Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Prove it United States

Problem:
As shown, $U$ and $C$ are points on the sides of triangle $M N H$ such that $M U = s$ , $U N = 6$ , $N C = 20$ , $C H = s$ , $H M = 25$ . If triangle $U N C$ and quadrilateral $M U C H$ have equal areas, what is $s$ ?

Solution

Solution:
Using brackets to denote areas, we have $[M C H] = + [M U C H] = 2$ . On the other hand, triangles with equal altitudes have their areas in the same ratio as their bases, so
$2 = \frac{[M N H]}{} = \frac{[M N H]}{[M N C]} \cdot \frac{[M N C]}{} = \frac{N H}{N C} \cdot \frac{M N}{U N} = \frac{s+20}{20} \cdot \frac{s+6}{6}.$
Clearing the denominator gives $(s+20)(s+6) = 240$ , and solving the quadratic gives $s = 4$ or $-30$ . Since $s > 0$ , we must have $s = 4$ .

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