Kvant Math Problem 1101

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Verdicts: FAIL + FAIL
Solve time: 10m37s
Source on kvant.digital

Problem

On the lateral sides $AB$ and $AC$ of the isosceles triangle $ABC$, there are points $D$ and $E$ respectively such that $AD=BC=EC$ and the triangle $ADE$ are isosceles. What can the vertex angle at $A$ be?

V. Kiriak (Romania)

Solution

Let $ABC$ be an isosceles triangle with $AB=AC=s$ and $BC=b$. Let $D\in AB$ and $E\in AC$ satisfy

$AD=BC=EC=b$.

Since $D$ lies on $AB$ and $E$ lies on $AC$, one has $b\le s$. In an isosceles triangle,

$b=2s\sin\frac{\angle A}{2}$,

$2s\sin\frac{\angle A}{2}\le s$, hence $\sin\frac{\angle A}{2}\le \frac12$ and $\angle A\le 60^\circ$.

Thus all admissible configurations satisfy an acute or right vertex angle bounded above by $60^\circ$.

Coordinate model

Place $A$ at the origin. Let $AB$ lie on the positive $x$-axis and let $AC$ form angle $\theta=\angle A$ with $AB$. Then

$B=(s,0)$, $C=(s\cos\theta,s\sin\theta)$.

Compute

$t=BC=2s\sin\frac{\theta}{2}$.

Since $AD=BC=EC=t$, the points are

$D=(t,0)$ and $E=((s-t)\cos\theta,(s-t)\sin\theta)$.

Scaling does not affect angle conditions, so set $s=1$. Then

$t=2\sin\frac{\theta}{2}$ and $E=((1-t)\cos\theta,(1-t)\sin\theta)$.

The condition that $\triangle ADE$ is isosceles becomes one of

$AD=AE$, $AD=DE$, $AE=DE$.

Case 1: $AD=AE$

Here $AD=t$ and $AE=1-t$, hence $t=1-t$ and $t=\frac12$.

Then $2\sin\frac{\theta}{2}=\frac12$, so $\sin\frac{\theta}{2}=\frac14$ and

$\angle A=2\arcsin\frac14$.

Case 2: $AD=DE$

The distance computation gives

$DE^2=t^2+(1-t)^2-2t(1-t)\cos\theta$.

The condition $AD=DE$ becomes

$t^2=t^2+(1-t)^2-2t(1-t)\cos\theta$, hence

$(1-t)^2=2t(1-t)\cos\theta$.

At this point two possibilities occur. If $1-t\ne0$, division is valid and gives

$1-t=2t\cos\theta$.

If $1-t=0$, then $t=1$. This case is not excluded by the geometry and must be checked separately. From $t=2\sin\frac{\theta}{2}=1$ one gets $\sin\frac{\theta}{2}=\frac12$, hence $\theta=60^\circ$. Then $AB=AC=BC$, so $ABC$ is equilateral. Consequently $D=B$ and $E=C$, and triangle $ADE$ coincides with $ABC$, hence is isosceles. This produces the valid vertex angle $\angle A=60^\circ$.

Assume now $1-t\ne0$. Substitute $t=2x$ where $x=\sin\frac{\theta}{2}$ and $\cos\theta=1-2x^2$. The equation becomes

$1-2x=4x(1-2x^2)$,

which reduces to

$8x^3-6x+1=0$.

This cubic has a unique root in $(0,\frac12)$, giving

$\angle A=2\arcsin x$, where $x$ satisfies $8x^3-6x+1=0$.

Case 3: $AE=DE$

Here $AE=1-t$ and the same expression for $DE^2$ gives

$(1-t)^2=t^2+(1-t)^2-2t(1-t)\cos\theta$,

$t^2=2t(1-t)\cos\theta$.

Since $t\ne0$, divide by $t$ to obtain

$t=2(1-t)\cos\theta$.

Substitute $t=2x$ and $\cos\theta=1-2x^2$, yielding

$2x=2(1-2x)(1-2x^2)$,

which simplifies to

$4x^3-2x^2-3x+1=0$.

This factors as

$(x-1)(4x^2+2x-1)=0$.

The admissible root in $(0,\frac12)$ is $x=\frac{\sqrt5-1}{4}$, hence

$\angle A=2\arcsin\frac{\sqrt5-1}{4}$.

Conclusion

All possible vertex angles $\angle A$ are

$60^\circ$, $2\arcsin\frac14$, $2\arcsin\frac{\sqrt5-1}{4}$, and $2\arcsin x$, where $x$ is the unique root in $(0,\frac12)$ of $8x^3-6x+1=0$.