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2025 AMC 12A Problems/Problem 11: Difference between revisions

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Let the intersection of the heights and their corresponding sides be <imath>H_1</imath> and <imath>H_2</imath>, respectively, and let the intersection of line <imath>AH_1</imath> and <imath>H_2B</imath> be <imath>E</imath>. Since <imath>m \angle BH_1E= m \angle BH_2C = 90 \deg</imath>
Let the intersection of the heights and their corresponding sides be <imath>H_1</imath> and <imath>H_2</imath>, respectively, and let the intersection of line <imath>AH_1</imath> and <imath>H_2B</imath> be <imath>E</imath>. Since <imath>m \angle BH_1E= m \angle BH_2C = 90 \deg</imath>,

Revision as of 10:03, 7 November 2025

Problem

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 $C(18,27)$?

$\textbf{(A)}~5\qquad\textbf{(B)}~17\qquad\textbf{(C)}~10+4\sqrt{17} +2\sqrt{13}\qquad\textbf{(D)}~\frac{113}{3}\qquad\textbf{(E)}~54$

Solution 1

Altitudes are perpendicular to sides and pass through a vertex. We have the coordinates of the vertices and the slope of the sides can be found, meaning we have all the information needed to find the equations of the lines of the altitudes. Then, we only need to find the intersections of the lines

Since $B$ and $C$ form a horizontal line, the altitude to $BC$ from $A$ is a vertical line, so its equation must be $x=2$. Then, we need to find the equation of one more altitude to solve a system of equations, giving us the coordinates. Suppose we choose the second altitude to be the one from $C$ to segment $AB$. The slope of segment $AB$ is $-\frac{2}{3}$, so the slope of the altitude, which is perpendicular to $AB$, is $\frac{3}{2}$. Since the altitude passes through $C$, by point-slope form, we have the equation of the altitude to be $y-27=\frac{3}{2}(x-18)$. Solving the two equations gives the coordinates of the orthocenter to be $(2, 3)$. The answer is therefore $2+3=\boxed{\text{(A) }5}$.

~dg6665 (edits for motivation by ~Logibyte)

Solution 2

[asy] size(200); pair A = (2,31); pair B = (8, 27); pair C = (18, 27); pair H1 = (2, 27); pair H2 = (8.5882352941176, 29.3529411764706); pair D = (2,3); label("$A$", A, NW); label("$B$", (8.5, 27), S); label("$C$", C, SE); label("$H_1$", H1, SW); label("$H_2$", H2, NE); label("$E$", E, S); draw(A--B--C--cycle); draw(A--E, dashed); draw(H1--B, dashed); draw(H2--E, dashed); [/asy]

Let the intersection of the heights and their corresponding sides be $H_1$ and $H_2$, respectively, and let the intersection of line $AH_1$ and $H_2B$ be $E$. Since $m \angle BH_1E= m \angle BH_2C = 90 \deg$,

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