Mixing
Maximizing the area of a cyclic trapezoid whose long base is the circumdiameter. Non-trigonometric solution?
$begingroup$
A half circle with a radius of R encompasses an isosceles trapezoid such that the large base of the trapezoid is the diameter of the circle encompassing it.
In terms of R, what is the length of the smaller base of all the possible trapezoids as described, whose area is maximal?
After some attempts at the problem, I managed to solve it using unit circle trigonometry, but I am curious if there are purely geometric solutions for this (which is what I was trying to find when I first attempted the problem).
Here is my solution:
Let $x$ be $measuredangle AOD$
Let $h$ be the height of the trapezoid
Assume $0 < x < 90^circ$
$$h = AOsin x = Rsin x$$
$$AB = 2AOcos x = 2Rcos x$$
Trapezoid area formula: $$frac{AB + DC}{2} cdot h $$
$$downarrow$$
$$S_{(x)} = frac{2Rcos x + 2R}{2} cdot Rsin x$$
From here, we find our function's derivative, get the $x$ for which there is a maxima, and plug it into our definition of AB to get it in terms of R, which would be AB = R.
Is there an alternative?
derivatives trigonometry euclidean-geometry
edited Jan 6 at 23:34
Blue
48k870153
asked Jan 6 at 23:25
daedsidogdaedsidog
29017
$endgroup$
add a comment |
$begingroup$
A half circle with a radius of R encompasses an isosceles trapezoid such that the large base of the trapezoid is the diameter of the circle encompassing it.
In terms of R, what is the length of the smaller base of all the possible trapezoids as described, whose area is maximal?
After some attempts at the problem, I managed to solve it using unit circle trigonometry, but I am curious if there are purely geometric solutions for this (which is what I was trying to find when I first attempted the problem).
Here is my solution:
Let $x$ be $measuredangle AOD$
Let $h$ be the height of the trapezoid
Assume $0 < x < 90^circ$
$$h = AOsin x = Rsin x$$
$$AB = 2AOcos x = 2Rcos x$$
Trapezoid area formula: $$frac{AB + DC}{2} cdot h $$
$$downarrow$$
$$S_{(x)} = frac{2Rcos x + 2R}{2} cdot Rsin x$$
From here, we find our function's derivative, get the $x$ for which there is a maxima, and plug it into our definition of AB to get it in terms of R, which would be AB = R.
Is there an alternative?
derivatives trigonometry euclidean-geometry
edited Jan 6 at 23:34
Blue
48k870153
asked Jan 6 at 23:25
daedsidogdaedsidog
29017
$endgroup$
$begingroup$
Yes, there is :)
$endgroup$
– Oldboy
Jan 7 at 10:10
add a comment |
$begingroup$
A half circle with a radius of R encompasses an isosceles trapezoid such that the large base of the trapezoid is the diameter of the circle encompassing it.
In terms of R, what is the length of the smaller base of all the possible trapezoids as described, whose area is maximal?
After some attempts at the problem, I managed to solve it using unit circle trigonometry, but I am curious if there are purely geometric solutions for this (which is what I was trying to find when I first attempted the problem).
Here is my solution:
Let $x$ be $measuredangle AOD$
Let $h$ be the height of the trapezoid
Assume $0 < x < 90^circ$
$$h = AOsin x = Rsin x$$
$$AB = 2AOcos x = 2Rcos x$$
Trapezoid area formula: $$frac{AB + DC}{2} cdot h $$
$$downarrow$$
$$S_{(x)} = frac{2Rcos x + 2R}{2} cdot Rsin x$$
From here, we find our function's derivative, get the $x$ for which there is a maxima, and plug it into our definition of AB to get it in terms of R, which would be AB = R.
Is there an alternative?
derivatives trigonometry euclidean-geometry
edited Jan 6 at 23:34
Blue
48k870153
asked Jan 6 at 23:25
daedsidogdaedsidog
29017
$endgroup$
A half circle with a radius of R encompasses an isosceles trapezoid such that the large base of the trapezoid is the diameter of the circle encompassing it.
In terms of R, what is the length of the smaller base of all the possible trapezoids as described, whose area is maximal?
After some attempts at the problem, I managed to solve it using unit circle trigonometry, but I am curious if there are purely geometric solutions for this (which is what I was trying to find when I first attempted the problem).
