42nd Balkan Mathematical Olympiad (BMO 2025 )

Problem 2.

In an acute-angled triangle $ABC$ , $H$ be the orthocenter of it and $D$ be any point on the side $BC$ . The points $E, F$ are on the segments $AB, AC$ , respectively, such that the points $A, B, D, F$ and $A, C, D, E$ are cyclic. The segments $BF$ and $CE$ intersect at $P.$ $L$ is a point on $HA$ such that $LC$ is tangent to the circumcircle of triangle $PBC$ at $C.$ $BH$ and $CP$ intersect at $X$ . Prove that the points $D, X,$ and $L$ lie on the same line.

Solution 1
We first claim the following:

This claim wrote:

The reflection of $P$ over $BC$ is the second intersection of $AD$ and $(ABC)$ .

Proof. Let $P'$ be the second intersection of $AD$ and $(ABC)$ . Then, since

$\begin{align*} \measuredangle PBC &= \measuredangle FBD \\ &= \measuredangle FAD \\ &= \measuredangle CAP' \\ &= \measuredangle CBP' \\ &= -\measuredangle P'BC \end{align*}$

then $BP$ and $BP'$ are reflections over $BC$ . Note that since $\measuredangle AEP = \measuredangle ADC = \measuredangle ADB = \measuredangle AFP$ , then $AEPF$ is cyclic, implying that $\measuredangle BPC = \measuredangle FPE = -\measuredangle BAC = \measuredangle BP'C$ , so $P$ and $P'$ are indeed reflections over $BC$ .

Now we reflect everything except $A$ over $BC$ , without overlaying the new diagram with the old one. We can also do barycentrics on $\triangle ABC$ now.
Let $a$ , $b$ , $c$ , $A$ , $B$ , $C$ , $S_A$ , $S_B$ , $S_C$ denote $BC$ , $CA$ , $AB$ , $(1, 0, 0)$ , $(0, 1, 0)$ , $(0, 0, 1)$ , $\frac{-a^2+b^2+c^2}{2}$ , $\frac{a^2-b^2+c^2}{2}$ , $\frac{a^2+b^2-c^2}{2}$ respectively. (I know that's a lot but they're just common notation anyway)
We first calculate $H$ . Let $H=(t:S_C:S_B)$ . Then, $\begin{align*} -a^2S_BS_C - b^2S_Ct - c^2tS_B &= 0 \\ \iff t &= -\frac{a^2S_BS_C}{b^2S_B+c^2S_C} \end{align*}$ so $H=(-a^2S_BS_C: S_C(b^2S_B+c^2S_C): S_B(b^2S_B+c^2S_C))$ . Let $P = (x:y:z)$ . Then obviously $-a^2yz-b^2zx-c^2xy=0$ . We can also calculate $X=(-a^2S_BS_Cx : -a^2S_BS_Cy : S_Bx(b^2S_B+c^2S_C))$ , $D=(0:y:z)$ and $L=(-a^2S_C:b^2S_C:b^2S_B)$ . Finally, $\begin{align*} \begin{vmatrix} 0&y&z\\ -a^2S_C&b^2S_C&b^2S_B\\ -a^2S_BS_Cx&-a^2S_BS_Cy&S_Bx(b^2S_B+c^2S_C)\\ \end{vmatrix} &= -a^2S_C \begin{vmatrix} 0&y&z\\ 1&b^2S_C&b^2S_B\\ S_Bx&-a^2S_BS_Cy&S_Bx(b^2S_B+c^2S_C)\\ \end{vmatrix} \\ &=-a^2S_BS_C \begin{vmatrix} 0&y&z\\ 1&b^2S_C&b^2S_B\\ x&-a^2S_Cy&x(b^2S_B+c^2S_C)\\ \end{vmatrix} \\ &= -a^2S_BS_C((-a^2S_Cyz - xy(b^2S_B+c^2S_C))+(b^2S_Bxy-b^2S_Cxz)) \\ &= -a^2S_BS_C(-a^2S_Cyz-b^2S_Czx-c^2S_Cxy) \\ &= 0 \end{align*}$ so we're done.

