A triangular pyramid $ABCD$ is given. A sphere $\omega_A$ is tangent to the face $BCD$ and to the planes of other faces in points don't lying on faces. Similarly, sphere $\omega_B$ is tangent to the face $ACD$ and to the planes of other faces in points don't lying on faces. Let $K$ be the point where $\omega_A$ is tangent to $ACD$, and let $L$ be the point where $\omega_B$ is tangent to $BCD$. The points $X$ and $Y$ are chosen on the prolongations of $AK$ and $BL$ over $K$ and $L$ such that $\angle CKD = \angle CXD + \angle CBD$ and $\angle CLD = \angle CYD +\angle CAD$. Prove that the distances from the points $X$, $Y$ to the midpoint of $CD$ are the same.
HIDE: thanks Thanks to the user Vlados021 for translating the problem.Problem
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Tags: geometry
JAnatolGT_00
18.09.2022 16:58
Amazing problem. Let insphere $\omega$ of tetrahedron is tangent to face $BCD$ at $A_1$ and define $B_1,C_1,D_1$ analogously. Let $\omega_A$ is tangent to face $BCD$ at $K'$ and $\omega_B$ is tangent to face $ACD$ at $L'.$ By homothety it follows $A\in B_1K,B\in A_1L.$ Claim. $A_1CYD,B_1CXD$ are cyclic quadrilaterals. Proof. We'll prove that $X\in \odot (B_1CD).$ As well-known $B_1,L'$ are isogonal conjugates wrt $\triangle ACD,$ therefore $$\angle CB_1D=\pi+\angle CAD-\angle CL'D=\pi+\angle CAD-\angle CLD=\pi -\angle CYD\text{ } \Box$$ Let $f$ be a rotation wrt $CD,$ which maps $A$ onto plane $BCD,$ to the same semiplane wrt $CD$ as $B.$ Clearly $f(B_1)=A_1$ and $$2\angle AB_1C=\angle AB_1C+\angle AD_1C=4\pi -\angle AC_1B-\angle BA_1C-\angle AC_1D-\angle CA_1D=$$$$=\angle BA_1D+\angle BC_1D=2\angle BA_1D\implies \angle f(A)A_1C=\angle AB_1C=\angle BA_1D.$$By claim points $Y,f(X)$ lie on $\odot (A_1CD),$ so they are symmetric wrt perpendicular bisector of $CD.$ This completes proof.