrightarrow{grad}(f)(M)=frac{overrightarrow{AM}}{AM}.$$
See Gradient of distance vector length and Appendix 2.
Thus the gradients in $X$ of functions $n$ and $d$ are resp.
$$overrightarrow{grad}(n)(X)=frac{overrightarrow{AX}}{AX}+frac{overrightarrow{BX}}{BX},$$
$$text{and} overrightarrow{grad}(d)(X)=frac{overrightarrow{CX}}{CX}+frac{overrightarrow{DX}}{DX}+frac{overrightarrow{EX}}{EX} tag{1}$$
The tangency condition will then be transferred as a gradients' proportionality. Moreover, we will switch to a complex function treatment that will give a more compact formulation :
$$arg(r(z-a)+r(z-b))=arg(r(z-c)+r(z-d)+r(z-e))+k pitag{2}$$
where
$$r(z):=frac{z}{|z|}$$
where $a,b,c,d,e$ stand for the complex numbers associated with $A,B,C,D,E$ resp. Indeed, having the same angle with $k$ even or opposite angles with $k$ odd means proportionality ; a further step is to replace (2) by a criterium on the imaginary part of the complex logarithm function (see line 8 in program below) :
$$operatorname{Im}(log(f(z))) text{with}$$
$$f(z):=frac{r(z-a)+r(z-b)}{r(z-c)+r(z-d)+r(z-e)} tag{3}$$
(recall : $log(re^{i theta})=log(r)+i theta$ for the principal branch of complex logarithm function). Why is this formulation interesting ? Because we have converted the search of minima into the search of zeroes of a certain function (more exactly, the minima constitute a subset of this set of zeroes : see explanation upwards about necessary and sufficient condition).
Let us use formulation (3) to obtain a graphical representation of the points under consideration : see figure $3$ below (which does not constitute - of course - a mathematical proof) where the circular arc AB appears as the solution.
Fig. 3. In black, all points $X$ having almost proportional gradients. See Appendix 1.
Appendix 1 : The Matlab program that has generated Fig. 3. reminder : this program generates all points for which this expression is close to $0$.
i=complex(0,1);
b=exp(i*pi/5);a=b^9;c=b^3;d=-1;e=b^7;% pentagon vertices
r=@(z)(z./abs(z));
dg=@(z)(r(z-c)+r(z-d)+r(z-e));
ng=@(z)(r(z-a)+r(z-b));
f=@(z)(ng(z)./dg(z));
[X,Y]=meshgrid(-2:0.005:2);
I=imag(log(f(X+i*Y)));% = imag(log(ng(X+i*Y))-log(dg(X+i*Y)));% if zero : gradients' proportionality
J=abs(I)>0.01;
imshow(J);
Appendix 2 : the gradient at point $X$ of function $n(.)$ for example is the inward or outward normal to the blue curve passing through $X$ (think to the blue curves as isobaric curves and to the gradient as depicting the wind's direction).
answered Jan 24 at 0:40
Jean MarieJean Marie
30.3k42153
$endgroup$
add a comment |
$begingroup$
If $X$ is on $AB$, $AX+BX$ is minimum. And if $X=A$ , $CX+DX+EX$ is maximum. So cesareo's answer follows.
answered Jan 22 at 8:53
Takahiro WakiTakahiro Waki
2,099620
$endgroup$
$begingroup$
This answer appears to be incorrect. It isn't necessary to minimize the numerator of the expression. In fact, the gradient of the numerator is zero on $AB$, whereas the gradient of the denominator is not, so it's "free" to increase the denominator by moving at least a small distance away from $AB$. You may want to delete this answer so that the question still appears as unanswered, which should help to attract more attention.
$endgroup$
– Chris Culter
Jan 22 at 19:15
$begingroup$
@ChrisCulter I don't know what you say, denominator is maximum on X=A or X=B.
$endgroup$
– Takahiro Waki
Jan 23 at 2:08
$begingroup$
But $X$ is not restricted to the inside of the pentagon. There is no global maximum of the denominator.
$endgroup$
– Chris Culter
Jan 23 at 5:31
$begingroup$
@ChrisCulter That's bug on question. The original question shows "on the plane of pentagon...". I think this doesn't mean "Find X in the plane".
$endgroup$
– Takahiro Waki
Jan 23 at 5:48
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We use Ptolemy inequality : For an arbitrary quadrilateral $PRQS$ we have: $$PQcdot RS leq PRcdot QS+PScdot RQ$$
Let $AB = s$ and $AC=m$ and $XA =a$, $XB =b$... Now we use this inequality on $3$ different quadrilaterals:
$bullet$ $AXBE$: $$ esleq bs+am$$
$bullet$ $AXBC$: $$ csleq as+bm$$
$bullet$ $AXBD$: $$ dsleq am+bm$$
Adding these three we get $$ s(c+d+e)leq (a+b)(2m+s)$$so we have $$ {sover 2m+s}leq {a+bover c+d+e}$$
So your expression is always $geq {sover 2m+s}$ and this value is achieved if we put $X=A$ or $X=B$ or at any point on arc between $AB$ on circumcircle for $ABCDE$ (Remember that equality is true iff all inequalities become equalities and that is if $AXBC$, $AXBD$ and $AXBE$ are cyclic with that order of points on circle).
edited Feb 1 at 0:08
Jean Marie
30.3k42153
answered Jan 23 at 19:09
greedoidgreedoid
45k1157112
$endgroup$
1
$begingroup$
[+1] Very neat solution for points $A$ and $B$ ! It remains the case of points of arc $AB...$
$endgroup$
– Jean Marie
Jan 23 at 20:48
1
$begingroup$
Well that can be easly done with Ptolomy theorem. :)
$endgroup$
– greedoid
Jan 23 at 20:49
$begingroup$
OK. Clearly, nobody can compete with your solution ! I am going to present modestly an intuitive reasoning that could be helpful for this kind of issues.