Solution 2

Let $\triangle ABC$ be inscribed in the unit circle, and let $AD$ meet the unit circle again at $U$ , so that
$|a|=|b|=|c|=|u|=1$
$h = a+b+c$
$d = \frac{au(b+c) - bc(a+u)}{au-bc}$

Let lines $BF,CE$ intersect the unit circle again at $V,W$ respectively. Now since $A,B,D,F$ are concyclic, we have
$\frac{(a-d)(b-f)}{(a-f)(b-d)} \in \mathbb{R} \implies \frac{(a-u)(b-v)}{(a-c)(b-c)} = \frac{c^2(a-u)(b-v)}{uv(a-c)(b-c)} \implies c^2 = uv$

So
$v = \frac{c^2}u$

and similarly

$w = \frac{b^2}u$

Then

$\begin{align*} p &= \frac{bv(c+w) - cw(b+v)}{bv-cw} \\ &= \frac{\frac{bc^2}u\left(c+\frac{b^2}u\right) - \frac{b^2c}u\left(b+\frac{c^2}u\right)}{\frac{bc^2}u-\frac{b^2c}u} \\ &= \frac{bc^3u+b^3c^2-b^3cu-b^2c^3}{bc^2u-b^2cu} \\ &= \frac{c^2u+b^2c-b^2u-bc^2}{cu-bu} \\ &= \frac{bu+cu-bc}u \end{align*}$

(So $P$ is the reflection of $U$ over line $BC$ .) Now since $L$ lies on line $HA$ , we have

$\overline{\ell} = \frac{a\ell + bc - a^2}{abc}$

And since $LC$ is tangent to the circumcircle of $\triangle PBC$ , we have

$\frac{(c-\ell)(b-p)}{(b-c)(c-p)} \in \mathbb{R}$ $\frac{c(c-\ell)(b-u)}{b(b-c)(c-u)} = \frac{\frac1c\left(\frac1c-\frac{a\ell + bc - a^2}{abc}\right)\left(\frac1b-\frac1u\right)}{\frac1b\left(\frac1b-\frac1c\right)\left(\frac1c-\frac1u\right)} = -\frac{(a^2+ab-a\ell-bc)(b-u)}{a(b-c)(c-u)}$ $ac(c-\ell) = -b(a^2+ab-a\ell-bc) \implies \ell = \frac{a^2b+ab^2+ac^2-b^2c}{a(b+c)}$

Now we find the coordinate of $X$ . Since $X$ lies on line $BH$ , we have

$\overline{x} = \frac{bx+ac-b^2}{abc}$

Since $X$ lies on line $CP$ , we have

$\frac{c-x}{c-p} \in \mathbb{R}$

$\frac{u(c-x)}{b(c-u)} = \frac{\frac1u\left(\frac1c - \frac{bx+ac-b^2}{abc}\right)}{\frac1b\left(\frac1c-\frac1u\right)} = -\frac{ab-ac+b^2-bx}{a(c-u)}$ $au(c-x) = -b(ab-ac+b^2-bx) \implies x = \frac{ab^2-abc+acu+b^3}{au+b^2}$

Now we find the vectors

$\begin{align*} d-\ell &= \frac{au(b+c) - bc(a+u)}{au-bc} - \frac{a^2b+ab^2+ac^2-b^2c}{a(b+c)} \\ &= \frac{a(b+c)(abu+acu-abc-bcu) - (au-bc)(a^2b+ab^2+ac^2-b^2c)}{a(b+c)(au-bc)} \\ &= \frac{-a^3bu-a^2bc^2+2a^2bcu+ab^3c+abc^3-abc^2u-b^3c^2}{a(b+c)(au-bc)} \\ &= -\frac{b(a-c)(a^2u+ac^2-acu-b^2c)}{a(b+c)(au-bc)} \end{align*}$

and

$\begin{align*} x-\ell &= \frac{ab^2-abc+acu+b^3}{au+b^2} - \frac{a^2b+ab^2+ac^2-b^2c}{a(b+c)} \\ &= \frac{a(b+c)(ab^2-abc+acu+b^3) - (au+b^2)(a^2b+ab^2+ac^2-b^2c)}{a(b+c)(au+b^2)} \\ &= \frac{-a^3bu-a^2b^2u-a^2bc^2+a^2bcu+ab^3c-ab^2c^2+ab^2cu+b^4c}{a(b+c)(au+b^2)} \\ &= -\frac{b(a+b)(a^2u+ac^2-acu-b^2c)}{a(b+c)(au+b^2)} \end{align*}$

Then

$\frac{d-\ell}{x-\ell} = \frac{(a-c)(au+b^2)}{(a+b)(au-bc)}$

which is real. $\blacksquare$