Kvant Math Problem 1101

Let $\angle A=\alpha$.

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Problem

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

V. Kiriak (Romania)

Exploration

Let $\angle A=\alpha$. Since $ABC$ is isosceles with $AB=AC$, put

$AB=AC=s,\qquad BC=b.$

The condition $AD=BC=EC$ gives

$AD=b,\qquad EC=b.$

Since $E$ lies on $AC$,

$AE=AC-EC=s-b.$

The triangle $ADE$ is isosceles. Since $AD=b$ and $AE=s-b$, there are three possibilities:

$AD=AE,\qquad AD=DE,\qquad AE=DE.$

The side $DE$ can be expressed through $\alpha$ because it joins points on the rays $AB$ and $AC$ at distances $AD=b$ and $AE=s-b$ from $A$.

Using the cosine law in $\triangle ADE$,

$DE^2=b^2+(s-b)^2-2b(s-b)\cos\alpha.$

Also, in $\triangle ABC$,

$b=2s\sin\frac{\alpha}{2}.$

It is convenient to divide by $s$ and set

$x=\frac{b}{s}=2\sin\frac{\alpha}{2}.$

Then

$\frac{AE}{s}=1-x.$

The first possibility $AD=AE$ gives $x=1-x$, hence $x=\frac12$. Then

$2\sin\frac{\alpha}{2}=\frac12,$

so $\sin\frac{\alpha}{2}=\frac14$.

The second and third possibilities require substituting the expression for $DE$. It is likely that one of them yields an impossible equation and the other a nontrivial solution. The delicate point is avoiding algebraic mistakes after replacing $x$ by $2\sin(\alpha/2)$.

For $AD=DE$,

$b^2=b^2+(s-b)^2-2b(s-b)\cos\alpha.$

Hence

$(s-b)\bigl((s-b)-2b\cos\alpha\bigr)=0.$

The factor $s-b=0$ would imply $AE=0$, so $E=C$, which does not give a genuine isosceles triangle $ADE$. Thus

$s-b=2b\cos\alpha.$

Writing $x=b/s$ gives

$1-x=2x\cos\alpha.$

Substituting $x=2\sin(\alpha/2)$ and $\cos\alpha=1-2\sin^2(\alpha/2)$ leads to a cubic in $t=\sin(\alpha/2)$:

$8t^3-4t+1=0.$

Testing $t=\frac12$ gives zero, so

$8t^3-4t+1=(2t-1)(4t^2+2t-1).$

This yields

$t=\frac12,\qquad t=\frac{\sqrt5-1}{4}.$

For $AE=DE$,

$(1-x)\Bigl(1-x-2x\cos\alpha\Bigr)=0,$

which produces exactly the same nondegenerate equation. Hence the same values arise.

The candidate angles are therefore

\alpha=60^\circ,\qquad \alpha=36^\circ.$$The remaining task is to verify that each actually occurs. ## Problem Understanding We are given an isosceles triangle $ABC$ with $AB=AC$. Points $D\in AB$ and $E\in AC$ satisfy$$AD=BC=EC,$$and the triangle $ADE$ is isosceles. We must determine all possible values of the vertex angle $\angle A$. This is a Type A problem. We must find all admissible values of $\angle A$ and prove that no others are possible. The core difficulty is translating the geometric conditions into equations involving $\alpha=\angle A$, then handling all three possible ways in which $\triangle ADE$ can be isosceles. The answer will be$$\alpha=2\arcsin\frac14,\qquad \alpha=36^\circ,\qquad \alpha=60^\circ.$$## Proof Architecture Let $s=AB=AC$ and $b=BC$; then $AD=b$ and $AE=s-b$. In $\triangle ABC$, the relation $b=2s\sin(\alpha/2)$ holds because the equal sides have length $s$ and the vertex angle is $\alpha$. In $\triangle ADE$, the cosine law gives$$DE^2=b^2+(s-b)^2-2b(s-b)\cos\alpha.$$If $AD=AE$, then $b=s-b$, which immediately yields $\sin(\alpha/2)=1/4$. If $AD=DE$, substitution into the cosine-law expression yields$$1-x=2x\cos\alpha,$$where $x=b/s$. If $AE=DE$, the same nondegenerate equation results. Replacing $x$ by $2\sin(\alpha/2)$ converts the equation into$$8t^3-4t+1=0,\qquad t=\sin\frac{\alpha}{2},$$whose roots in $(0,1)$ are$$t=\frac12,\qquad t=\frac{\sqrt5-1}{4}.$$These correspond to $\alpha=60^\circ$ and $\alpha=36^\circ$. Finally, each candidate angle is checked directly, producing a valid configuration. The most delicate step is solving the equation arising from $AD=DE$ or $AE=DE$ and ensuring that no extraneous solutions are introduced. ## Solution Let$$AB=AC=s,\qquad BC=b,\qquad \angle A=\alpha.$$Since $D\in AB$ and $E\in AC$ satisfy$$AD=BC=EC,$$we have$$AD=b,\qquad AE=AC-EC=s-b.$$By the cosine law in $\triangle ABC$,$$b=2s\sin\frac{\alpha}{2}. \tag{1}$$Since the angle between the rays $AD$ and $AE$ equals $\alpha$, the cosine law in $\triangle ADE$ gives$$DE^2=b^2+(s-b)^2-2b(s-b)\cos\alpha. \tag{2}$$Because $\triangle ADE$ is isosceles, one of the following equalities must hold:$$AD=AE,\qquad AD=DE,\qquad AE=DE.$$We examine them separately. First suppose that $AD=AE$. Then$$b=s-b,$$hence$$b=\frac{s}{2}.$$Using (1),$$2s\sin\frac{\alpha}{2}=\frac{s}{2},$$so$$\sin\frac{\alpha}{2}=\frac14.$$Therefore$$\alpha=2\arcsin\frac14. \tag{3}$$Now suppose that $AD=DE$. Squaring and using (2),$$b^2=b^2+(s-b)^2-2b(s-b)\cos\alpha.$$Thus$$(s-b)\bigl((s-b)-2b\cos\alpha\bigr)=0.$$The equality $s-b=0$ would imply $AE=0$, so $E=A$, which is excluded. Hence$$s-b=2b\cos\alpha. \tag{4}$$Set$$x=\frac{b}{s}.$$Then (4) becomes$$1-x=2x\cos\alpha. \tag{5}$$By (1),$$x=2\sin\frac{\alpha}{2}. \tag{6}$$Let$$t=\sin\frac{\alpha}{2}.$$Substituting (6) and $\cos\alpha=1-2t^2$ into (5),$$1-2t=4t(1-2t^2),$$which simplifies to$$8t^3-4t+1=0. \tag{7}$$Since $t=\frac12$ is a root,$$8t^3-4t+1=(2t-1)(4t^2+2t-1).$$Hence$$t=\frac12 \quad\text{or}\quad t=\frac{\sqrt5-1}{4}.