$endgroup$
– Jean Marie
Jan 23 at 22:10
1
$begingroup$
Great solution , but aren’t we yet to prove that all the points which satisfy the equality condition lie on arc $AB$ ? We seem to have proved that the points lying on arc $AB$ satisfy the equality condition
$endgroup$
– Sinπ
Jan 24 at 4:01
add a comment |
$begingroup$
We use Ptolemy inequality : For an arbitrary quadrilateral $PRQS$ we have: $$PQcdot RS leq PRcdot QS+PScdot RQ$$
Let $AB = s$ and $AC=m$ and $XA =a$, $XB =b$... Now we use this inequality on $3$ different quadrilaterals:
$bullet$ $AXBE$: $$ esleq bs+am$$
$bullet$ $AXBC$: $$ csleq as+bm$$
$bullet$ $AXBD$: $$ dsleq am+bm$$
Adding these three we get $$ s(c+d+e)leq (a+b)(2m+s)$$so we have $$ {sover 2m+s}leq {a+bover c+d+e}$$
So your expression is always $geq {sover 2m+s}$ and this value is achieved if we put $X=A$ or $X=B$ or at any point on arc between $AB$ on circumcircle for $ABCDE$ (Remember that equality is true iff all inequalities become equalities and that is if $AXBC$, $AXBD$ and $AXBE$ are cyclic with that order of points on circle).
edited Feb 1 at 0:08
Jean Marie
30.3k42153
answered Jan 23 at 19:09
greedoidgreedoid
45k1157112
$endgroup$
1
$begingroup$
[+1] Very neat solution for points $A$ and $B$ ! It remains the case of points of arc $AB...$
$endgroup$
– Jean Marie
Jan 23 at 20:48
1
$begingroup$
Well that can be easly done with Ptolomy theorem. :)
$endgroup$
– greedoid
Jan 23 at 20:49
$begingroup$
OK. Clearly, nobody can compete with your solution ! I am going to present modestly an intuitive reasoning that could be helpful for this kind of issues.
$endgroup$
– Jean Marie
Jan 23 at 22:10
1
$begingroup$
Great solution , but aren’t we yet to prove that all the points which satisfy the equality condition lie on arc $AB$ ? We seem to have proved that the points lying on arc $AB$ satisfy the equality condition
$endgroup$
– Sinπ
Jan 24 at 4:01
add a comment |
$begingroup$
We use Ptolemy inequality : For an arbitrary quadrilateral $PRQS$ we have: $$PQcdot RS leq PRcdot QS+PScdot RQ$$
Let $AB = s$ and $AC=m$ and $XA =a$, $XB =b$... Now we use this inequality on $3$ different quadrilaterals:
$bullet$ $AXBE$: $$ esleq bs+am$$
$bullet$ $AXBC$: $$ csleq as+bm$$
$bullet$ $AXBD$: $$ dsleq am+bm$$
Adding these three we get $$ s(c+d+e)leq (a+b)(2m+s)$$so we have $$ {sover 2m+s}leq {a+bover c+d+e}$$
So your expression is always $geq {sover 2m+s}$ and this value is achieved if we put $X=A$ or $X=B$ or at any point on arc between $AB$ on circumcircle for $ABCDE$ (Remember that equality is true iff all inequalities become equalities and that is if $AXBC$, $AXBD$ and $AXBE$ are cyclic with that order of points on circle).
edited Feb 1 at 0:08
Jean Marie
30.3k42153
answered Jan 23 at 19:09
greedoidgreedoid
45k1157112
$endgroup$
We use Ptolemy inequality : For an arbitrary quadrilateral $PRQS$ we have: $$PQcdot RS leq PRcdot QS+PScdot RQ$$
Let $AB = s$ and $AC=m$ and $XA =a$, $XB =b$... Now we use this inequality on $3$ different quadrilaterals:
$bullet$ $AXBE$: $$ esleq bs+am$$
$bullet$ $AXBC$: $$ csleq as+bm$$
$bullet$ $AXBD$: $$ dsleq am+bm$$
Adding these three we get $$ s(c+d+e)leq (a+b)(2m+s)$$so we have $$ {sover 2m+s}leq {a+bover c+d+e}$$
So your expression is always $geq {sover 2m+s}$ and this value is achieved if we put $X=A$ or $X=B$ or at any point on arc between $AB$ on circumcircle for $ABCDE$ (Remember that equality is true iff all inequalities become equalities and that is if $AXBC$, $AXBD$ and $AXBE$ are cyclic with that order of points on circle).
edited Feb 1 at 0:08
Jean Marie
30.3k42153
answered Jan 23 at 19:09
greedoidgreedoid
45k1157112
edited Feb 1 at 0:08
Jean Marie
30.3k42153
edited Feb 1 at 0:08
Jean Marie
30.3k42153
edited Feb 1 at 0:08
Jean Marie
30.3k42153
30.3k42153
answered Jan 23 at 19:09
greedoidgreedoid
45k1157112
answered Jan 23 at 19:09
greedoidgreedoid
45k1157112
answered Jan 23 at 19:09
greedoidgreedoid
45k1157112
45k1157112
1
$begingroup$
[+1] Very neat solution for points $A$ and $B$ ! It remains the case of points of arc $AB...$
$endgroup$
– Jean Marie
Jan 23 at 20:48
1
$begingroup$
Well that can be easly done with Ptolomy theorem. :)
$endgroup$
– greedoid
Jan 23 at 20:49
$begingroup$
OK. Clearly, nobody can compete with your solution ! I am going to present modestly an intuitive reasoning that could be helpful for this kind of issues.
