AMC12 2025 A

AMC12 2025 A · Q11

AMC12 2025 A · Q11. It mainly tests Triangles (properties), Coordinate geometry.

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 $C(18,27)$?

三角形的垂心是三条（可能延长）高线的并发交点。顶点为 $A(2,31)$、$B(8,27)$ 和 $C(18,27)$ 的三角形的垂心的坐标之和是多少？

(A) 5 5

(B) 17 17

(D) \frac{113}{3} \frac{113}{3}

(E) 54 54

Answer

Correct choice: (A)

正确答案：（A）

Solution

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 $2+3=\boxed{\text{(A) }5}$.

高线垂直于边并通过顶点。我们有顶点坐标，可以求出边的斜率，从而得到高线方程。然后，只需求两条高线的交点。由于 $B$ 和 $C$ 构成一条水平线，从 $A$ 到 $BC$ 的高线是垂直线，因此其方程为 $x=2$。然后，我们需要求另一条高线的方程来解方程组。假设选择从 $C$ 到线段 $AB$ 的高线。线段 $AB$ 的斜率为 $-\frac{2}{3}$，因此垂直于 $AB$ 的高线的斜率为 $\frac{3}{2}$。由于高线通过 $C$，利用点斜式，其方程为 $y-27=\frac{3}{2}(x-18)$。求解这两个方程得到垂心坐标为 $(2, 3)$。因此答案是 $2+3=\boxed{\text{(A) }5}$。

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