Here is my solution:
Let $x$ be $measuredangle AOD$
Let $h$ be the height of the trapezoid
Assume $0 < x < 90^circ$
$$h = AOsin x = Rsin x$$
$$AB = 2AOcos x = 2Rcos x$$
Trapezoid area formula: $$frac{AB + DC}{2} cdot h $$
$$downarrow$$
$$S_{(x)} = frac{2Rcos x + 2R}{2} cdot Rsin x$$
From here, we find our function's derivative, get the $x$ for which there is a maxima, and plug it into our definition of AB to get it in terms of R, which would be AB = R.
Is there an alternative?
derivatives trigonometry euclidean-geometry
derivatives trigonometry euclidean-geometry
edited Jan 6 at 23:34
Blue
48k870153
asked Jan 6 at 23:25
daedsidogdaedsidog
29017
edited Jan 6 at 23:34
Blue
48k870153
asked Jan 6 at 23:25
daedsidogdaedsidog
29017
edited Jan 6 at 23:34
Blue
48k870153
edited Jan 6 at 23:34
Blue
48k870153
edited Jan 6 at 23:34
Blue
48k870153
48k870153
asked Jan 6 at 23:25
daedsidogdaedsidog
29017
asked Jan 6 at 23:25
daedsidogdaedsidog
29017
asked Jan 6 at 23:25
daedsidogdaedsidog
29017
29017
$begingroup$
Yes, there is :)
$endgroup$
– Oldboy
Jan 7 at 10:10
add a comment |
$begingroup$
Yes, there is :)
$endgroup$
– Oldboy
Jan 7 at 10:10
$begingroup$
Yes, there is :)
$endgroup$
– Oldboy
Jan 7 at 10:10
$begingroup$
Yes, there is :)
$endgroup$
– Oldboy
Jan 7 at 10:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You don't need any trigonometry here.
Ellegant approach: Try to think outside of the box! Ask yourself a more general question:
Of all quadrilaterals $ABCD$ inscribed in a semicricle with diameter AB, which one has the maximum area?
We'll prove that for optimal quadrilateral:
$$BC=CD=DA=Rtag{1}$$
Suppose that (1) is not true. In other words, suppose that the optimal quadrilateral $ABCD$ looks like in the picture above and $ADneq CD$. Find point $D'$ on arc AC such that $AD'=CD'$.
Note that triangle $triangle ACD'$ has bigger area than $triangle ACD$ because $h_{D'}>h_{D}$. So obviously:
$$P_{ABCD}=P_{ABC}+P_{ACD}<P_{ABC}+P_{ACD'}<P_{ABCD'}$$
So quadrialteral $ABCD$ has smaller area compared to $ABCD'$ and cannot be optimal. In other words, if (1) is not true, we can always find a quadrilateral with a bigger area. Consequentially, the quadrialteral with the biggest area must have sides $BC$, $CD$ and $DA$ of equal lengths which is possible only if (1) is true.
And such quadrialteral is also a trapezoid. Any other quadrilateral, being it a trapezoid or not must have a smaller area.
No so ellegant approach, but still without trigonometry:
Denote the height of the trapezod with $h$ and the length of the smaller base with $b$:
You have to maximize the following expression:
$$A=frac{(2R+b)h}2$$
...or:
$$B=A^2=left(R+frac b2right)^2h^2tag{1}$$
...with the following constraint:
$$h^2=R^2-left(frac b2right)^2tag{2}$$
Replace (2) into (1) and you get:
$$B=left(R+frac b2right)^2left(R^2-left(frac b2right)^2right)$$
$$B=left(R+frac b2right)^3left(R-frac b2right)$$
For simplicity, introduce expression $c=R+frac b2$:
$$B(c)=c^3(2R-c)$$
$$B'(c)=3c^2(2R-c)-c^3=0implies c=frac{3R}2implies b=R$$
answered Jan 7 at 8:34
OldboyOldboy
7,6571935
$endgroup$
$begingroup$
Marvelous answer. Thank you.
$endgroup$
– daedsidog
Jan 7 at 12:08
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You don't need any trigonometry here.
Ellegant approach: Try to think outside of the box! Ask yourself a more general question:
Of all quadrilaterals $ABCD$ inscribed in a semicricle with diameter AB, which one has the maximum area?
We'll prove that for optimal quadrilateral:
$$BC=CD=DA=Rtag{1}$$
Suppose that (1) is not true. In other words, suppose that the optimal quadrilateral $ABCD$ looks like in the picture above and $ADneq CD$. Find point $D'$ on arc AC such that $AD'=CD'$.
