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Difference between revisions of "2021 JMC 10 Problems/Problem 24"

(Created page with "==Problem== In cyclic convex hexagon <math>AZBXCY,</math> diagonals <math>\overline{AX}</math>, <math>\overline{BY}</math>, and <math>\overline{CZ}</math> concur at the circu...")
 
(→‎Solution)
 
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Using analogous reasoning and summing congruent areas, we see that <math>[\triangle{ABC}]=\frac{[AZBXCY]}{2} =1</math>. Similarly, <math>[\triangle{XYZ}]=1</math>. Notice that <math>[\triangle{ABC}]+[\triangle{XYZ}]</math> counts the central hexagon two times and the smaller triangles sharing a side with that hexagon once. Therefore, adding in one copy of <math>[AZBXCY]</math> will equate to the sum of all three rectangles;
+
Using analogous reasoning and summing congruent areas, we see that <math>[\triangle{ABC}]=\tfrac{[AZBXCY]}{2} =1</math>. Similarly, <math>[\triangle{XYZ}]=1</math>. Notice that <math>[\triangle{ABC}]+[\triangle{XYZ}]</math> counts the central hexagon two times and the smaller triangles sharing a side with that hexagon once. Therefore, adding in one copy of <math>[AZBXCY]</math> will equate to the sum of all three rectangles;
 
<cmath>[AZBXCY] + [\triangle{ABC}] +[\triangle{XYZ}] = 4 = [BCYZ] +[CAZX] +[ABXY],</cmath> so we have <math>[CAZX] +[ABXY]=3,</math> as desired.
 
<cmath>[AZBXCY] + [\triangle{ABC}] +[\triangle{XYZ}] = 4 = [BCYZ] +[CAZX] +[ABXY],</cmath> so we have <math>[CAZX] +[ABXY]=3,</math> as desired.
  

Latest revision as of 15:35, 1 April 2021

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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