# Difference between revisions of "1984 IMO Problems/Problem 4"

## Problem

Let $ABCD$ be a convex quadrilateral with the line $CD$ being tangent to the circle on diameter $AB$. Prove that the line $AB$ is tangent to the circle on diameter $CD$ if and only if the lines $BC$ and $AD$ are parallel.

## Solution

First, we prove that if $BC$ and $AD$ are parallel then the claim is true: Let $AB$ and $CD$ intersect at $E$ (assume $E$ is closer to $AD$, the other case being analogous). Let $M,N$ be the midpoints of $AB,CD$ respectively. Let the length of the perpendicular from $N$ to $AB$ be $r$. It is known that the length of the perpendicular from $M$ to $CD$ is $\frac{1}{2}AB$. Let the foot of the perpendicular from $C$ to $AB$ be $H$, and similarly define $G$ for side $CD$. Then, since triangles $MNE$ and $BCE$ are similar, we have $\frac{CH}{r}=\frac{BG}{\frac{1}{2}AB}$. This gives an expression for $r$: $r=\frac{1}{2}AB\cdot\frac{CH}{BG}$

Noticing that $CH=BC\sin EBC,BG=BC\sin ECB$ simplifies the expression to $r=\frac{1}{2}AB\cdot\frac{\sin EBC}{\sin ECB}$

By the Law of Sines, $\frac{\sin EBC}{\sin ECB}=\frac{EC}{EB}$. Since triangles $EDA,ECB$ are similar, we have $\frac{EC}{EB}=\frac{CD}{AB}$ and thus we have $r=\frac{1}{2}AB\cdot\frac{CD}{AB}=\frac{1}{2}CD$

and we are done.

Now to prove the converse. Suppose we have the quadrilateral with $BC$ parallel to $AD$, and with all conditions satisfied. We shall prove that there exists no point $T$ on $CD$ such that $T$ is a midpoint of a side $CD^\prime$ of a quadrilateral $ABCD^\prime$ which also satisfies the condition. Suppose there was such a $T$. Like before, define the points $E,M,N$ for quadrilateral $ABCD$. Let $t$ be the length of the perpendicular from $T$ to $AB$. Then, using similar triangles, $\frac{ET}{t}=\frac{EN}{\frac{1}{2}CD}$. This gives $t=\frac{\frac{1}{2}CD\cdot ET}{EN}$

But, we must have $t=DT$. Thus, we have $DT=\frac{\frac{1}{2}CD\cdot ET}{EN}$ $\Rightarrow \frac{1}{2}CD\cdot EN+NT\cdot EN=\frac{1}{2}CD\cdot (EN+NT)$ $\Rightarrow NT\cdot EN=\frac{1}{2}CD\cdot NT$

Since $\frac{1}{2}CD\ne EN$, we have $NT=0$ as desired.