2025 AMC 12A Problem 11

Below is the professionally curated solution for Problem 11 of the 2025 AMC 12A, 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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Concepts:coordinate geometry altitude slope

Difficulty rating: 1570

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 $A(2, 31),$ $B(8, 27),$ and $C(18, 27)?$

$5$

$17$

$10 + 4\sqrt{17} + 2\sqrt{13}$

$\dfrac{113}{3}$

$54$

Solution:

Since $B$ and $C$ both have $y = 27,$ side $BC$ is horizontal and the altitude from $A$ is the vertical line $x = 2.$

Side $AC$ has slope $\dfrac{27 - 31}{18 - 2} = -\dfrac{1}{4},$ so the altitude from $B$ has slope $4$ : $y - 27 = 4(x - 8).$

At $x = 2,$ $y = 27 + 4(2 - 8) = 3.$ The orthocenter is $(2, 3),$ with coordinate sum $5.$

Thus, the correct answer is A.

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