Mixing
Show that $CDperp AB $.
$begingroup$
Let $triangle ABC $ with all angles $<90°$. Let $A_1$ the middle of $ [BC] $.
If $angle BAA_1=30°$ and $Din [AB] $ s.t. $CD=AB $ show that $CDperp AB $.
My idea : I draw $A_1Tperp AB $, $Tin [AB]$.
I have to show that $A_1T $ is the middle line in $triangle CBD $ $Leftrightarrow$ $A_1T=frac {CD}{2}=frac {AB}{2} $.
But $A_1T=frac{AA_1}{2} $. So I need to prove that $ AA_1=AB $. Now I am stuck.
geometry euclidean-geometry circle
asked Jan 4 at 11:14
rafarafa
598212
$endgroup$
add a comment |
$begingroup$
Let $triangle ABC $ with all angles $<90°$. Let $A_1$ the middle of $ [BC] $.
If $angle BAA_1=30°$ and $Din [AB] $ s.t. $CD=AB $ show that $CDperp AB $.
My idea : I draw $A_1Tperp AB $, $Tin [AB]$.
I have to show that $A_1T $ is the middle line in $triangle CBD $ $Leftrightarrow$ $A_1T=frac {CD}{2}=frac {AB}{2} $.
But $A_1T=frac{AA_1}{2} $. So I need to prove that $ AA_1=AB $. Now I am stuck.
geometry euclidean-geometry circle
asked Jan 4 at 11:14
rafarafa
598212
$endgroup$
$begingroup$
Is this supposed to be true for, for example, an equilateral triangle?
$endgroup$
– random
Jan 4 at 12:25
$begingroup$
@random In an equilateral triangle there is no point $D $ with that property
$endgroup$
– rafa
Jan 4 at 13:05
add a comment |
$begingroup$
Let $triangle ABC $ with all angles $<90°$. Let $A_1$ the middle of $ [BC] $.
If $angle BAA_1=30°$ and $Din [AB] $ s.t. $CD=AB $ show that $CDperp AB $.
My idea : I draw $A_1Tperp AB $, $Tin [AB]$.
I have to show that $A_1T $ is the middle line in $triangle CBD $ $Leftrightarrow$ $A_1T=frac {CD}{2}=frac {AB}{2} $.
But $A_1T=frac{AA_1}{2} $. So I need to prove that $ AA_1=AB $. Now I am stuck.
geometry euclidean-geometry circle
asked Jan 4 at 11:14
rafarafa
598212
$endgroup$
Let $triangle ABC $ with all angles $<90°$. Let $A_1$ the middle of $ [BC] $.
If $angle BAA_1=30°$ and $Din [AB] $ s.t. $CD=AB $ show that $CDperp AB $.
My idea : I draw $A_1Tperp AB $, $Tin [AB]$.
I have to show that $A_1T $ is the middle line in $triangle CBD $ $Leftrightarrow$ $A_1T=frac {CD}{2}=frac {AB}{2} $.
But $A_1T=frac{AA_1}{2} $. So I need to prove that $ AA_1=AB $. Now I am stuck.
geometry euclidean-geometry circle
geometry euclidean-geometry circle
asked Jan 4 at 11:14
rafarafa
598212
asked Jan 4 at 11:14
rafarafa
598212
edited Jan 4 at 11:33
rafa
asked Jan 4 at 11:14
rafarafa
598212
asked Jan 4 at 11:14
rafarafa
598212
asked Jan 4 at 11:14
rafarafa
598212
598212
$begingroup$
Is this supposed to be true for, for example, an equilateral triangle?
$endgroup$
– random
Jan 4 at 12:25
$begingroup$
@random In an equilateral triangle there is no point $D $ with that property
$endgroup$
– rafa
Jan 4 at 13:05
add a comment |
$begingroup$
Is this supposed to be true for, for example, an equilateral triangle?
$endgroup$
– random
Jan 4 at 12:25
$begingroup$
@random In an equilateral triangle there is no point $D $ with that property
$endgroup$
– rafa
Jan 4 at 13:05
$begingroup$
Is this supposed to be true for, for example, an equilateral triangle?