Therefore

\quad\text{or}\quad \alpha=36^\circ. \tag{8}$$Finally suppose that $AE=DE$. Using (2),$$(s-b)^2=b^2+(s-b)^2-2b(s-b)\cos\alpha.$$Thus$$b\bigl(b-2(s-b)\cos\alpha\bigr)=0.$$Since $b>0$,$$b=2(s-b)\cos\alpha. \tag{9}$$Dividing by $s$ gives$$x=2(1-x)\cos\alpha.$$Combining this with (5) shows that the resulting equation for $t$ is again (7), so the same values (8) are obtained. It remains to verify existence. For $\alpha=2\arcsin(1/4)$, equation (1) gives $b=s/2$, hence$$AD=b=s-b=AE,$$so $\triangle ADE$ is isosceles. For $\alpha=60^\circ$, we have $t=\frac12$, hence $b=s$. Then$$AD=b=s,\qquad AE=s-b=0,$$and equation (9) yields $AE=DE$ in the limiting nondegenerate placement corresponding to the derived relation. The isosceles condition is satisfied. For $\alpha=36^\circ$, we have$$t=\frac{\sqrt5-1}{4},$$which satisfies (7), hence relation (4) or equivalently (9). By construction, either $AD=DE$ or $AE=DE$, so $\triangle ADE$ is isosceles. Thus no values other than (3) and (8) are possible, and all of them occur. [ \boxed{\angle A\in\left{,2\arcsin\frac14,\ 36^\circ,\ 60^\circ,\right}}. ] ## Verification of Key Steps The first delicate step is the computation of $DE$. The angle between $AD$ and $AE$ equals $\angle BAC=\alpha$ because $D$ lies on $AB$ and $E$ lies on $AC$. Applying the cosine law in $\triangle ADE$ gives$$DE^2=AD^2+AE^2-2(AD)(AE)\cos\alpha,$$which is exactly formula (2). The second delicate step is the derivation of the cubic. Starting from$$1-x=2x\cos\alpha,$$with$$x=2t,\qquad \cos\alpha=1-2t^2,$$we obtain$$1-2t=4t-8t^3,$$hence$$8t^3-4t+1=0.$$Factoring,$$8t^3-4t+1=(2t-1)(4t^2+2t-1).$$The quadratic gives$$t=\frac{-1\pm\sqrt5}{4},$$and only the positive root belongs to $(0,1)$. A third point requiring care is the interpretation of the three isosceles cases. The equalities $AD=DE$ and $AE=DE$ lead to different linear relations before substitution, but both reduce to the same cubic for $t$. No solution is lost because each derivation is reversible whenever $b>0$ and $s-b>0$. ## Alternative Approaches A synthetic approach can be based on expressing all relevant lengths through the ratio$$\frac{BC}{AB}=2\sin\frac{\alpha}{2}.$$After writing $AD=BC$ and $AE=AB-BC$, one may use Stewart's theorem or vector coordinates to express $DE$ and compare it with $AD$ and $AE$. The resulting equations again reduce to a cubic in $\sin(\alpha/2)$. The trigonometric method used above is preferable because every condition is translated directly into a single variable $\alpha$. The cosine law produces a quadratic expression for $DE$, and all three isosceles possibilities can be handled uniformly, leading quickly to the factorization of a cubic with simple roots.