Vector geometry with unusual points

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Source: Romania NMO 2025 9.2

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#1 h Apr 2, 2025, 12:28 PM

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Let $\triangle ABC$ be an acute-angled triangle, with circumcenter $O$ , circumradius $R$ and orthocenter $H$ . Let $A_1$ be a point on $BC$ such that $A_1H+A_1O=R$ . Define $B_1$ and $C_1$ similarly.
If $\overrightarrow{AA_1} + \overrightarrow{BB_1} + \overrightarrow{CC_1} = \overrightarrow{0}$ , prove that $\triangle ABC$ is equilateral.

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#2 h Apr 6, 2025, 12:13 PM

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Solution. In the complex plane, denote with lowercase letters the affixes of the respective points.
We will use the following lemma:
Lemma. Consider points $M$ , $N$ on the unit circle. If $P$ is an arbitrary point in the complex plane and $Q$ is the reflection of $P$ over the line $MN$ , then the affix of point $Q$ is given by: $q = m + n - \bar{p} \cdot mn.$ Proof. Consider point $T$ on line $MN$ such that $PT \perp MN$ . Since $Q$ is the reflection of point $P$ over the line $MN$ , we have that $T$ is the midpoint of $[PQ]$ . Therefore, $t = \dfrac{p + q}{2} \Rightarrow q = 2t - p$ .
The condition $T \in MN$ now becomes
$\dfrac{t - m}{t - n} \in \mathbb{R} \iff \dfrac{t - m}{t - n} = \dfrac{ \bar{t} - \bar{m} }{ \bar{t} - \bar{n} } \iff t(\bar{n} - \bar{m}) - \bar{t}(n - m) + (n \bar{m} - \bar{n}m) = 0 \iff t + \bar{t}mn - (m + n) = 0, \; (1).$ The condition $PT \perp MN$ is equivalent to $\dfrac{p - t}{m - n} \in i \mathbb{R} \iff \dfrac{p - t}{m - n} = - \Bigg(\overline{ \dfrac{p - t}{m - n} } \Bigg) = - \dfrac{\bar{p} - \bar{t}}{\bar{m} - \bar{n}} = \dfrac{(\bar{p} - \bar{t})mn}{m - n} \iff p - t = (\bar{p} - \bar{t})mn, \; (2).$ From $(1)$ and $(2)$ , we have $2t = m + n + p - \bar{p}mn$ , yielding $q = m + n - \bar{p} \cdot mn$ .

In the given problem, let $O$ be the origin of the complex plane. Without loss of generality, let circle $(ABC)$ be the unit circle.
Denote $H_A$ the reflection of $H$ with respect to $BC$ . Then $H_A \in \mathcal{C}$ (since $H$ is the orthocenter of $\triangle ABC$ ), where $\mathcal{C}$ denotes the circumcircle of $\triangle ABC$ . By Sylvester's Theorem, we have $h = a + b + c$ and $\bar{h} = \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}$ .
Using the above lemma, we have $h_a = \dfrac{bc}{a}$ . Since $A_1H = A_1H_A$ and $OA_1 + A_1H = R$ , we also have $A_1H_A + A_1O = R$ , so the points $O$ , $A_1$ and $H_A$ are collinear. Therefore, we have that $A_1 \in BC$ implies that
$\dfrac{a_1 - b}{a_1 - c} \in \mathbb{R} \iff a_1(\bar{b} - \bar{c}) - \bar{a_1}(b - c) + (b \bar{c} - \bar{b}c) = 0 \iff a_1 + \bar{a_1}bc = b + c.$ Since $O$ , $H_A$ and $A_1$ are collinear,
$\dfrac{h_a}{a_1} \in \mathbb{R} \iff a_1 = h_a^2 \bar{a_1} = \Bigg( \dfrac{bc}{a} \Bigg)^2 \bar{a_1}.$ We obtain $a_1 = \dfrac{bc(b + c)}{a^2 + bc}$ and the similar expressions for $b_1$ and $c_1$ .

Claim. If $m \overset{\text{not}}{=} \dfrac{A_1B}{A_1C}$ and $n$ and $p$ are defined similarly, then
$\sum \overrightarrow{AA_1} = \vec{0} \Rightarrow m = n = p.$ Proof. We have $\overrightarrow{AA_1} = \dfrac{\overrightarrow{AB} + m\overrightarrow{AC}}{1 + m}$ , $\overrightarrow{BB_1} = \dfrac{\overrightarrow{BC} + n\overrightarrow{BA}}{1 + n}$ , $\overrightarrow{CC_1} = \dfrac{\overrightarrow{CA} + p\overrightarrow{CB}}{1 + p}$ .
Since $\sum \overrightarrow{AA_1} = \vec{0}$ , we have
$\Bigg( \dfrac{1}{1 + m} - \dfrac{n}{1 + n} \Bigg) \overrightarrow{AB} + \Bigg( \dfrac{1}{1 + n} - \dfrac{p}{1 + p} \Bigg) \overrightarrow{BC} + \Bigg( \dfrac{m}{1 + m} - \dfrac{1}{1 + p} \Bigg) \overrightarrow{AC} = \vec{0}.$ However, $\overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB}$ , and the above equation becomes:
$\Bigg( \dfrac{1}{1 + m} + \dfrac{p}{1 + p} - 1\Bigg) \overrightarrow{AB} + \Bigg( \dfrac{m}{1 + m} + \dfrac{1}{1 + n} - 1 \Bigg) \overrightarrow{AC} = \vec{0}.$ Because the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are not collinear, we have:
$\dfrac{1}{1 + m} + \dfrac{p}{1 + p} - 1 = 0,$ and
$\dfrac{m}{1 + m} + \dfrac{1}{1 + n} - 1 = 0.$ Consequently, from these two, $m = n = p$ .

We will now compute $m$ , $n$ , $p$ . We have $a_1 = \dfrac{bc(b + c)}{a^2 + bc}$ , and $m = \dfrac{a_1 - b}{c - a_1} = \dfrac{\frac{bc(b + c)}{a^2 + bc} - b}{c - \frac{bc(b + c)}{a^2 + bc}} = \dfrac{b(c - a)(c + a)}{c(a - b)(a + b)}$ . Analogously, $n = \dfrac{c(a - b)(a + b)}{a(b - c)(b + c)}$ and $p = \dfrac{a(b - c)(b + c)}{b(c - a)(c + a)}$ .
Observing that $mnp = 1$ , and combining this with $m = n = p$ , we have $m = n = p = 1$ . From this, we have
$CA \cdot |c + a| = AB\cdot |a + b| = BC \cdot |b + c|.$ Since $R = 1$ , we have $CA = 2 \sin B$ , $AB = 2 \sin C$ , $BC = 2 \sin A$ , and $|c + a| = \cos B$ , $|a + b| = \cos C$ , $|b + c| = \cos A$ .
We have now $2 \sin A \cos A = 2 \sin B \cos B = 2 \sin C \cos C \iff \sin(2A) = \sin(2B) = \sin(2C),$ and because the triangle $ABC$ is acute, we obtain $\angle A = \angle B = \angle C$ , or that $\triangle ABC$ is equilateral.
Indeed, $\sin(2B) = \sin(2C) \iff 2 \sin(B - C) \cdot \cos(B + C) = 0 \iff \sin(B - C) = 0 \iff B = C + 2k \pi = C \iff B = C$ .

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This post has been edited 2 times. Last edited by ericdimc, Apr 19, 2025, 6:03 PM

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