Two Orthocenters and an Invariant Point

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Two Orthocenters and an Invariant Point

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Source: 2025 Nepal Mock TST Day 1 Problem 3

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#1 h Apr 6, 2025, 1:30 PM • 3 Y

Y by RANDOM__USER, PikaPika999, khan.academy

Let $\triangle{ABC}$ be a triangle, and let $P$ be an arbitrary point on line $AO$ , where $O$ is the circumcenter of $\triangle{ABC}$ . Define $H_1$ and $H_2$ as the orthocenters of triangles $\triangle{APB}$ and $\triangle{APC}$ . Prove that $H_1H_2$ passes through a fixed point which is independent of the choice of $P$ .

(Kritesh Dhakal, Nepal)

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#2 h Apr 6, 2025, 2:14 PM • 1 Y

Y by PikaPika999

Let us linearly move $P$ on the line $AO$ , then $H_1$ and $H_2$ move linearly on the perpendicular lines from $B$ and $C$ onto $AO$ respectively. Additionally the lines on which $H_1$ and $H_2$ move are parallel to each other, because the perpendicular lines from $B$ and $C$ onto $AO$ are parallel.

Claim: If $D$ is the base of the altitude from $A$ onto $BC$ , then $D \in H_1H_2$ .
Proof: Because of the properties stated above, we only need to verify this statement for two positions of $P$ to prove it holds for a general position of $P$ . When $P$ is the antipode of $A$ in $(ABC)$ the statement is more than trivial, and when $P \in BC$ , $H_1$ and $H_2$ lie on $AD$ .

Thus $H_1H_2$ always passes through $D$ . $\blacksquare$

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#3 h Apr 6, 2025, 7:05 PM • 1 Y

Y by PikaPika999

Complex bash with $\triangle ABC$ inscribed in the unit circle, so that
$|a|=|b|=|c|=1$ $o=0$ $p = \lambda a$ for some real number $\lambda$ . Then the circumcenter of $\triangle APB$ has coordinate
$o_1 = \frac{ab(p\overline{p}-1)}{p-a-b+ab\overline{p}} = \frac{ab(1+\lambda)}{a+b}$ and so its orthocenter has coordinate
$h_1 = a + b + p - 2o_1 = \frac{a^2+b^2+\lambda a^2-\lambda ab}{a+b}$ Similarly,
$h_2 = \frac{a^2+c^2+\lambda a^2-\lambda ac}{a+c}$ To guess the fixed point, we pick some convenient values of $\lambda$ . When $\lambda = -1$ , we have $h_1 = b$ and $h_2 = c$ . When $\lambda = 0$ , we have $h_1 = \frac{a^2+b^2}{a+b}$ and $h_2 = \frac{a^2+c^2}{a+c}$ . Thus, we let $Q$ be the intersection of these two lines. Since $Q$ is on line $BC$ , we have
$\overline{q} = \frac{b+c-q}{bc}$ Since $Q$ lies on the line through $\frac{a^2+b^2}{a+b}$ and $\frac{a^2+c^2}{a+c}$ , we have
$\frac{\frac{a^2+b^2}{a+b} - q}{\frac{a^2+b^2}{a+b} - \frac{a^2+c^2}{a+c}} \in \mathbb{R}$ We set this equal to its conjugate, and use the fact that $\overline{q} = \frac{b+c-q}{bc}$ , to get
$\frac{(a+c)(a^2+b^2-aq-bq)}{(b-c)(-a^2+ab+ac+bc)} = \frac{(a+c)(b^2c-a^2b-ab^2-abc+a^2q+abq)}{a(c-b)(a^2+ab+ac-bc)}$ $a(a^2+b^2-aq-bq)(a^2+ab+ac-bc) = (-a^2+ab+ac+bc)(a^2b+ab^2+abc-b^2c-a^2q-abq)$ $q = \frac{a(a^2+b^2)(a^2+ab+ac-bc) - b(a^2+ab+ac-bc)(-a^2+ab+ac+bc)}{a(a+b)(a^2+ab+ac-bc) - a(a+b)(-a^2+ab+ac+bc)}$ We simplify this to
$q = \frac{(a^2+ab+ac-bc)(a^3+a^2b-abc-b^2c)}{a(a+b)(2a^2-2bc)} = \frac{a^2+ab+ac-bc}{2a}$ So $Q$ is the foot of the altitude from $A$ onto line $BC$ . It remains to show that $Q$ always lies on line $H_1H_2$ as $\lambda$ varies. Indeed, we find
$h_1-q = \frac{2a(a^2+b^2+\lambda a^2-\lambda ab) - (a+b)(a^2+ab+ac-bc)}{2a(a+b)} = \frac{(a-b)(2\lambda a^2+a^2-ab-ac-bc)}{2a(a+b)}$ and similarly
$h_2-q = \frac{(a-c)(2\lambda a^2+a^2-ab-ac-bc)}{2a(a+c)}$ So the factor involving $\lambda$ cancels out, and
$\frac{h_1-q}{h_2-q} = \frac{(a-b)(a+c)}{(a-c)(a+b)}$ which is real. $\blacksquare$

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