Problem 12.2. A circle through the vertex $A$ of $\triangle A B C, A B \neq A C$, meets the sides $A B$ and $A C$ at points $M$ and $N$, respectively, and the side $B C$ at points $P$ and $Q$, where $Q$ lies between $B$ and $P$. Find $\Varangle B A C$, if $M P \| A C$, $N Q \| A B$ and $\frac{B P}{C Q}=\frac{A B}{A C}$

Oleg Mushkarov, Nikolai Nikolov
12.2. Set $B C=a, C A=b$ and $A B=c$. Then $B A \cdot B M=B P . B Q$, and

$$
\frac{B M}{c}=\frac{B P}{a}
$$

Hence $B Q=\frac{c^{2}}{a}$ and analogously $C P=\frac{b^{2}}{a}$. Then $B P=\frac{a^{2}-b^{2}}{a}, C Q=\frac{a^{2}-c^{2}}{a}$ and the condition $\frac{B P}{C Q}=\frac{A B}{A C}$ becomes

![](https://cdn.mathpix.com/cropped/2024_06_03_f8ccd71b4db367e7c2a9g-132.jpg?height=419&width=555&top_left_y=127&top_left_x=1090)

$$
b\left(a^{2}-b^{2}\right)=c\left(a^{2}-c^{2}\right) \text {, i.e. }(b-c)\left(a^{2}-b^{2}-c^{2}-b c\right)=0
$$

Since $b \neq c$, we get $a^{2}-b^{2}-c^{2}-b c=0$ and the Cosine theorem gives $\cos \Varangle B A C=-\frac{1}{2}$. Therefore $\Varangle B A C=120^{\circ}$.

Second solution. Since the quadrilateral $A M P N$ is a cyclic trapezoid, it follows that $A M=N P$. Also, if $T=M P \cap N Q$, then AMTN is a parallelogram and $A M=N T$. Then $N P=N T$ and in the same way $M Q=M T$. Hence $T P N$ and $T Q M$ are similar isosceles triangles and we have

$$
\frac{T P}{T Q}=\frac{T N}{T M}
$$

Using the Sine theorem, we obtain $\frac{M P}{\sin \beta}=\frac{B P}{\sin \alpha}, \frac{N Q}{\sin \gamma}=\frac{C Q}{\sin \alpha}$ and since $\frac{B P}{C Q}=\frac{A B}{A C}=\frac{\sin \gamma}{\sin \beta}$, we conclude that $M P=N Q$. From here and (1) it follows that

$$
T M+T Q \frac{T N}{T M}=T N+T Q \Longleftrightarrow(T M-T N)(T M-T Q)=0
$$

Assume that $T M=T N$. We have $M Q=N P$ and $N A=M A$. The first of these identities shows that $M N \| P Q$ and by the second one we obtain $A C=$ $A B$, a contradiction.

Therefore $T M=T Q$, i.e. $\triangle M T Q$ is equilateral. Hence

$$
\Varangle B A C=\Varangle M T N=120^{\circ}
$$