AMC 10 · 2025 · #11

Grade 10 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Line Intersection Point as the Answer

coordinate-geometryperpendicular-slopesslope-interceptsystems-of-equations identify-subproblems ↑ Prerequisites: coordinate-geometryslope-intercept

📏 Short solution 💡 2 insights

Problem

A triangle has vertices A(2, 31), B(8, 27), and C(18, 27). Its orthocenter is the single point where all three altitudes meet, where each altitude is the line through one vertex perpendicular to the opposite side. Find the sum of the x- and y-coordinates of that orthocenter.

Pick an answer.

(A)

(B)

(C)

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

(D)

$\frac{113}{3}$

(E)

How to solve

Strategy Identify Subproblems

The orthocenter is where three altitudes meet, but two lines already fix a single point, so the job splits into: find two altitude lines, then intersect them. Picking the two easiest altitudes keeps the algebra tiny. One side of this triangle is horizontal, which hands over a free vertical altitude, so only one more altitude needs real work before the two lines are crossed.

1STEP 1

Use the horizontal side for a free altitude

BC is horizontal (both y = 27), so the altitude from A is vertical through A: the line x = 2.

B_y = C_y = 27 → BC horizontal → altitude from A: x = 2

2STEP 2

Take on the altitude from B

The altitude from B is perpendicular to AC. Slope of AC between A(2, 31) and C(18, 27) is $\frac{27 - 31}{18 - 2}$ = - $\frac{1}{4}$ .

m_AC = (27 - 31)/(18 - 2) = -4/16 = -1/4

3STEP 3

Turn the altitude into an equation

Perpendicular to slope - $\frac{1}{4}$ means slope 4. Through B(8, 27): y - 27 = 4(x - 8), which simplifies to y = 4x - 5.

m_ ⊥ = -1/-1/4 = 4, y - 27 = 4(x - 8) → y = 4x - 5

4STEP 4

Intersect and add the coordinates

Solve x = 2 with y = 4x - 5 to get y = 3, so the orthocenter is (2, 3); the third altitude from C confirms it and the sum is 2 + 3 = 5.

x = 2 ; y = 4x - 5 → (2, 3), 2 + 3 = 5

Answer

All three altitudes were made to agree: x = 2 from A, y = 4x - 5 from B, and y = (

\frac{3}{2}

)x from C all pass through (2, 3), so the orthocenter is consistent. The point sits well outside the triangle (whose vertices have y-values 27 and 31), which is expected because the triangle is obtuse, and an obtuse triangle's orthocenter falls outside it. The sum 5 is choice (A); the other choices either mismatch (17,

\frac{113}{3}

, 54) or carry irrelevant radicals, so 5 is the clean, exact fit.

💡Key takeaway

You only need two altitudes to find the orthocenter, so grab the easy vertical one off the flat side, build one more with a perpendicular slope, and cross them.

Use the horizontal side for a free altitude
Take on the altitude from B
Turn the altitude into an equation
Intersect and add the coordinates

Use the horizontal side for a free altitude

Take on the altitude from B

Turn the altitude into an equation

Intersect and add the coordinates

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