2025 AMC 10A Problem 20

Attempt Problem 20 of the 2025 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 10A solutions, or check the answer key.

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20.

A silo (right circular cylinder) with diameter 2020 meters stands in a field. MacDonald is located 2020 meters west and 1515 meters south of the center of the silo. McGregor is located 2020 meters east and g>0g \gt 0 meters south of the center of the silo. The line of sight between MacDonald and McGregor is tangent to the silo. The value of gg can be written as abcd,\dfrac{a\sqrt{b} - c}{d}, where a,b,c,a, b, c, and dd are positive integers, bb is not divisible by the square of any prime, and dd is relatively prime to the greatest common divisor of aa and c.c. What is a+b+c+d?a + b + c + d?

119119

120120

121121

122122

123123

Answer: A
Concepts:tangent linecoordinate geometryquadratic
Difficulty rating: 2080
Solution:

Put the silo's center at the origin with radius 10.10. Then MacDonald is at D=(20,15)D = (-20, -15) and McGregor at G=(20,g).G = (20, -g). The tangent length from DD is DT=DS2102DT = \sqrt{DS^2 - 10^2} =252100= \sqrt{25^2 - 100} =525,= \sqrt{525}, and from GG it is TG=g2+202102.TG = \sqrt{g^2 + 20^2 - 10^2}. The tangent point TT sits between the two men, so DG=DT+TG.DG = DT + TG. But also DG=402+(15g)2.DG = \sqrt{40^2 + (15 - g)^2}. Squaring twice and simplifying gives 3g2+150g925=0.3g^2 + 150g - 925 = 0. Its positive root, which satisfies the original tangent-length equation, is g=2021753.g = \frac{20\sqrt{21} - 75}{3}. Therefore a+b+c+da + b + c + d =20+21+75+3= 20 + 21 + 75 + 3 =119.= 119. Thus, A is the correct answer.

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