$endgroup$
– Jean Marie
Jan 23 at 22:10
1
$begingroup$
Great solution , but aren’t we yet to prove that all the points which satisfy the equality condition lie on arc $AB$ ? We seem to have proved that the points lying on arc $AB$ satisfy the equality condition
$endgroup$
– Sinπ
Jan 24 at 4:01
add a comment |
1
$begingroup$
[+1] Very neat solution for points $A$ and $B$ ! It remains the case of points of arc $AB...$
$endgroup$
– Jean Marie
Jan 23 at 20:48
1
$begingroup$
Well that can be easly done with Ptolomy theorem. :)
$endgroup$
– greedoid
Jan 23 at 20:49
$begingroup$
OK. Clearly, nobody can compete with your solution ! I am going to present modestly an intuitive reasoning that could be helpful for this kind of issues.
$endgroup$
– Jean Marie
Jan 23 at 22:10
1
$begingroup$
Great solution , but aren’t we yet to prove that all the points which satisfy the equality condition lie on arc $AB$ ? We seem to have proved that the points lying on arc $AB$ satisfy the equality condition
$endgroup$
– Sinπ
Jan 24 at 4:01
1
1
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[+1] Very neat solution for points $A$ and $B$ ! It remains the case of points of arc $AB...$
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– Jean Marie
Jan 23 at 20:48
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[+1] Very neat solution for points $A$ and $B$ ! It remains the case of points of arc $AB...$
$endgroup$
– Jean Marie
Jan 23 at 20:48
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Well that can be easly done with Ptolomy theorem. :)
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– greedoid
Jan 23 at 20:49
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Well that can be easly done with Ptolomy theorem. :)
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– greedoid
Jan 23 at 20:49
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OK. Clearly, nobody can compete with your solution ! I am going to present modestly an intuitive reasoning that could be helpful for this kind of issues.
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– Jean Marie
Jan 23 at 22:10
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OK. Clearly, nobody can compete with your solution ! I am going to present modestly an intuitive reasoning that could be helpful for this kind of issues.
$endgroup$
– Jean Marie
Jan 23 at 22:10
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1
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Great solution , but aren’t we yet to prove that all the points which satisfy the equality condition lie on arc $AB$ ? We seem to have proved that the points lying on arc $AB$ satisfy the equality condition
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– Sinπ
Jan 24 at 4:01
$begingroup$
Great solution , but aren’t we yet to prove that all the points which satisfy the equality condition lie on arc $AB$ ? We seem to have proved that the points lying on arc $AB$ satisfy the equality condition
$endgroup$
– Sinπ
Jan 24 at 4:01
add a comment |
$begingroup$
What follows is not a solution (an excellent one has been given by @greedoid) but a heuristic method that provides an efficient angle of attack of this locus and loci of the same "flavour". This presentation has a handwaving side, but I think that this kind of intuitive reasoning can have some interest, in particular for students in mathematics in order to convey the spirit of discoverers of infinitesimal calculus.
Let :
$$q(X):=frac{n(X)}{d(X)}, text{with}$$
$$n(X):=XA+XB text{and } $$
$$d(X):=XC+XD+XE$$
Remark :
$lim q(X)=frac23$ when $|overrightarrow{OX}|$ tends to $infty$.
Consider figure $1$ that depicts some contour lines of functions $n$ (in blue) and $d$ (in red) :
Fig. 1. : To each point $X$ in the plane is attached two curves, a red one and a blue one giving the locus of points having resp. the same numerator $n(X)$ and the same denominator $d(X)$. One shouldn't be surprized that the blue curves are confocal ellipses (with foci $A$ and $B$).
What usage can we do of these two families of curves ?
A first result is that a point $X$ can be a minimum of function $q$ (or at least a relative minimum), if the blue and the red contour lines passing through $X$ are tangent ; otherwise, if these curves are transverse in $X$ (transverse = non tangent), there are two directions (see Fig. 2) one can take that are more advantageous.
Fig. 2 : A minimum of function $q$ cannot occur in a point $X$ with transverse intersection of the blue and red contour lines passing through this point : indeed, moving $X$ in one of the given directions decreases the value of $q$.
Caveat : tangency in $X$ of the red and the blue curves passing through $X$ is a necessary condition, not a sufficient one ; let us take the (counter) example of a point $X$ belonging to the interval $I=(-1,cos(pi/5))$ of the $x$ axis; moving $X$ to the right into $X'$ still in $I$, will also give $q(X')<q(X)$ (a particulary favorable case where $n(X)$ decreases while $d(X)$ increases!).
More generally, it is possible to eliminate all points of the $x$-axis (except point $(1,0)$) although blue and red curves are tangent in these points.
What are the remaining candidate points for the minimum of $q(.)$ ? A concrete example is provided by the tangency of blue curve indexed by $1.2$ (the most elongated ellipse) and the red curve indexed by $5.1$ [just at the right of vertical line segment $AB$]. A detailed look show that all solution points $X$ are in this area and constitute a continuum of points ; a more detailed analysis is needed to prove that this "continuum" is the small circular arc $AB$ of the unit circle. I haven't done it.
Besides, and this is a second idea, one can have access to a numerical solution using gradient computations.
Let us first recall that the value of the gradient of distance function defined by $f(M)=AM$, where $A$ is any fixed point of the plane, is the unit norm vector defined in point $M$ as :
$$overrightarrow{grad}(f)(M)=frac{overrightarrow{AM}}{AM}.$$
See Gradient of distance vector length and Appendix 2.
