2017 AIME I Problem 4

Below is the professionally curated solution for Problem 4 of the 2017 AIME I, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AIME I solutions, or check the answer key.

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Concepts:pyramidcircumcircle, circumcenter, and circumradiusPythagorean Theoremvolume

Difficulty rating: 2390

4.

A pyramid has a triangular base with side lengths 20,20, 20,20, and 24.24. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length 25.25. The volume of the pyramid is mn,m\sqrt{n}, where mm and nn are positive integers, and nn is not divisible by the square of any prime. Find m+n.m + n.

Solution:

Since the apex is equidistant from all three base vertices, its foot is the circumcenter of the base. The base is isosceles with sides 20,20,24:20, 20, 24: its altitude to the side of length 2424 is 202122=16,\sqrt{20^2 - 12^2} = 16, so its area is K=122416=192,K = \frac{1}{2} \cdot 24 \cdot 16 = 192, and its circumradius is R=abc4K=2020244192=252.R = \frac{abc}{4K} = \frac{20 \cdot 20 \cdot 24}{4 \cdot 192} = \frac{25}{2}.

The height of the pyramid is 252(252)2=2532,\sqrt{25^2 - \left(\frac{25}{2}\right)^2} = \frac{25\sqrt{3}}{2}, so the volume is 131922532=8003.\frac{1}{3} \cdot 192 \cdot \frac{25\sqrt{3}}{2} = 800\sqrt{3}. Then m+n=800+3=803.m + n = 800 + 3 = 803.

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