Note that triangle $triangle ACD'$ has bigger area than $triangle ACD$ because $h_{D'}>h_{D}$. So obviously:
$$P_{ABCD}=P_{ABC}+P_{ACD}<P_{ABC}+P_{ACD'}<P_{ABCD'}$$
So quadrialteral $ABCD$ has smaller area compared to $ABCD'$ and cannot be optimal. In other words, if (1) is not true, we can always find a quadrilateral with a bigger area. Consequentially, the quadrialteral with the biggest area must have sides $BC$, $CD$ and $DA$ of equal lengths which is possible only if (1) is true.
And such quadrialteral is also a trapezoid. Any other quadrilateral, being it a trapezoid or not must have a smaller area.
No so ellegant approach, but still without trigonometry:
Denote the height of the trapezod with $h$ and the length of the smaller base with $b$:
You have to maximize the following expression:
$$A=frac{(2R+b)h}2$$
...or:
$$B=A^2=left(R+frac b2right)^2h^2tag{1}$$
...with the following constraint:
$$h^2=R^2-left(frac b2right)^2tag{2}$$
Replace (2) into (1) and you get:
$$B=left(R+frac b2right)^2left(R^2-left(frac b2right)^2right)$$
$$B=left(R+frac b2right)^3left(R-frac b2right)$$
For simplicity, introduce expression $c=R+frac b2$:
$$B(c)=c^3(2R-c)$$
$$B'(c)=3c^2(2R-c)-c^3=0implies c=frac{3R}2implies b=R$$
answered Jan 7 at 8:34
OldboyOldboy
7,6571935
$endgroup$
$begingroup$
Marvelous answer. Thank you.
$endgroup$
– daedsidog
Jan 7 at 12:08
add a comment |
$begingroup$
You don't need any trigonometry here.
Ellegant approach: Try to think outside of the box! Ask yourself a more general question:
Of all quadrilaterals $ABCD$ inscribed in a semicricle with diameter AB, which one has the maximum area?
We'll prove that for optimal quadrilateral:
$$BC=CD=DA=Rtag{1}$$
Suppose that (1) is not true. In other words, suppose that the optimal quadrilateral $ABCD$ looks like in the picture above and $ADneq CD$. Find point $D'$ on arc AC such that $AD'=CD'$.
Note that triangle $triangle ACD'$ has bigger area than $triangle ACD$ because $h_{D'}>h_{D}$. So obviously:
$$P_{ABCD}=P_{ABC}+P_{ACD}<P_{ABC}+P_{ACD'}<P_{ABCD'}$$
So quadrialteral $ABCD$ has smaller area compared to $ABCD'$ and cannot be optimal. In other words, if (1) is not true, we can always find a quadrilateral with a bigger area. Consequentially, the quadrialteral with the biggest area must have sides $BC$, $CD$ and $DA$ of equal lengths which is possible only if (1) is true.
And such quadrialteral is also a trapezoid. Any other quadrilateral, being it a trapezoid or not must have a smaller area.
No so ellegant approach, but still without trigonometry:
Denote the height of the trapezod with $h$ and the length of the smaller base with $b$:
You have to maximize the following expression:
$$A=frac{(2R+b)h}2$$
...or:
$$B=A^2=left(R+frac b2right)^2h^2tag{1}$$
...with the following constraint:
$$h^2=R^2-left(frac b2right)^2tag{2}$$
Replace (2) into (1) and you get:
$$B=left(R+frac b2right)^2left(R^2-left(frac b2right)^2right)$$
$$B=left(R+frac b2right)^3left(R-frac b2right)$$
For simplicity, introduce expression $c=R+frac b2$:
$$B(c)=c^3(2R-c)$$
$$B'(c)=3c^2(2R-c)-c^3=0implies c=frac{3R}2implies b=R$$
answered Jan 7 at 8:34
OldboyOldboy
7,6571935
$endgroup$
$begingroup$
Marvelous answer. Thank you.
$endgroup$
– daedsidog
Jan 7 at 12:08
add a comment |
$begingroup$
You don't need any trigonometry here.
Ellegant approach: Try to think outside of the box! Ask yourself a more general question:
Of all quadrilaterals $ABCD$ inscribed in a semicricle with diameter AB, which one has the maximum area?
We'll prove that for optimal quadrilateral:
$$BC=CD=DA=Rtag{1}$$
Suppose that (1) is not true. In other words, suppose that the optimal quadrilateral $ABCD$ looks like in the picture above and $ADneq CD$. Find point $D'$ on arc AC such that $AD'=CD'$.
