Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Find the answer

5. Given four points in the Cartesian coordinate system $A(-2,4)$ , $B(-2,0)$ , $C(2,-3)$ , and $D(2,0)$ . Let $P$ be a point on the $x$ -axis, and after connecting $P A$ and $P C$ , the two triangles formed with $A B$ , $C D$ , and the $x$ -axis (i.e., $\triangle P A B$ and $\triangle P C D$ ) are similar. Then the number of all points $P$ that satisfy the above conditions is:

Pick one

Solution

5.C.
(1) $A B$ and $C D$ are corresponding sides.
(i) $P$ is the intersection point of $A C$ and the $x$ -axis. The equation of $A C$ is $y=-\frac{7}{4} x+\frac{1}{2}$ , giving $x=\frac{2}{7}, y=0$ .
(ii) Take the point $C$ 's reflection about the $x$ -axis, $C^{\prime}(2,3)$ , $P$ is the intersection point of $A C^{\prime}$ and the $x$ -axis. The equation of $A C^{\prime}$ is $y=-\frac{1}{4} x+\frac{7}{2}$ , giving $x=14, y=0$ .
(2) $A B$ and $C D$ are not corresponding sides, i.e., $\frac{P B}{A B}=\frac{C D}{P D}$ .
Let point $P(x, 0)$ . Then
$\begin{array}{l} \frac{|x+2|}{4}=\frac{3}{|x-2|} \\ \Rightarrow\left|x^{2}-4\right|=12 \Rightarrow x= \pm 4 . \end{array}$

Therefore, there are 4 points $P$ that meet the conditions, which are
$\left(\frac{2}{7}, 0\right),(14,0),(4,0),(-4,0) \text {. }$

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.