2023 SSMO Accuracy Round Problems/Problem 4

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 $\frac{1}{2} \cdot 10 \cdot 8 = \boxed{40}.$

$[asy] size(7cm); pair a = (10,10); pair b = (0,10); pair c = (0,0); pair d = (10,0); pair e = (2,2); pair f = (8,2); draw((5,0)--(5,10), dashed+magenta); draw(a--c, blue); draw(e--f, blue); draw(a--b--c--d--cycle, blue+linewidth(1)); fill(a--b--c--d--cycle, lightcyan); fill(circumcircle(a,b,e), lightgreen); fill(circumcircle(c,d,e), lightgreen); fill(b--f--c--cycle, lightblue); dot("$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$", f, dir(0)); [/asy]$

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2023 SSMO Accuracy Round Problems/Problem 4

Problem

Solution