2021 JMC 10 Problems/Problem 24

Problem

In cyclic convex hexagon $AZBXCY,$ diagonals $\overline{AX}$ , $\overline{BY}$ , and $\overline{CZ}$ concur at the circumcenter of the hexagon, and quadrilateral $BCYZ$ has area $1.$ If the sum of the areas of $\triangle{ABC}$ and the original hexagon is equal to $3,$ what is the sum of the areas of quadrilaterals $XZAC$ and $YXBA?$

$\textbf{(A) }\dfrac{3}{2}\qquad\textbf{(B) }2\qquad\textbf{(C) }\dfrac{8}{3}\qquad\textbf{(D) }3\qquad\textbf{(E) }4$

Solution

Note that $[BCYZ] +[CAZX] +[ABXY]$ is equivalent to counting the central hexagon three times, the smaller triangles sharing a side with that hexagon twice, and the outer triangles just once. If $H$ is the orthocenter of triangle $ABC$ , it is well-known that $H$ and $X$ are reflections of each other across the midpoint of $BC$ . Therefore, $\triangle{BHC}$ and $\triangle{CXB}$ are congruent.

Using analogous reasoning and summing congruent areas, we see that $[\triangle{ABC}]=\tfrac{[AZBXCY]}{2} =1$ . Similarly, $[\triangle{XYZ}]=1$ . Notice that $[\triangle{ABC}]+[\triangle{XYZ}]$ counts the central hexagon two times and the smaller triangles sharing a side with that hexagon once. Therefore, adding in one copy of $[AZBXCY]$ will equate to the sum of all three rectangles; $[AZBXCY] + [\triangle{ABC}] +[\triangle{XYZ}] = 4 = [BCYZ] +[CAZX] +[ABXY],$ so we have $[CAZX] +[ABXY]=3,$ as desired.

$[asy]size(4cm); draw(circle((0,0),1)); pair A=(0,1); pair X=(0,-1); pair Z =(-0.8660254038,0.5); pair C =(0.8660254038,-0.5); pair Y=(0.9659258263,0.2588190451); pair B=(-0.9659258263,-0.2588190451); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$Z$", Z, NW); label("$Y$", Y, NE); label("$X$", X, S); draw(A--B--C--cycle, purple+dashed); draw(X--Y--Z--cycle, dashed); draw(A--Z--B--X--C--Y--cycle);[/asy]$

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2021 JMC 10 Problems/Problem 24

Problem

Solution