Difference between revisions of "2024 SSMO Relay Round 1 Problems/Problem 1"
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==Solution== | ==Solution== | ||
| + | From the Pythagorean Theorem, we have <math>DC = 15.</math> From similar triangles, <math>\frac{DC}{AD} = \frac{EC}{DE} = \frac{3}{4}\implies AD = 20.</math> So, <math>BD = \sqrt{29^2-20^2} = 21.</math> In conclusion, <cmath>[ABC] = \frac{1}{2}(BC)AD = \frac{1}{2}(BD+DC)(AD) = \frac{1}{2}(21+15)(20) = \boxed{360}.</cmath> | ||
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| + | ~SMO_Team | ||
Latest revision as of 14:42, 10 September 2025
Problem
Let 
be an altitude of triangle 
and let 
be an altitude of triangle 
If 
and 
what is the area of triangle 
![[asy] size(200); pair A,B,C,D,EE; A = (0,20); B = (-21,0); C = (15,0); D = (0,0); EE = 0.36*A+0.64*C; draw(A--B--C--cycle); draw(A--D--EE); dot("\(A\)", A, N); dot("\(B\)", B, W); dot("\(C\)", C, E); dot("\(D\)", D, S); dot("\(E\)", EE, dir(EE)); [/asy]](http://latex.artofproblemsolving.com/5/6/9/5698f4b3dca327e78898c650e94b6d413c0f49fb.png)
Solution
From the Pythagorean Theorem, we have 
From similar triangles, 
So, 
In conclusion, ![\[[ABC] = \frac{1}{2}(BC)AD = \frac{1}{2}(BD+DC)(AD) = \frac{1}{2}(21+15)(20) = \boxed{360}.\]](http://latex.artofproblemsolving.com/4/e/8/4e89c41f8ee33d38d0fd5c629461ed37b57068ac.png)
~SMO_Team
