AMC10 2020 B

AMC10 2020 B · Q8

AMC10 2020 B · Q8. It mainly tests Circle theorems, Area & perimeter.

Points $P$ and $Q$ lie in a plane with $PQ = 8$. How many locations for point $R$ in this plane are there such that the triangle with vertices $P$, $Q$, and $R$ is a right triangle with area 12 square units?

点$P$和$Q$在平面内，$PQ = 8$。在这个平面中有多少个点$R$的位置使得顶点为$P$、$Q$和$R$的三角形是直角三角形且面积为12平方单位？

(A) 2 2

(B) 4 4

(D) 8 8

(E) 12 12

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): In order for $\triangle PQR$ to have area 12, point $R$ must lie on one of the two lines parallel to and 3 units away from line $\overline{PQ}$. If $\overline{PQ}$ is to be a leg of the right triangle, then there are 2 possible right triangles if $P$ is the right angle, and there are an additional 2 possible right triangles if $Q$ is the right angle. These 4 triangles are shown in the figure on the left. If instead $\overline{PQ}$ is the hypotenuse of the right triangle, then $R$ must lie on the circle with diameter $\overline{PQ}$. Because the radius of this circle is 4, each of the two lines parallel to and 3 units from $\overline{PQ}$ intersects the circle in exactly 2 points. These 4 triangles are shown in the figure on the right. The figures below show the 8 possible locations for $R$ that make $\triangle PQR$ a right triangle.

答案（D）：为了使$\triangle PQR$的面积为12，点$R$必须位于与直线$\overline{PQ}$平行且相距3个单位的两条直线中的一条上。如果$\overline{PQ}$作为直角三角形的一条直角边，那么当$P$为直角顶点时有2种可能的直角三角形；当$Q$为直角顶点时还会再有2种可能。因此共有4个三角形，如左图所示。如果$\overline{PQ}$作为直角三角形的斜边，那么$R$必须在以$\overline{PQ}$为直径的圆上。由于该圆的半径为4，与$\overline{PQ}$平行且距$\overline{PQ}$为3个单位的两条直线中的每一条都与该圆恰好相交于2点。因此共有4个三角形，如右图所示。下方的图形展示了使$\triangle PQR$成为直角三角形时点$R$的8个可能位置。

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