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Difference between revisions of "2023 SSMO Accuracy Round Problems/Problem 4"
 
Line 8: Line 8:
 
<asy>
 
<asy>
 
size(7cm);
 
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);
+
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(a--c, blue);
 
draw(e--f, blue);
 
draw(e--f, blue);
−
filldraw(a--b--c--d--cycle, opacity(0.2)+palecyan, blue);
+
draw(a--b--c--d--cycle, blue+linewidth(1));
−
 
 
−
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);
+
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("$A$", a, dir(30));

Latest revision as of 14:55, 9 September 2025

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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