2023 SSMO Accuracy Round Problems/Problem 4

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Problem

In square $ABCD,$ point $E$ is selected on diagonal $AC.$ Let $F$ be the intersection of the circumcircles of triangles $ABE$ and $CDE.$ Given that $AB = 10$ and $EF = 6,$ find the maximum possible area of triangle $BEC.$ (A circumcircle of some triangle $\triangle ABC$ is the circle containing $A$, $B$, and $C$)

Solution

Note that by symmetry, $F$ is the reflection of $E$ over the perpendicular bisector of $CD$. Define coordinates $A = (10, 10)$, $B = (0, 10)$, $C = (0, 0)$, and $D = (10, 0)$. Since $EF = 6$, it follows that either $E$ is $(8, 8)$ or $(2, 2)$, so $F$ is either $(2, 8)$ or $(8, 2)$. The second case produces the larger solution of \[\frac{1}{2} \cdot 10 \cdot 8 = \boxed{40}.\]

size(7cm);
point a, b, c, d, e, f;
a = (10,10);
b = (0,10);
c = (0,0);
d = (10,0);
e = (2, 2);
f = (8, 2);

draw(line((5,0),(5,10)), dashed+magenta);

draw(a--c, blue);
draw(e--f, blue);
filldraw(a--b--c--d--cycle, opacity(0.2)+palecyan, blue);

filldraw(circumcircle(a,b,e), opacity(0.2)+palegreen, green);
filldraw(circumcircle(c,d,e), opacity(0.2)+palegreen, green);

filldraw(b--f--c--cycle, opacity(0.2)+lightblue, blue);

dot("$A$", a, dir(30));
dot("$B$", b, dir(150));
dot("$C$", c, dir(220));
dot("$D$", d, dir(340));
dot("$E$", e, dir(180));
dot("$F$", f, dir(0));
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