## Task 6 - 220936

For a triangle $A B C$, let the side lengths be $A B=c, A C=b$ and $B C=a$. The angle bisectors of $\angle C A B$ and $\angle A B C$ intersect at $P$. Through $P$, a line parallel to $A B$ is drawn. It intersects $A C$ at $Q$ and $B C$ at $R$.

Determine the length $Q R$ in terms of the three given side lengths!
Certainly! Here is the translation of the provided text into English, preserving the original formatting:

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![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1046.jpg?height=362&width=577&top_left_y=1281&top_left_x=248)

$C A: C Q = A B: A R$,
Let $x = P Q$, $y = P R$. Then, by the given condition, $\angle P A Q = \angle B A P = \angle Q P A$ (alternate interior angles at the parallel lines $A B$ and $P Q$), so triangle $A P Q$ is isosceles with $A Q = P Q = x$. Similarly, $B R = P R = y$.

Since $Q R \parallel A B$, it follows from the intercept theorem that

$$
C A: C B = Q A: R B, \quad b: a = x: y, \quad b y = a x
$$

Substituting (1) into (2), we get $b x + a x = b c - c x$ and

$$
x = \frac{b c}{a + b + c}
$$

From this and (1), it follows that $y = \frac{a x}{b} = \frac{a c}{a + b + c}$. Therefore,

$$
Q R = x + y = \frac{(a + b) c}{a + b + c}
$$

Solution adopted from [5]

### 6.25 XXIII. Olympiad 1983

### 6.25.1 First Round 1983, Grade 9

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