Note that triangle $triangle ACD'$ has bigger area than $triangle ACD$ because $h_{D'}>h_{D}$. So obviously:
$$P_{ABCD}=P_{ABC}+P_{ACD}<P_{ABC}+P_{ACD'}<P_{ABCD'}$$
So quadrialteral $ABCD$ has smaller area compared to $ABCD'$ and cannot be optimal. In other words, if (1) is not true, we can always find a quadrilateral with a bigger area. Consequentially, the quadrialteral with the biggest area must have sides $BC$, $CD$ and $DA$ of equal lengths which is possible only if (1) is true.
And such quadrialteral is also a trapezoid. Any other quadrilateral, being it a trapezoid or not must have a smaller area.
No so ellegant approach, but still without trigonometry:
Denote the height of the trapezod with $h$ and the length of the smaller base with $b$:
You have to maximize the following expression:
$$A=frac{(2R+b)h}2$$
...or:
$$B=A^2=left(R+frac b2right)^2h^2tag{1}$$
...with the following constraint:
$$h^2=R^2-left(frac b2right)^2tag{2}$$
Replace (2) into (1) and you get:
$$B=left(R+frac b2right)^2left(R^2-left(frac b2right)^2right)$$
$$B=left(R+frac b2right)^3left(R-frac b2right)$$
For simplicity, introduce expression $c=R+frac b2$:
$$B(c)=c^3(2R-c)$$
$$B'(c)=3c^2(2R-c)-c^3=0implies c=frac{3R}2implies b=R$$
answered Jan 7 at 8:34
OldboyOldboy
7,6571935
$endgroup$
You don't need any trigonometry here.
Ellegant approach: Try to think outside of the box! Ask yourself a more general question:
Of all quadrilaterals $ABCD$ inscribed in a semicricle with diameter AB, which one has the maximum area?
We'll prove that for optimal quadrilateral:
$$BC=CD=DA=Rtag{1}$$
Suppose that (1) is not true. In other words, suppose that the optimal quadrilateral $ABCD$ looks like in the picture above and $ADneq CD$. Find point $D'$ on arc AC such that $AD'=CD'$.
Note that triangle $triangle ACD'$ has bigger area than $triangle ACD$ because $h_{D'}>h_{D}$. So obviously:
$$P_{ABCD}=P_{ABC}+P_{ACD}<P_{ABC}+P_{ACD'}<P_{ABCD'}$$
So quadrialteral $ABCD$ has smaller area compared to $ABCD'$ and cannot be optimal. In other words, if (1) is not true, we can always find a quadrilateral with a bigger area. Consequentially, the quadrialteral with the biggest area must have sides $BC$, $CD$ and $DA$ of equal lengths which is possible only if (1) is true.
And such quadrialteral is also a trapezoid. Any other quadrilateral, being it a trapezoid or not must have a smaller area.
No so ellegant approach, but still without trigonometry:
Denote the height of the trapezod with $h$ and the length of the smaller base with $b$:
You have to maximize the following expression:
$$A=frac{(2R+b)h}2$$
...or:
$$B=A^2=left(R+frac b2right)^2h^2tag{1}$$
...with the following constraint:
$$h^2=R^2-left(frac b2right)^2tag{2}$$
Replace (2) into (1) and you get:
$$B=left(R+frac b2right)^2left(R^2-left(frac b2right)^2right)$$
$$B=left(R+frac b2right)^3left(R-frac b2right)$$
For simplicity, introduce expression $c=R+frac b2$:
$$B(c)=c^3(2R-c)$$
$$B'(c)=3c^2(2R-c)-c^3=0implies c=frac{3R}2implies b=R$$
answered Jan 7 at 8:34
OldboyOldboy
7,6571935
edited Jan 7 at 10:09
answered Jan 7 at 8:34
OldboyOldboy
7,6571935
answered Jan 7 at 8:34
OldboyOldboy
7,6571935
answered Jan 7 at 8:34
OldboyOldboy
7,6571935
7,6571935
$begingroup$
Marvelous answer. Thank you.
$endgroup$
– daedsidog
Jan 7 at 12:08
add a comment |
$begingroup$
Marvelous answer. Thank you.
$endgroup$
– daedsidog
Jan 7 at 12:08
$begingroup$
Marvelous answer. Thank you.
$endgroup$
– daedsidog
Jan 7 at 12:08
$begingroup$
Marvelous answer. Thank you.
$endgroup$
– daedsidog
Jan 7 at 12:08
add a comment |
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$begingroup$
Yes, there is :)
$endgroup$
– Oldboy
Jan 7 at 10:10