Thus the gradients in $X$ of functions $n$ and $d$ are resp.
$$overrightarrow{grad}(n)(X)=frac{overrightarrow{AX}}{AX}+frac{overrightarrow{BX}}{BX},$$
$$text{and} overrightarrow{grad}(d)(X)=frac{overrightarrow{CX}}{CX}+frac{overrightarrow{DX}}{DX}+frac{overrightarrow{EX}}{EX} tag{1}$$
The tangency condition will then be transferred as a gradients' proportionality. Moreover, we will switch to a complex function treatment that will give a more compact formulation :
$$arg(r(z-a)+r(z-b))=arg(r(z-c)+r(z-d)+r(z-e))+k pitag{2}$$
where
$$r(z):=frac{z}{|z|}$$
where $a,b,c,d,e$ stand for the complex numbers associated with $A,B,C,D,E$ resp. Indeed, having the same angle with $k$ even or opposite angles with $k$ odd means proportionality ; a further step is to replace (2) by a criterium on the imaginary part of the complex logarithm function (see line 8 in program below) :
$$operatorname{Im}(log(f(z))) text{with}$$
$$f(z):=frac{r(z-a)+r(z-b)}{r(z-c)+r(z-d)+r(z-e)} tag{3}$$
(recall : $log(re^{i theta})=log(r)+i theta$ for the principal branch of complex logarithm function). Why is this formulation interesting ? Because we have converted the search of minima into the search of zeroes of a certain function (more exactly, the minima constitute a subset of this set of zeroes : see explanation upwards about necessary and sufficient condition).
Let us use formulation (3) to obtain a graphical representation of the points under consideration : see figure $3$ below (which does not constitute - of course - a mathematical proof) where the circular arc AB appears as the solution.
Fig. 3. In black, all points $X$ having almost proportional gradients. See Appendix 1.
Appendix 1 : The Matlab program that has generated Fig. 3. reminder : this program generates all points for which this expression is close to $0$.
i=complex(0,1);
b=exp(i*pi/5);a=b^9;c=b^3;d=-1;e=b^7;% pentagon vertices
r=@(z)(z./abs(z));
dg=@(z)(r(z-c)+r(z-d)+r(z-e));
ng=@(z)(r(z-a)+r(z-b));
f=@(z)(ng(z)./dg(z));
[X,Y]=meshgrid(-2:0.005:2);
I=imag(log(f(X+i*Y)));% = imag(log(ng(X+i*Y))-log(dg(X+i*Y)));% if zero : gradients' proportionality
J=abs(I)>0.01;
imshow(J);
Appendix 2 : the gradient at point $X$ of function $n(.)$ for example is the inward or outward normal to the blue curve passing through $X$ (think to the blue curves as isobaric curves and to the gradient as depicting the wind's direction).
answered Jan 24 at 0:40
Jean MarieJean Marie
30.3k42153
$endgroup$
add a comment |
$begingroup$
What follows is not a solution (an excellent one has been given by @greedoid) but a heuristic method that provides an efficient angle of attack of this locus and loci of the same "flavour". This presentation has a handwaving side, but I think that this kind of intuitive reasoning can have some interest, in particular for students in mathematics in order to convey the spirit of discoverers of infinitesimal calculus.
Let :
$$q(X):=frac{n(X)}{d(X)}, text{with}$$
$$n(X):=XA+XB text{and } $$
$$d(X):=XC+XD+XE$$
Remark :
$lim q(X)=frac23$ when $|overrightarrow{OX}|$ tends to $infty$.
Consider figure $1$ that depicts some contour lines of functions $n$ (in blue) and $d$ (in red) :
Fig. 1. : To each point $X$ in the plane is attached two curves, a red one and a blue one giving the locus of points having resp. the same numerator $n(X)$ and the same denominator $d(X)$. One shouldn't be surprized that the blue curves are confocal ellipses (with foci $A$ and $B$).
What usage can we do of these two families of curves ?
A first result is that a point $X$ can be a minimum of function $q$ (or at least a relative minimum), if the blue and the red contour lines passing through $X$ are tangent ; otherwise, if these curves are transverse in $X$ (transverse = non tangent), there are two directions (see Fig. 2) one can take that are more advantageous.
Fig. 2 : A minimum of function $q$ cannot occur in a point $X$ with transverse intersection of the blue and red contour lines passing through this point : indeed, moving $X$ in one of the given directions decreases the value of $q$.
Caveat : tangency in $X$ of the red and the blue curves passing through $X$ is a necessary condition, not a sufficient one ; let us take the (counter) example of a point $X$ belonging to the interval $I=(-1,cos(pi/5))$ of the $x$ axis; moving $X$ to the right into $X'$ still in $I$, will also give $q(X')<q(X)$ (a particulary favorable case where $n(X)$ decreases while $d(X)$ increases!).
More generally, it is possible to eliminate all points of the $x$-axis (except point $(1,0)$) although blue and red curves are tangent in these points.
What are the remaining candidate points for the minimum of $q(.)$ ? A concrete example is provided by the tangency of blue curve indexed by $1.2$ (the most elongated ellipse) and the red curve indexed by $5.1$ [just at the right of vertical line segment $AB$]. A detailed look show that all solution points $X$ are in this area and constitute a continuum of points ; a more detailed analysis is needed to prove that this "continuum" is the small circular arc $AB$ of the unit circle. I haven't done it.
Besides, and this is a second idea, one can have access to a numerical solution using gradient computations.
Let us first recall that the value of the gradient of distance function defined by $f(M)=AM$, where $A$ is any fixed point of the plane, is the unit norm vector defined in point $M$ as :
$$overrightarrow{grad}(f)(M)=frac{overrightarrow{AM}}{AM}.$$
See Gradient of distance vector length and Appendix 2.
