USA IMO

First Solution. The key observation is the following Lemma. Lemma. Segment $D_{1}Q$ is a diameter of circle $\omega$. Proof. Let $I$ be the center of circle $\omega$, that is, $I$ is the incenter of triangle $ABC$. Extend segment $D_{1}I$ through $I$ to intersect circle $\omega$ again at $Q^{\prime }$, and extend segment $A{Q}^{\prime }$ through $Q^{\prime }$ to intersect segment $BC$ at $D^{\prime }$. We show that $D_2=D^{\prime }$, which in turn implies that $Q=Q^{\prime }$, that is, $D_{1}Q$ is a diameter of $\omega$. Let $\ell$ be the line tangent to circle $\omega$ at $Q^{\prime }$, and let $\ell$ intersect segments $AB$ and $AC$ at $B_1$ and $C_1$, respectively. Then $\omega$ is an excircle of triangle $A{B}_1{C}_1$. Let ${\mathbf{H}}_1$ denote the dilation with center $A$ and ratio $A{D}^{\prime }/A{Q}^{\prime }$. Since $\ell \perp D_1{Q}^{\prime }$ and $BC\perp D_{1}Q$, $\ell \parallel BC$. Hence, $AB/A{B}_1=AC/A{C}_1=A{D}^{\prime }/A{Q}^{\prime }$. Thus, ${\mathbf{H}}_1(Q^{\prime })=D^{\prime }$, ${\mathbf{H}}_1(B_1)=B$, and ${\mathbf{H}}_1(C_1)=C$. It also follows that the excircle $\Omega$ of triangle $ABC$ opposite vertex $A$ is tangent to side $BC$ at $D^{\prime }$.

It is well known that $$C{D}_1=\frac{1}{2}(BC+CA-AB).(1)$$ We compute $B{D}^{\prime }$. Let $X$ and $Y$ denote the points of tangency of circle $\Omega$ with rays $AB$ and $AC$, respectively. Then by equal tangents, $AX=AY$, $B{D}^{\prime }=BX$, and $D^{\prime }C=YC$. Hence, $$\begin{array}{rl}AX& =AY=\frac{1}{2}(AX+AY)\\ & =\frac{1}{2}(AB+BX+YC+CA)\\ & =\frac{1}{2}(AB+BC+CA).\end{array}$$ It follows that $$B{D}^{\prime }=BX=AX-AB=\frac{1}{2}(BC+CA-AB).(2)$$ Combining (1) and (2) yields $B{D}^{\prime }=C{D}_1$. Thus, $$B{D}_2=B{D}_1-D_2{D}_1=D_{2}C-D_2{D}_1=D_{1}C=B{D}^{\prime },$$ that is, $D^{\prime }=D_2$, as desired. ■

Now we prove our main result. Let $M_1$ and $M_2$ be the midpoints of segments $BC$ and $CA$, respectively. Then $M_1$ is also the midpoint of segment $D_1{D}_2$, from which it follows that $I{M}_1$ is a midline of triangle $D_{1}Q{D}_2$. Hence, $$Q{D}_2=2I{M}_1(3)$$ and $A{D}_2\parallel M_{1}I$. Similarly, we can prove that $B{E}_2\parallel M_{2}I$. Let $G$ be the centroid of triangle $ABC$. Thus, segments $A{M}_1$ and $B{M}_2$ intersect at $G$. Define transformation ${\mathbf{H}}_2$ as the dilation with center $G$ and ratio $-1/2$. Then ${\mathbf{H}}_2(A)=M_1$ and ${\mathbf{H}}_2(B)=M_2$. Under the dilation, parallel lines go to parallel lines and the intersection of two lines goes to the intersection of their images. Since $A{D}_2\parallel M_{1}I$ and $B{E}_2\parallel M_{2}I$, $\mathbf{H}$ maps lines $A{D}_2$ and $B{E}_2$ to lines $M_{1}I$ and $M_{2}I$, respectively. It also follows that ${\mathbf{H}}_2(P)=I$ and that $$\frac{I{M}_1}{AP}=\frac{G{M}_1}{AG}=\frac{1}{2}$$ or $$AP=2I{M}_1.(4)$$ Combining (3) and (4) yields $$AQ=AP-QP=2I{M}_1-QP=Q{D}_2-QP=P{D}_2,$$ as desired.

Second Solution. From the Lemma, we have $$\frac{AQ}{A{D}_2}=\frac{r}{r_a},$$ where $r$ and $r_a$ are the radii of circles $\omega$ and $\Omega$, respectively. Note that $$r(AB+BC+CA)=2[ABC]$$ and that $$\begin{array}{rl}{r}_a(AB+AC-BC)& =2[I_{a}AB]+2[I_{a}AC]-2[I_{a}BC]\\ & =2[I_{a}BAC]-2[I_{a}BC]=2[ABC],\end{array}$$ where $I_a$ is the center of $\Omega$ and $[\mathcal{R}]$ is the area of region $\mathcal{R}$. Thus, $$\frac{AQ}{A{D}_2}=\frac{AB+AC-BC}{AB+BC+CA}.(5)$$ Applying Menelaus's Theorem to triangle $A{D}_{2}C$ and line $B{E}_2$ gives $$\frac{AP}{P{D}_2}\cdot \frac{D_{2}B}{BC}\cdot \frac{C{E}_2}{E_{2}A}=1,$$ or $$\begin{gathered} \frac{AP}{P{D}_2}=\frac{BC\cdot E_{2}A}{D_{2}B\cdot C{E}_2}=\frac{BC\cdot C{E}_1}{C{D}_1\cdot A{E}_1} \\ =\frac{BC}{A{E}_1}=\frac{2BC}{AB+AC-BC}. \end{gathered}$$ Hence, $$\frac{A{D}_2}{P{D}_2}=1+\frac{AP}{P{D}_2}=\frac{AB+AC+BC}{AB+AC-BC},$$ or $$\frac{P{D}_2}{A{D}_2}=\frac{AB+AC-BC}{AB+AC+BC}.(6)$$ The desired result now follows from (5) and (6).