Difference between revisions of "2024 SSMO Accuracy Round Problems/Problem 4"
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| + | Since <math>BDC</math> and <math>BEC</math> are isosceles right triangles, we have <math>DC = BC = EC</math>. Since <math>AD+DC+CE = EA = 13AD,</math> we have <math>DC+CE = 12AD \implies DC = CE = 6AD.</math> From here, we see that <math>AC = AD+DC = 7AD.</math> By the Pythagorean Theorem, <cmath>AC^2+BC^2 = AB^2 = 1\implies (7AD)^2+(6AD)^2 = 1\implies AD^2 = \frac{1}{85}.</cmath> The area of triangle <math>ABC</math> is <cmath>\frac{AC\cdot BC}{2} = 21AD^2 = \frac{21}{85}\implies 21+85 = \boxed{106}.</cmath> | ||
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| + | ~SMO_Team | ||
Latest revision as of 14:31, 10 September 2025
Problem
Right triangle 
has a right angle at 
and hypotenuse 
. Let points 
and 
lie on 
such that 
. 
and are colinear in that order. Given that 
, the area of 
can be expressed as 
for relatively prime 
and 
. Find 
.
Solution
Since 
and 
are isosceles right triangles, we have 
. Since 
we have 
From here, we see that 
By the Pythagorean Theorem, ![\[AC^2+BC^2 = AB^2 = 1\implies (7AD)^2+(6AD)^2 = 1\implies AD^2 = \frac{1}{85}.\]](http://latex.artofproblemsolving.com/6/b/5/6b5cb34bbab0631823cfb13ced638353203f7342.png)
The area of triangle is ![\[\frac{AC\cdot BC}{2} = 21AD^2 = \frac{21}{85}\implies 21+85 = \boxed{106}.\]](http://latex.artofproblemsolving.com/0/c/3/0c360ee5c228ab1b5174fafcae989cebe8559cb6.png)
~SMO_Team