Thus the gradients in $X$ of functions $n$ and $d$ are resp.
$$overrightarrow{grad}(n)(X)=frac{overrightarrow{AX}}{AX}+frac{overrightarrow{BX}}{BX},$$
$$text{and} overrightarrow{grad}(d)(X)=frac{overrightarrow{CX}}{CX}+frac{overrightarrow{DX}}{DX}+frac{overrightarrow{EX}}{EX} tag{1}$$
The tangency condition will then be transferred as a gradients' proportionality. Moreover, we will switch to a complex function treatment that will give a more compact formulation :
$$arg(r(z-a)+r(z-b))=arg(r(z-c)+r(z-d)+r(z-e))+k pitag{2}$$
where
$$r(z):=frac{z}{|z|}$$
where $a,b,c,d,e$ stand for the complex numbers associated with $A,B,C,D,E$ resp. Indeed, having the same angle with $k$ even or opposite angles with $k$ odd means proportionality ; a further step is to replace (2) by a criterium on the imaginary part of the complex logarithm function (see line 8 in program below) :
$$operatorname{Im}(log(f(z))) text{with}$$
$$f(z):=frac{r(z-a)+r(z-b)}{r(z-c)+r(z-d)+r(z-e)} tag{3}$$
(recall : $log(re^{i theta})=log(r)+i theta$ for the principal branch of complex logarithm function). Why is this formulation interesting ? Because we have converted the search of minima into the search of zeroes of a certain function (more exactly, the minima constitute a subset of this set of zeroes : see explanation upwards about necessary and sufficient condition).
Let us use formulation (3) to obtain a graphical representation of the points under consideration : see figure $3$ below (which does not constitute - of course - a mathematical proof) where the circular arc AB appears as the solution.
Fig. 3. In black, all points $X$ having almost proportional gradients. See Appendix 1.
Appendix 1 : The Matlab program that has generated Fig. 3. reminder : this program generates all points for which this expression is close to $0$.
i=complex(0,1);
b=exp(i*pi/5);a=b^9;c=b^3;d=-1;e=b^7;% pentagon vertices
r=@(z)(z./abs(z));
dg=@(z)(r(z-c)+r(z-d)+r(z-e));
ng=@(z)(r(z-a)+r(z-b));
f=@(z)(ng(z)./dg(z));
[X,Y]=meshgrid(-2:0.005:2);
I=imag(log(f(X+i*Y)));% = imag(log(ng(X+i*Y))-log(dg(X+i*Y)));% if zero : gradients' proportionality
J=abs(I)>0.01;
imshow(J);
Appendix 2 : the gradient at point $X$ of function $n(.)$ for example is the inward or outward normal to the blue curve passing through $X$ (think to the blue curves as isobaric curves and to the gradient as depicting the wind's direction).
answered Jan 24 at 0:40
Jean MarieJean Marie
30.3k42153
$endgroup$
add a comment |
$begingroup$
What follows is not a solution (an excellent one has been given by @greedoid) but a heuristic method that provides an efficient angle of attack of this locus and loci of the same "flavour". This presentation has a handwaving side, but I think that this kind of intuitive reasoning can have some interest, in particular for students in mathematics in order to convey the spirit of discoverers of infinitesimal calculus.
Let :
$$q(X):=frac{n(X)}{d(X)}, text{with}$$
$$n(X):=XA+XB text{and } $$
$$d(X):=XC+XD+XE$$
Remark :
$lim q(X)=frac23$ when $|overrightarrow{OX}|$ tends to $infty$.
Consider figure $1$ that depicts some contour lines of functions $n$ (in blue) and $d$ (in red) :
Fig. 1. : To each point $X$ in the plane is attached two curves, a red one and a blue one giving the locus of points having resp. the same numerator $n(X)$ and the same denominator $d(X)$. One shouldn't be surprized that the blue curves are confocal ellipses (with foci $A$ and $B$).
What usage can we do of these two families of curves ?
A first result is that a point $X$ can be a minimum of function $q$ (or at least a relative minimum), if the blue and the red contour lines passing through $X$ are tangent ; otherwise, if these curves are transverse in $X$ (transverse = non tangent), there are two directions (see Fig. 2) one can take that are more advantageous.
Fig. 2 : A minimum of function $q$ cannot occur in a point $X$ with transverse intersection of the blue and red contour lines passing through this point : indeed, moving $X$ in one of the given directions decreases the value of $q$.
Caveat : tangency in $X$ of the red and the blue curves passing through $X$ is a necessary condition, not a sufficient one ; let us take the (counter) example of a point $X$ belonging to the interval $I=(-1,cos(pi/5))$ of the $x$ axis; moving $X$ to the right into $X'$ still in $I$, will also give $q(X')<q(X)$ (a particulary favorable case where $n(X)$ decreases while $d(X)$ increases!).
More generally, it is possible to eliminate all points of the $x$-axis (except point $(1,0)$) although blue and red curves are tangent in these points.
What are the remaining candidate points for the minimum of $q(.)$ ? A concrete example is provided by the tangency of blue curve indexed by $1.2$ (the most elongated ellipse) and the red curve indexed by $5.1$ [just at the right of vertical line segment $AB$]. A detailed look show that all solution points $X$ are in this area and constitute a continuum of points ; a more detailed analysis is needed to prove that this "continuum" is the small circular arc $AB$ of the unit circle. I haven't done it.
Besides, and this is a second idea, one can have access to a numerical solution using gradient computations.
Let us first recall that the value of the gradient of distance function defined by $f(M)=AM$, where $A$ is any fixed point of the plane, is the unit norm vector defined in point $M$ as :
$$overrightarrow{grad}(f)(M)=frac{overrightarrow{AM}}{AM}.$$
See Gradient of distance vector length and Appendix 2.