$endgroup$
– random
Jan 4 at 12:25
$begingroup$
Is this supposed to be true for, for example, an equilateral triangle?
$endgroup$
– random
Jan 4 at 12:25
$begingroup$
@random In an equilateral triangle there is no point $D $ with that property
$endgroup$
– rafa
Jan 4 at 13:05
$begingroup$
@random In an equilateral triangle there is no point $D $ with that property
$endgroup$
– rafa
Jan 4 at 13:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $A = (1,0),$ $B=(0,0),$ and $C = left(2 - frac2{sqrt3},frac23right).$
Since $2 - frac2{sqrt3} approx 0.845299,$
angles $angle CAB$ and $angle CBA$ are acute,
and since the distance from $C$ to $left(frac12,0right)$ is greater than
$frac12,$ the angle $angle ABC$ also is acute.
Then $A_1 = left(1 - frac1{sqrt3},frac13right).$
Line $AA_1$ has the equation $y = frac1{sqrt3}(1 - x),$
which you can confirm by plugging in the coordinates of $A$ and $A_1,$
and the slope of $AA_1$ is
$-frac1{sqrt3} = tan(-30^circ),$
so $angle BAA_1 = 30^circ.$
The circle of radius $AB = 1$ around $C$ has equation
$$left(x + frac2{sqrt3} - 2right)^2 + left(y - frac23right)^2 = 1.$$
Put $y = 0$ in this equation and we find that
$$left(x + frac2{sqrt3} - 2right)^2 + frac49 = 1,$$
$$x + frac2{sqrt3} - 2 = pm sqrt{frac59},$$
$$x = 2 - frac2{sqrt3} pm frac{sqrt5}3.$$
We have $frac{sqrt5}3 approx 0.745356,$
so one of the solutions for $x$ is
$x_1 = 2 - frac2{sqrt3} - frac{sqrt5}3 approx 0.099943.$
Let $D = (x_1, 0).$ Then $D$ lies on the segment $AB$ and $CD = AB = 1,$
but $CD$ is not perpendicular to $AB.$
In fact, if $C$ is any point $(x,y)$ that satisfies $y = frac1{sqrt3}(2 - x)$
and $0 < x < 1,$ then all three angles of the triangle are acute and
$angle BAA_1 = 30^circ.$
But the only way we have $D$ on the segment $AB$ with $CD = AB$ and $CD perp AB$
is if $C = left(1 - frac{sqrt3}2,1right).$
It seems some necessary condition is missing from your problem statement.
answered Jan 4 at 14:12
David KDavid K
53.2k341115
$endgroup$
$begingroup$
I came to the same conclusion by drawing a counterexample.
$endgroup$
– Oldboy
Jan 4 at 14:15
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $A = (1,0),$ $B=(0,0),$ and $C = left(2 - frac2{sqrt3},frac23right).$
Since $2 - frac2{sqrt3} approx 0.845299,$
angles $angle CAB$ and $angle CBA$ are acute,
and since the distance from $C$ to $left(frac12,0right)$ is greater than
$frac12,$ the angle $angle ABC$ also is acute.
Then $A_1 = left(1 - frac1{sqrt3},frac13right).$
Line $AA_1$ has the equation $y = frac1{sqrt3}(1 - x),$
which you can confirm by plugging in the coordinates of $A$ and $A_1,$
and the slope of $AA_1$ is
$-frac1{sqrt3} = tan(-30^circ),$
so $angle BAA_1 = 30^circ.$
The circle of radius $AB = 1$ around $C$ has equation
$$left(x + frac2{sqrt3} - 2right)^2 + left(y - frac23right)^2 = 1.$$
Put $y = 0$ in this equation and we find that
$$left(x + frac2{sqrt3} - 2right)^2 + frac49 = 1,$$
$$x + frac2{sqrt3} - 2 = pm sqrt{frac59},$$
$$x = 2 - frac2{sqrt3} pm frac{sqrt5}3.$$
We have $frac{sqrt5}3 approx 0.745356,$
so one of the solutions for $x$ is
$x_1 = 2 - frac2{sqrt3} - frac{sqrt5}3 approx 0.099943.$
Let $D = (x_1, 0).$ Then $D$ lies on the segment $AB$ and $CD = AB = 1,$
but $CD$ is not perpendicular to $AB.$
In fact, if $C$ is any point $(x,y)$ that satisfies $y = frac1{sqrt3}(2 - x)$
and $0 < x < 1,$ then all three angles of the triangle are acute and
$angle BAA_1 = 30^circ.$
But the only way we have $D$ on the segment $AB$ with $CD = AB$ and $CD perp AB$
is if $C = left(1 - frac{sqrt3}2,1right).$
It seems some necessary condition is missing from your problem statement.
