AMC12 2019 B

AMC12 2019 B · Q6

AMC12 2019 B · Q6. It mainly tests Circle theorems, Geometry misc.

In a given plane, points $A$ and $B$ are 10 units apart. How many points $C$ are there in the plane such that the perimeter of $\triangle ABC$ is 50 units and the area of $\triangle ABC$ is 100 square units?

在给定平面中，点 $A$ 和 $B$ 相距 10 个单位单位。有多少点 $C$ 在平面中，使得 $\triangle ABC$ 的周长为 50 个单位且面积为 100 平方单位？

(A) 0 0

(B) 2 2

(D) 8 8

(E) infinitely many 无限多个

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): In order for the area of $\triangle ABC$ to be 100, the altitude to the base $AB$ must be 20. Thus $C$ must lie on one of the two lines parallel to and 20 units from line $AB$. In order for the perimeter of $\triangle ABC$ to be 50, the sum of the lengths of the other two sides must be 40, which implies that point $C$ lies on an ellipse whose foci are $A$ and $B$ and whose semi-minor axis has length $\frac{1}{2}\sqrt{40^2-10^2}=\sqrt{375}$, which is less than 20. Therefore the ellipse does not intersect either of the parallel lines, and there are no such points $C$.

答案（A）：为了使$\triangle ABC$的面积为100，以$AB$为底的高必须为20。因此点$C$必须在与直线$AB$平行且距离$AB$为20的两条直线中的一条上。为了使$\triangle ABC$的周长为50，另外两边的长度之和必须为40，这意味着点$C$在一个以$A$和$B$为焦点的椭圆上，并且该椭圆的半短轴长为$\frac{1}{2}\sqrt{40^2-10^2}=\sqrt{375}$，这小于20。因此该椭圆与这两条平行线都不相交，所以不存在这样的点$C$。

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