Track / Stage 7 / 163 of 300 #1563 of 2000

Problem 1563

National olympiad second round; IMO P1/P4

Geometry Difficulty 7.3 Prove it

Given a parallelogram $ABCD$ . The line perpendicular to $AC$ passing through $C$ and the line perpendicular to $BD$ passing through $A$ intersect at point $P$ . The circle centered at point $P$ and radius $PC$ intersects the line $BC$ at point $X$ , ( $X \ne C$ ) and the line $DC$ at point $Y$ , ( $Y \ne C$ ). Prove that the line $AX$ passes through the point $Y$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

1. Setup the problem with coordinates:
Assume that the coordinates of the vertices of the parallelogram $ABCD$ are $A(a+1, b)$ , $B(1, b)$ , $C(0, 0)$ , and $D(a, 0)$ with $a, b \neq 0$ . This setup ensures that $BC \parallel AD$ and $AB \parallel CD$ .

2. **Find the equations of the lines $AP$ and $CP$ :**
- The line $AP$ is perpendicular to $BD$ and passes through $A$ . The slope of $BD$ is $\frac{b}{1-a}$ , so the slope of $AP$ is $-\frac{1-a}{b}$ . The equation of $AP$ is:
$y - b = -\frac{1-a}{b}(x - (a+1)) \Rightarrow (a-1)x - by - a^2 + b^2 + 1 = 0$
- The line $CP$ is perpendicular to $AC$ and passes through $C$ . The slope of $AC$ is $\frac{b}{a+1}$ , so the slope of $CP$ is $-\frac{a+1}{b}$ . The equation of $CP$ is:
$y = -\frac{a+1}{b}x \Rightarrow (a+1)x + by = 0$

3. **Find the coordinates of point $P$ :**
Solve the system of equations for $AP$ and $CP$ :
$\begin{cases} (a-1)x - by - a^2 + b^2 + 1 = 0 \\ (a+1)x + by = 0 \end{cases}$
Adding these equations, we get:
$(a-1)x + (a+1)x = a^2 - b^2 - 1 \Rightarrow 2ax = a^2 - b^2 - 1 \Rightarrow x = \frac{a^2 - b^2 - 1}{2a}$
Substituting $x$ back into the equation of $CP$ :
$(a+1)\left(\frac{a^2 - b^2 - 1}{2a}\right) + by = 0 \Rightarrow y = -\frac{(a+1)(a^2 - b^2 - 1)}{2ab}$
Therefore, $P\left(\frac{a^2 - b^2 - 1}{2a}, -\frac{(a+1)(a^2 - b^2 - 1)}{2ab}\right)$ .

4. **Find the coordinates of point $Y$ :**
Since $PY = PC$ , point $Y$ lies on the line $DC$ and has the same distance from $P$ as $C$ . The coordinates of $Y$ are:
$Y\left(\frac{a^2 - b^2 - 1}{a}, 0\right)$

5. **Prove the similarity of triangles $\triangle ADY$ and $\triangle ADC$ :**
- Since $DY = \left|\frac{b^2 + 1}{a}\right|$ and $DC = |a|$ , we have:
$DY \cdot DC = \left|\frac{b^2 + 1}{a}\right| \cdot |a| = b^2 + 1 = DY^2$
- Since $\angle ADY = \angle ADC$ , the triangles $\triangle ADY$ and $\triangle ADC$ are similar by the AA (Angle-Angle) similarity criterion.

6. **Show that $AX$ passes through $Y$ :**
- Let the tangent from $X$ to the circle $(PYC)$ intersect $AC$ at $M$ and $CP$ intersect $(PYC)$ at $N$ . Since $AC$ is tangent to $(PYC)$ , we have:
$\angle XYC = \angle XNC = \angle XCM = \angle ACB = \angle CAD = \angle DYA$
- Since $D, Y, C$ are collinear and $A$ and $X$ are not on the same side divided by $CD$ , it follows that $A, Y, X$ are collinear.

$\blacksquare$

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