answered Jan 4 at 14:12
David KDavid K
53.2k341115
$endgroup$
$begingroup$
I came to the same conclusion by drawing a counterexample.
$endgroup$
– Oldboy
Jan 4 at 14:15
add a comment |
$begingroup$
Let $A = (1,0),$ $B=(0,0),$ and $C = left(2 - frac2{sqrt3},frac23right).$
Since $2 - frac2{sqrt3} approx 0.845299,$
angles $angle CAB$ and $angle CBA$ are acute,
and since the distance from $C$ to $left(frac12,0right)$ is greater than
$frac12,$ the angle $angle ABC$ also is acute.
Then $A_1 = left(1 - frac1{sqrt3},frac13right).$
Line $AA_1$ has the equation $y = frac1{sqrt3}(1 - x),$
which you can confirm by plugging in the coordinates of $A$ and $A_1,$
and the slope of $AA_1$ is
$-frac1{sqrt3} = tan(-30^circ),$
so $angle BAA_1 = 30^circ.$
The circle of radius $AB = 1$ around $C$ has equation
$$left(x + frac2{sqrt3} - 2right)^2 + left(y - frac23right)^2 = 1.$$
Put $y = 0$ in this equation and we find that
$$left(x + frac2{sqrt3} - 2right)^2 + frac49 = 1,$$
$$x + frac2{sqrt3} - 2 = pm sqrt{frac59},$$
$$x = 2 - frac2{sqrt3} pm frac{sqrt5}3.$$
We have $frac{sqrt5}3 approx 0.745356,$
so one of the solutions for $x$ is
$x_1 = 2 - frac2{sqrt3} - frac{sqrt5}3 approx 0.099943.$
Let $D = (x_1, 0).$ Then $D$ lies on the segment $AB$ and $CD = AB = 1,$
but $CD$ is not perpendicular to $AB.$
In fact, if $C$ is any point $(x,y)$ that satisfies $y = frac1{sqrt3}(2 - x)$
and $0 < x < 1,$ then all three angles of the triangle are acute and
$angle BAA_1 = 30^circ.$
But the only way we have $D$ on the segment $AB$ with $CD = AB$ and $CD perp AB$
is if $C = left(1 - frac{sqrt3}2,1right).$
It seems some necessary condition is missing from your problem statement.
answered Jan 4 at 14:12
David KDavid K
53.2k341115
$endgroup$
$begingroup$
I came to the same conclusion by drawing a counterexample.
$endgroup$
– Oldboy
Jan 4 at 14:15
add a comment |
$begingroup$
Let $A = (1,0),$ $B=(0,0),$ and $C = left(2 - frac2{sqrt3},frac23right).$
Since $2 - frac2{sqrt3} approx 0.845299,$
angles $angle CAB$ and $angle CBA$ are acute,
and since the distance from $C$ to $left(frac12,0right)$ is greater than
$frac12,$ the angle $angle ABC$ also is acute.