Thus the gradients in $X$ of functions $n$ and $d$ are resp.
$$overrightarrow{grad}(n)(X)=frac{overrightarrow{AX}}{AX}+frac{overrightarrow{BX}}{BX},$$
$$text{and} overrightarrow{grad}(d)(X)=frac{overrightarrow{CX}}{CX}+frac{overrightarrow{DX}}{DX}+frac{overrightarrow{EX}}{EX} tag{1}$$
The tangency condition will then be transferred as a gradients' proportionality. Moreover, we will switch to a complex function treatment that will give a more compact formulation :
$$arg(r(z-a)+r(z-b))=arg(r(z-c)+r(z-d)+r(z-e))+k pitag{2}$$
where
$$r(z):=frac{z}{|z|}$$
where $a,b,c,d,e$ stand for the complex numbers associated with $A,B,C,D,E$ resp. Indeed, having the same angle with $k$ even or opposite angles with $k$ odd means proportionality ; a further step is to replace (2) by a criterium on the imaginary part of the complex logarithm function (see line 8 in program below) :
$$operatorname{Im}(log(f(z))) text{with}$$
$$f(z):=frac{r(z-a)+r(z-b)}{r(z-c)+r(z-d)+r(z-e)} tag{3}$$
(recall : $log(re^{i theta})=log(r)+i theta$ for the principal branch of complex logarithm function). Why is this formulation interesting ? Because we have converted the search of minima into the search of zeroes of a certain function (more exactly, the minima constitute a subset of this set of zeroes : see explanation upwards about necessary and sufficient condition).
Let us use formulation (3) to obtain a graphical representation of the points under consideration : see figure $3$ below (which does not constitute - of course - a mathematical proof) where the circular arc AB appears as the solution.
Fig. 3. In black, all points $X$ having almost proportional gradients. See Appendix 1.
Appendix 1 : The Matlab program that has generated Fig. 3. reminder : this program generates all points for which this expression is close to $0$.
i=complex(0,1);
b=exp(i*pi/5);a=b^9;c=b^3;d=-1;e=b^7;% pentagon vertices
r=@(z)(z./abs(z));
dg=@(z)(r(z-c)+r(z-d)+r(z-e));
ng=@(z)(r(z-a)+r(z-b));
f=@(z)(ng(z)./dg(z));
[X,Y]=meshgrid(-2:0.005:2);
I=imag(log(f(X+i*Y)));% = imag(log(ng(X+i*Y))-log(dg(X+i*Y)));% if zero : gradients' proportionality
J=abs(I)>0.01;
imshow(J);
Appendix 2 : the gradient at point $X$ of function $n(.)$ for example is the inward or outward normal to the blue curve passing through $X$ (think to the blue curves as isobaric curves and to the gradient as depicting the wind's direction).
answered Jan 24 at 0:40
Jean MarieJean Marie
30.3k42153
$endgroup$
What follows is not a solution (an excellent one has been given by @greedoid) but a heuristic method that provides an efficient angle of attack of this locus and loci of the same "flavour". This presentation has a handwaving side, but I think that this kind of intuitive reasoning can have some interest, in particular for students in mathematics in order to convey the spirit of discoverers of infinitesimal calculus.
Let :
$$q(X):=frac{n(X)}{d(X)}, text{with}$$
$$n(X):=XA+XB text{and } $$
$$d(X):=XC+XD+XE$$
Remark :
$lim q(X)=frac23$ when $|overrightarrow{OX}|$ tends to $infty$.
Consider figure $1$ that depicts some contour lines of functions $n$ (in blue) and $d$ (in red) :
Fig. 1. : To each point $X$ in the plane is attached two curves, a red one and a blue one giving the locus of points having resp. the same numerator $n(X)$ and the same denominator $d(X)$. One shouldn't be surprized that the blue curves are confocal ellipses (with foci $A$ and $B$).
What usage can we do of these two families of curves ?
A first result is that a point $X$ can be a minimum of function $q$ (or at least a relative minimum), if the blue and the red contour lines passing through $X$ are tangent ; otherwise, if these curves are transverse in $X$ (transverse = non tangent), there are two directions (see Fig. 2) one can take that are more advantageous.
Fig. 2 : A minimum of function $q$ cannot occur in a point $X$ with transverse intersection of the blue and red contour lines passing through this point : indeed, moving $X$ in one of the given directions decreases the value of $q$.
Caveat : tangency in $X$ of the red and the blue curves passing through $X$ is a necessary condition, not a sufficient one ; let us take the (counter) example of a point $X$ belonging to the interval $I=(-1,cos(pi/5))$ of the $x$ axis; moving $X$ to the right into $X'$ still in $I$, will also give $q(X')<q(X)$ (a particulary favorable case where $n(X)$ decreases while $d(X)$ increases!).
More generally, it is possible to eliminate all points of the $x$-axis (except point $(1,0)$) although blue and red curves are tangent in these points.
What are the remaining candidate points for the minimum of $q(.)$ ? A concrete example is provided by the tangency of blue curve indexed by $1.2$ (the most elongated ellipse) and the red curve indexed by $5.1$ [just at the right of vertical line segment $AB$]. A detailed look show that all solution points $X$ are in this area and constitute a continuum of points ; a more detailed analysis is needed to prove that this "continuum" is the small circular arc $AB$ of the unit circle. I haven't done it.
Besides, and this is a second idea, one can have access to a numerical solution using gradient computations.
Let us first recall that the value of the gradient of distance function defined by $f(M)=AM$, where $A$ is any fixed point of the plane, is the unit norm vector defined in point $M$ as :
$$overrightarrow{grad}(f)(M)=frac{overrightarrow{AM}}{AM}.$$
See Gradient of distance vector length and Appendix 2.
