Consider two distinct points $P$ and $Q$ on the side $[B C]$ of a triangle $A B C$. Let $K$ and $L$ be the feet of the perpendiculars to $(A C)$ from $P$ and $Q$ respectively. On the line $(A B)$, place the points $M$ and $N$ such that $P M = P A$ and $Q N = Q A$. We assume that none of the points $K, L, M, N$ coincide with $A$. Furthermore, we assume that the circumcircles of $A K M$ and $A L N$ intersect at $R$, the midpoint of $B C$.

Calculate the ratios of the lengths of the sides: $\frac{B C}{C A}$ and $\frac{C A}{A B}$.

![](https://cdn.mathpix.com/cropped/2024_05_10_da6e2ad19b1b4e3d33dcg-100.jpg?height=838&width=1282&top_left_y=234&top_left_x=451)
The idea of the solution is to show that, given $ABC$, $x=CP$ is the solution of a first-degree equation. Since $Q$ is defined in the same way as $P$, $y=CQ$ is the solution of the same first-degree equation. A first-degree equation admitting two distinct solutions is necessarily: $0 \times x=0$. It follows that any point on $(BC)$ will satisfy the same property as $P$ and $Q$.

Let $a=BC, b=CA, c=AB$ be the lengths of the sides of the triangle, and $\widehat{A}, \widehat{B}, \widehat{C}$ the three angles of the same triangle $ABC$. We will calculate the power of $B$ and the power of $C$ in two different ways with respect to the circumcircle of $AKM$, which intersects $(BC)$ at $X$. The power of $B$ is $BR \times BX = BA \times BM$, and the midpoint of $MA$ is the orthogonal projection of $P$ onto $(AB)$, satisfying: $\frac{BM + c}{2} = (a - x) \cos \widehat{B}$, hence $BM = 2(a - x) \cos \widehat{B} - c$, and $BX \times BR = BM \times BA = 2c(a - x) \cos \widehat{B} - c^2 = \frac{a}{2} \times BX$. The power of $C$ is $CK \times CA = CR \times CX = bx \cos \widehat{C} = \frac{a}{2} \times CX$. Since $BX + CX = a$, by adding these two equations we get: $\frac{a^2}{2} = 2c(a - x) \cos \widehat{B} - c^2 + bx \cos \widehat{C}$, so that: $(4c \cos \widehat{B} - 2b \cos \widehat{C}) x = 4ca \cos \widehat{B} - 2c^2 - a^2$. Since the same relation $(4c \cos \widehat{B} - 2b \cos \widehat{C}) y = 4ca \cos \widehat{B} - 2c^2 - a^2$ is also satisfied by $y = CQ \neq x$, we necessarily have: $4c \cos \widehat{B} - 2b \cos \widehat{C} = 0$ and $4ca \cos \widehat{B} - 2c^2 - a^2 = 0$. According to the Law of Cosines, $4ca \cos \widehat{B} = 2a^2 + 2c^2 - 2b^2$, so $a^2 = 2b^2$, $a = b\sqrt{2}$. And since $a = c \cos \widehat{B} + b \cos \widehat{C}$, $b \cos \widehat{C} = 2c \cos \widehat{B}$ implies $a = 3c \cos \widehat{B} = \frac{3}{2} b \cos \widehat{C}$. Therefore, $\cos \widehat{C} = \frac{2}{3} \sqrt{2}$, and $\sin \widehat{C} = \frac{1}{3}$. Since $c \cos \widehat{B} = a - b \cos \widehat{C} = b \frac{\sqrt{2}}{3}$, $c \sin \widehat{B} = b \sin \widehat{C} = \frac{b}{3}$ (Law of Sines), we have: $c^2 = \left(\frac{2}{9} + \frac{1}{9}\right) b^2$, so $c = \frac{b}{\sqrt{3}}$. Therefore, $\frac{BC}{CA} = \frac{a}{b} = \sqrt{2}$ and $\frac{CA}{AB} = \frac{b}{c} = \sqrt{3}$.

## 6 Friday afternoon, 19th: Jean-Louis Tu

## Exercises