2024 SSMO Relay Round 1 Problems/Problem 1

Problem

Let $AD$ be an altitude of triangle $ABC$ and let $DE$ be an altitude of triangle $ACD.$ If $AB=29,CE=9,$ and $DE=12,$ what is the area of triangle $ABC?$

$[asy] size(200); pair A,B,C,D,EE; A = (0,20); B = (-21,0); C = (15,0); D = (0,0); EE = 0.36*A+0.64*C; draw(A--B--C--cycle); draw(A--D--EE); dot("$A$", A, N); dot("$B$", B, W); dot("$C$", C, E); dot("$D$", D, S); dot("$E$", EE, dir(EE)); [/asy]$

Solution

From the Pythagorean Theorem, we have $DC = 15.$ From similar triangles, $\frac{DC}{AD} = \frac{EC}{DE} = \frac{3}{4}\implies AD = 20.$ So, $BD = \sqrt{29^2-20^2} = 21.$ In conclusion, $[ABC] = \frac{1}{2}(BC)AD = \frac{1}{2}(BD+DC)(AD) = \frac{1}{2}(21+15)(20) = \boxed{360}.$

~SMO_Team

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2024 SSMO Relay Round 1 Problems/Problem 1

Problem

Solution