2023 USAJMO Problems/Problem 6

Problem

Isosceles triangle $ABC$ , with $AB=AC$ , is inscribed in circle $\omega$ . Let $D$ be an arbitrary point inside $BC$ such that $BD\neq DC$ . Ray $AD$ intersects $\omega$ again at $E$ (other than $A$ ). Point $F$ (other than $E$ ) is chosen on $\omega$ such that $\angle DFE = 90^\circ$ . Line $FE$ intersects rays $AB$ and $AC$ at points $X$ and $Y$ , respectively. Prove that $\angle XDE = \angle EDY$ .

Solution

All angle and side length names are defined as in the figures below. Figure 1 is the diagram of the problem while Figure 2 is the diagram of the Ratio Lemma. Do note that the point names defined in the Ratio Lemma are not necessarily the same defined points on Figure 1.

First, we claim the Ratio Lemma:

$\frac{\sin{\theta_2}}{\sin{\theta_2’}}=\frac{PB}{PC}\cdot\frac{b}{c}.$

We prove this as follows:

By the Law of Sines of $\triangle{ABP},$ we get: $\frac{\sin\theta_2}{BP}=\frac{\sin\theta_1}{c}=\frac{\sin B}{AP}.$ By the Law of Sines of $\triangle{ACP},$ we get: $\frac{\sin\theta_2’}{PC}=\frac{\sin\theta_1’}{b}=\frac{\sin C}{AP}.$ Dividing each of the corresponding equations, we get: $\frac{PC}{PB}\cdot\frac{\sin\theta_2}{\sin\theta_1}=\frac{b}{c}\cdot\frac{\sin\theta_1}{\sin\theta_1’}=\frac{\sin B}{\sin C}.$ Noting that $\sin \theta_1= \sin(180^\circ - \theta_1’)=\sin \theta_1’,$ we have: $\frac{\sin\theta_2}{\sin\theta_1}=\frac{PB}{PC}\cdot\frac{b}{c}.$

We transition into our problem.

By Power of a Point on circle $\omega,$ we have: $XK\cdot XD=XF\cdot XE$ $YL\cdot YD=YF\cdot YE.$ By power of a point on circle $\omega_1,$ we have: $XB\cdot XA=XF\cdot XE$ $YC\cdot YA=YF\cdot YE.$ Using the transitive property, we have: $XK\cdot XD=XB\cdot XA$ $YC\cdot YA=YL\cdot YD.$ Thus, by reverse Power of a Point, we have that quadrilaterals $ADKB$ and $ACLD$ are cyclic.

Now, we move to angle chasing.

$\angle{ABD}=\angle{AKD}=\alpha$ (from cyclic $ADKB$ )

$\angle{ABD}=\alpha=\angle{ACD}$ (from the isosceles triangle)

$\angle{ACD}=\alpha=\angle{ALD}$ (from cyclic $ACLD$ )

It is from here in which we focus on quadrilateral $AKEL.$

Let $\angle KDE=\beta$ and $\angle EDL=\gamma.$ Now realize that all we want to show, as given by the problem statement, is $\beta=\gamma.$

Thus, $\angle{KED}=90^\circ-\beta, \angle{DEL}=90^\circ-\gamma.$

We also have: \begin{align*} \angle LAD&=180^\circ-(\angle ADL+\alpha)\\ &=180^\circ-((180^\circ-\gamma)+\alpha))\\ &=\gamma-\alpha. \end{align*} Similarly, $\angle{KAD}=\beta-\alpha.$

Calling back to what we want to show - if we want to show that $\beta=\gamma,$ it suffices to show that $\angle KAD=\angle LAD$ and $\angle KED=\angle LED.$ But this is equivalent to showing $\triangle{AKE}\sim\triangle{ALE}.$

However, applying our Ratio Lemma to $\triangle{AKE},$ we have: $\frac{\sin\alpha}{\sin 90^\circ}=\frac{AD}{DE}\cdot\frac{KE}{KA}.$ Applying our Ratio Lemma to $\triangle{ALE},$ we have: $\frac{\sin\alpha}{\sin 90^\circ}=\frac{AD}{DE}\cdot\frac{LE}{LA}.$ From the transitive property, we have: $\frac{KE}{KA}=\frac{LE}{LA}.$ Now, realize that we already have a shared side $AE$ and equivalent angles $\angle{ALE}=90^\circ+\alpha=\angle{AKE}.$ Thus, we establish a stronger bound between these two triangles - a congruence.

Thus, these two triangles are similar, and our proof is complete.

mathboy282

(cred. to v_Enhance diagram)

2023 USAJMO (Problems • Resources)
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2023 USAJMO Problems/Problem 6

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