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Difference between revisions of "2023 SSMO Accuracy Round Problems/Problem 4"

(Created page with "==Problem== In square <math>ABCD,</math> point <math>E</math> is selected on diagonal <math>AC.</math> Let <math>F</math> be the intersection of the circumcircles of triangles...")
 
 
(One intermediate revision by the same user not shown)
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==Solution==
 
==Solution==
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 +
Note that by symmetry, <math>F</math> is the reflection of <math>E</math> over the perpendicular bisector of <math>CD</math>. Define coordinates <math>A = (10, 10)</math>, <math>B = (0, 10)</math>, <math>C = (0, 0)</math>, and <math>D = (10, 0)</math>. Since <math>EF = 6</math>, it follows that either <math>E</math> is <math>(8, 8)</math> or <math>(2, 2)</math>, so <math>F</math> is either <math>(2, 8)</math> or <math>(8, 2)</math>. The second case produces the larger solution of <cmath>\frac{1}{2} \cdot 10 \cdot 8 = \boxed{40}.</cmath>
 +
 +
<asy>
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size(7cm);
 +
 +
pair a = (10,10);
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pair b = (0,10);
 +
pair c = (0,0);
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pair d = (10,0);
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pair e = (2,2);
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pair f = (8,2);
 +
 +
draw((5,0)--(5,10), dashed+magenta);
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 +
draw(a--c, blue);
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draw(e--f, blue);
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draw(a--b--c--d--cycle, blue+linewidth(1));
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 +
fill(a--b--c--d--cycle, lightcyan);
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fill(circumcircle(a,b,e), lightgreen);
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fill(circumcircle(c,d,e), lightgreen);
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fill(b--f--c--cycle, lightblue);
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 +
dot("$A$", a, dir(30));
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dot("$B$", b, dir(150));
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dot("$C$", c, dir(220));
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dot("$D$", d, dir(340));
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dot("$E$", e, dir(180));
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dot("$F$", f, dir(0));
 +
</asy>

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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