2025 AMC 12A Problem 11

Attempt Problem 11 of the 2025 AMC 12A 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 12A solutions, or check the answer key.

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11.

The orthocenter of a triangle is the concurrent intersection of the three (possibly extended) altitudes. What is the sum of the coordinates of the orthocenter of the triangle whose vertices are A(2,31),A(2, 31), B(8,27),B(8, 27), and C(18,27)?C(18, 27)?

55

1717

10+417+21310 + 4\sqrt{17} + 2\sqrt{13}

1133\dfrac{113}{3}

5454

Answer: A
Concepts:coordinate geometryaltitudeslope
Difficulty rating: 1570
Solution:

Since BB and CC both have y=27,y = 27, side BCBC is horizontal and the altitude from AA is the vertical line x=2.x = 2.

Side ACAC has slope 2731182=14,\dfrac{27 - 31}{18 - 2} = -\dfrac{1}{4}, so the altitude from BB has slope 44: y27=4(x8).y - 27 = 4(x - 8).

At x=2,x = 2, y=27+4(28)=3.y = 27 + 4(2 - 8) = 3. The orthocenter is (2,3),(2, 3), with coordinate sum 5.5.

Thus, the correct answer is A.

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