Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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...")
 
Line 3: Line 3:
  
 
==Solution==
 
==Solution==
 +
 +
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>
 +
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));
 +
</asy>

Revision as of 14:54, 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 \[\frac{1}{2} \cdot 10 \cdot 8 = \boxed{40}.\] 

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));
 (Error making remote request. Unknown error_msg)
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_SSMO_Accuracy_Round_Problems/Problem_4&oldid=260136"