Thus the gradients in $X$ of functions $n$ and $d$ are resp.
$$overrightarrow{grad}(n)(X)=frac{overrightarrow{AX}}{AX}+frac{overrightarrow{BX}}{BX},$$
$$text{and} overrightarrow{grad}(d)(X)=frac{overrightarrow{CX}}{CX}+frac{overrightarrow{DX}}{DX}+frac{overrightarrow{EX}}{EX} tag{1}$$
The tangency condition will then be transferred as a gradients' proportionality. Moreover, we will switch to a complex function treatment that will give a more compact formulation :
$$arg(r(z-a)+r(z-b))=arg(r(z-c)+r(z-d)+r(z-e))+k pitag{2}$$
where
$$r(z):=frac{z}{|z|}$$
where $a,b,c,d,e$ stand for the complex numbers associated with $A,B,C,D,E$ resp. Indeed, having the same angle with $k$ even or opposite angles with $k$ odd means proportionality ; a further step is to replace (2) by a criterium on the imaginary part of the complex logarithm function (see line 8 in program below) :
$$operatorname{Im}(log(f(z))) text{with}$$
$$f(z):=frac{r(z-a)+r(z-b)}{r(z-c)+r(z-d)+r(z-e)} tag{3}$$
(recall : $log(re^{i theta})=log(r)+i theta$ for the principal branch of complex logarithm function). Why is this formulation interesting ? Because we have converted the search of minima into the search of zeroes of a certain function (more exactly, the minima constitute a subset of this set of zeroes : see explanation upwards about necessary and sufficient condition).
Let us use formulation (3) to obtain a graphical representation of the points under consideration : see figure $3$ below (which does not constitute - of course - a mathematical proof) where the circular arc AB appears as the solution.
Fig. 3. In black, all points $X$ having almost proportional gradients. See Appendix 1.
Appendix 1 : The Matlab program that has generated Fig. 3. reminder : this program generates all points for which this expression is close to $0$.
i=complex(0,1);
b=exp(i*pi/5);a=b^9;c=b^3;d=-1;e=b^7;% pentagon vertices
r=@(z)(z./abs(z));
dg=@(z)(r(z-c)+r(z-d)+r(z-e));
ng=@(z)(r(z-a)+r(z-b));
f=@(z)(ng(z)./dg(z));
[X,Y]=meshgrid(-2:0.005:2);
I=imag(log(f(X+i*Y)));% = imag(log(ng(X+i*Y))-log(dg(X+i*Y)));% if zero : gradients' proportionality
J=abs(I)>0.01;
imshow(J);
Appendix 2 : the gradient at point $X$ of function $n(.)$ for example is the inward or outward normal to the blue curve passing through $X$ (think to the blue curves as isobaric curves and to the gradient as depicting the wind's direction).
answered Jan 24 at 0:40
Jean MarieJean Marie
30.3k42153
edited Feb 1 at 0:13
answered Jan 24 at 0:40
Jean MarieJean Marie
30.3k42153
answered Jan 24 at 0:40
Jean MarieJean Marie
30.3k42153
answered Jan 24 at 0:40
Jean MarieJean Marie
30.3k42153
30.3k42153
add a comment |
add a comment |
$begingroup$
If $X$ is on $AB$, $AX+BX$ is minimum. And if $X=A$ , $CX+DX+EX$ is maximum. So cesareo's answer follows.
answered Jan 22 at 8:53
Takahiro WakiTakahiro Waki
2,099620
$endgroup$
$begingroup$
This answer appears to be incorrect. It isn't necessary to minimize the numerator of the expression. In fact, the gradient of the numerator is zero on $AB$, whereas the gradient of the denominator is not, so it's "free" to increase the denominator by moving at least a small distance away from $AB$. You may want to delete this answer so that the question still appears as unanswered, which should help to attract more attention.
$endgroup$
– Chris Culter
Jan 22 at 19:15
$begingroup$
@ChrisCulter I don't know what you say, denominator is maximum on X=A or X=B.
$endgroup$
– Takahiro Waki
Jan 23 at 2:08
$begingroup$
But $X$ is not restricted to the inside of the pentagon. There is no global maximum of the denominator.
$endgroup$
– Chris Culter
Jan 23 at 5:31
$begingroup$
@ChrisCulter That's bug on question. The original question shows "on the plane of pentagon...". I think this doesn't mean "Find X in the plane".
$endgroup$
– Takahiro Waki
Jan 23 at 5:48
add a comment |
$begingroup$
If $X$ is on $AB$, $AX+BX$ is minimum. And if $X=A$ , $CX+DX+EX$ is maximum. So cesareo's answer follows.
answered Jan 22 at 8:53
Takahiro WakiTakahiro Waki
2,099620
$endgroup$
$begingroup$
This answer appears to be incorrect. It isn't necessary to minimize the numerator of the expression. In fact, the gradient of the numerator is zero on $AB$, whereas the gradient of the denominator is not, so it's "free" to increase the denominator by moving at least a small distance away from $AB$. You may want to delete this answer so that the question still appears as unanswered, which should help to attract more attention.
$endgroup$
– Chris Culter
Jan 22 at 19:15
$begingroup$
@ChrisCulter I don't know what you say, denominator is maximum on X=A or X=B.
$endgroup$
– Takahiro Waki
Jan 23 at 2:08
$begingroup$
But $X$ is not restricted to the inside of the pentagon. There is no global maximum of the denominator.
$endgroup$
– Chris Culter
Jan 23 at 5:31
$begingroup$
@ChrisCulter That's bug on question. The original question shows "on the plane of pentagon...". I think this doesn't mean "Find X in the plane".