Then $A_1 = left(1 - frac1{sqrt3},frac13right).$
Line $AA_1$ has the equation $y = frac1{sqrt3}(1 - x),$
which you can confirm by plugging in the coordinates of $A$ and $A_1,$
and the slope of $AA_1$ is
$-frac1{sqrt3} = tan(-30^circ),$
so $angle BAA_1 = 30^circ.$
The circle of radius $AB = 1$ around $C$ has equation
$$left(x + frac2{sqrt3} - 2right)^2 + left(y - frac23right)^2 = 1.$$
Put $y = 0$ in this equation and we find that
$$left(x + frac2{sqrt3} - 2right)^2 + frac49 = 1,$$
$$x + frac2{sqrt3} - 2 = pm sqrt{frac59},$$
$$x = 2 - frac2{sqrt3} pm frac{sqrt5}3.$$
We have $frac{sqrt5}3 approx 0.745356,$
so one of the solutions for $x$ is
$x_1 = 2 - frac2{sqrt3} - frac{sqrt5}3 approx 0.099943.$
Let $D = (x_1, 0).$ Then $D$ lies on the segment $AB$ and $CD = AB = 1,$
but $CD$ is not perpendicular to $AB.$
In fact, if $C$ is any point $(x,y)$ that satisfies $y = frac1{sqrt3}(2 - x)$
and $0 < x < 1,$ then all three angles of the triangle are acute and
$angle BAA_1 = 30^circ.$
But the only way we have $D$ on the segment $AB$ with $CD = AB$ and $CD perp AB$
is if $C = left(1 - frac{sqrt3}2,1right).$
It seems some necessary condition is missing from your problem statement.
answered Jan 4 at 14:12
David KDavid K
53.2k341115
$endgroup$
Let $A = (1,0),$ $B=(0,0),$ and $C = left(2 - frac2{sqrt3},frac23right).$
Since $2 - frac2{sqrt3} approx 0.845299,$
angles $angle CAB$ and $angle CBA$ are acute,
and since the distance from $C$ to $left(frac12,0right)$ is greater than
$frac12,$ the angle $angle ABC$ also is acute.
Then $A_1 = left(1 - frac1{sqrt3},frac13right).$
Line $AA_1$ has the equation $y = frac1{sqrt3}(1 - x),$
which you can confirm by plugging in the coordinates of $A$ and $A_1,$
and the slope of $AA_1$ is
$-frac1{sqrt3} = tan(-30^circ),$
so $angle BAA_1 = 30^circ.$
The circle of radius $AB = 1$ around $C$ has equation
$$left(x + frac2{sqrt3} - 2right)^2 + left(y - frac23right)^2 = 1.$$
Put $y = 0$ in this equation and we find that
$$left(x + frac2{sqrt3} - 2right)^2 + frac49 = 1,$$
$$x + frac2{sqrt3} - 2 = pm sqrt{frac59},$$
$$x = 2 - frac2{sqrt3} pm frac{sqrt5}3.$$
We have $frac{sqrt5}3 approx 0.745356,$
so one of the solutions for $x$ is
$x_1 = 2 - frac2{sqrt3} - frac{sqrt5}3 approx 0.099943.$
Let $D = (x_1, 0).$ Then $D$ lies on the segment $AB$ and $CD = AB = 1,$
but $CD$ is not perpendicular to $AB.$
In fact, if $C$ is any point $(x,y)$ that satisfies $y = frac1{sqrt3}(2 - x)$
and $0 < x < 1,$ then all three angles of the triangle are acute and
$angle BAA_1 = 30^circ.$
But the only way we have $D$ on the segment $AB$ with $CD = AB$ and $CD perp AB$
is if $C = left(1 - frac{sqrt3}2,1right).$
It seems some necessary condition is missing from your problem statement.
answered Jan 4 at 14:12
David KDavid K
53.2k341115
edited Jan 4 at 14:33
answered Jan 4 at 14:12
David KDavid K
53.2k341115
answered Jan 4 at 14:12
David KDavid K
53.2k341115
answered Jan 4 at 14:12
David KDavid K
53.2k341115
53.2k341115
$begingroup$
I came to the same conclusion by drawing a counterexample.
$endgroup$
– Oldboy
Jan 4 at 14:15
add a comment |
$begingroup$
I came to the same conclusion by drawing a counterexample.
$endgroup$
– Oldboy
Jan 4 at 14:15
$begingroup$
I came to the same conclusion by drawing a counterexample.
$endgroup$
– Oldboy
Jan 4 at 14:15
$begingroup$
I came to the same conclusion by drawing a counterexample.
$endgroup$
– Oldboy
Jan 4 at 14:15
add a comment |
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$begingroup$
Is this supposed to be true for, for example, an equilateral triangle?
$endgroup$
– random
Jan 4 at 12:25
$begingroup$
@random In an equilateral triangle there is no point $D $ with that property
$endgroup$
– rafa
Jan 4 at 13:05