$endgroup$
– Takahiro Waki
Jan 23 at 5:48
add a comment |
$begingroup$
If $X$ is on $AB$, $AX+BX$ is minimum. And if $X=A$ , $CX+DX+EX$ is maximum. So cesareo's answer follows.
answered Jan 22 at 8:53
Takahiro WakiTakahiro Waki
2,099620
$endgroup$
If $X$ is on $AB$, $AX+BX$ is minimum. And if $X=A$ , $CX+DX+EX$ is maximum. So cesareo's answer follows.
answered Jan 22 at 8:53
Takahiro WakiTakahiro Waki
2,099620
answered Jan 22 at 8:53
Takahiro WakiTakahiro Waki
2,099620
answered Jan 22 at 8:53
Takahiro WakiTakahiro Waki
2,099620
answered Jan 22 at 8:53
Takahiro WakiTakahiro Waki
2,099620
2,099620
$begingroup$
This answer appears to be incorrect. It isn't necessary to minimize the numerator of the expression. In fact, the gradient of the numerator is zero on $AB$, whereas the gradient of the denominator is not, so it's "free" to increase the denominator by moving at least a small distance away from $AB$. You may want to delete this answer so that the question still appears as unanswered, which should help to attract more attention.
$endgroup$
– Chris Culter
Jan 22 at 19:15
$begingroup$
@ChrisCulter I don't know what you say, denominator is maximum on X=A or X=B.
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– Takahiro Waki
Jan 23 at 2:08
$begingroup$
But $X$ is not restricted to the inside of the pentagon. There is no global maximum of the denominator.
$endgroup$
– Chris Culter
Jan 23 at 5:31
$begingroup$
@ChrisCulter That's bug on question. The original question shows "on the plane of pentagon...". I think this doesn't mean "Find X in the plane".
$endgroup$
– Takahiro Waki
Jan 23 at 5:48
add a comment |
$begingroup$
This answer appears to be incorrect. It isn't necessary to minimize the numerator of the expression. In fact, the gradient of the numerator is zero on $AB$, whereas the gradient of the denominator is not, so it's "free" to increase the denominator by moving at least a small distance away from $AB$. You may want to delete this answer so that the question still appears as unanswered, which should help to attract more attention.
$endgroup$
– Chris Culter
Jan 22 at 19:15
$begingroup$
@ChrisCulter I don't know what you say, denominator is maximum on X=A or X=B.
$endgroup$
– Takahiro Waki
Jan 23 at 2:08
$begingroup$
But $X$ is not restricted to the inside of the pentagon. There is no global maximum of the denominator.
$endgroup$
– Chris Culter
Jan 23 at 5:31
$begingroup$
@ChrisCulter That's bug on question. The original question shows "on the plane of pentagon...". I think this doesn't mean "Find X in the plane".
$endgroup$
– Takahiro Waki
Jan 23 at 5:48
$begingroup$
This answer appears to be incorrect. It isn't necessary to minimize the numerator of the expression. In fact, the gradient of the numerator is zero on $AB$, whereas the gradient of the denominator is not, so it's "free" to increase the denominator by moving at least a small distance away from $AB$. You may want to delete this answer so that the question still appears as unanswered, which should help to attract more attention.
$endgroup$
– Chris Culter
Jan 22 at 19:15
$begingroup$
This answer appears to be incorrect. It isn't necessary to minimize the numerator of the expression. In fact, the gradient of the numerator is zero on $AB$, whereas the gradient of the denominator is not, so it's "free" to increase the denominator by moving at least a small distance away from $AB$. You may want to delete this answer so that the question still appears as unanswered, which should help to attract more attention.
$endgroup$
– Chris Culter
Jan 22 at 19:15
$begingroup$
@ChrisCulter I don't know what you say, denominator is maximum on X=A or X=B.
$endgroup$
– Takahiro Waki
Jan 23 at 2:08
$begingroup$
@ChrisCulter I don't know what you say, denominator is maximum on X=A or X=B.
$endgroup$
– Takahiro Waki
Jan 23 at 2:08
$begingroup$
But $X$ is not restricted to the inside of the pentagon. There is no global maximum of the denominator.
$endgroup$
– Chris Culter
Jan 23 at 5:31
$begingroup$
But $X$ is not restricted to the inside of the pentagon. There is no global maximum of the denominator.
$endgroup$
– Chris Culter
Jan 23 at 5:31
$begingroup$
@ChrisCulter That's bug on question. The original question shows "on the plane of pentagon...". I think this doesn't mean "Find X in the plane".
$endgroup$
– Takahiro Waki
Jan 23 at 5:48
$begingroup$
@ChrisCulter That's bug on question. The original question shows "on the plane of pentagon...". I think this doesn't mean "Find X in the plane".
$endgroup$
– Takahiro Waki
Jan 23 at 5:48
add a comment |
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$begingroup$
Is X inside of ABCDE? Othwerwise there is no such point...
$endgroup$
– Jakub Andruszkiewicz
Jan 19 at 8:44
$begingroup$
It was not stated in the problem. Why such point does not exist outside $ABCDE$?
$endgroup$
– SamHar
Jan 19 at 8:51
1
$begingroup$
The minimum value is $BA/(EA+DA+CA)$
$endgroup$
– Cesareo
Jan 19 at 11:57
4
$begingroup$
@i. m. soloveichik In fact, numerical simulations show that the whole "little" arc $AB$ is solution, i.e. one can take any $X (cos(a),sin(a))$ with $-frac{pi}{5} leq a leq frac{pi}{5}$.
$endgroup$
– Jean Marie
Jan 20 at 21:21
3
$begingroup$
@SamHar: What is the source of this problem? Knowing that would help readers gauge the intended level of difficulty.
$endgroup$
– Blue
Jan 22 at 5:52