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A geometry problem from the TOT

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Source: Tournament of towns Spring 2018 A-level P4

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#1 h Mar 10, 2020, 11:51 AM

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Let O be the center of the circumscribed circle of the triangle ABC. Let AH be the altitude in this triangle, and let P be the base of the perpendicular drawn from point A to the line CO. Prove that the line HP passes through the midpoint of the side AB. (6 points)

Egor Bakaev

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#2 h Mar 10, 2020, 12:20 PM

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Sorry for the lack of a diagram.
Solution

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#3 h Mar 16, 2020, 3:56 AM

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Let $HP$ intersect $AB$ at $M$ . Since $\angle CHA = \angle CPA = 90$ , $CHPA$ is cyclic, so $\angle HPC= \angle HAC = \angle OAB$ (well-known). Thus, $OPMA$ is cyclic, so $\angle OMA = \angle OPA = 90$ , so since the circumcenter $O$ lies on the perpendicular bisector of $AB$ , we must have $AM=MB$ .

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#4 h Mar 16, 2020, 6:15 AM

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Let the perpendicular bisector of $AB$ meet $BC$ at $Q$ . Then $A,O,P,Q$ are concyclic, therefore $HP$ , the Simson-line of $A$ w.r.t. the triangle $OCQ$ , passes through the midpoint of $AB$ .

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#5 h Mar 13, 2024, 9:03 AM

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Let the unit circle be $(ABC)$ , and $c=1$ . Then $m=\frac{1}{2} (a+b)$ , $h=\frac{1}{2} (a+b+1- \frac{b}{a})$ .
Since, $P$ , $O$ , $C$ collinear, $p=k$ for $k \in \mathbb{R}$ and since $AP \perp OC$ , $\frac{1}{k-a}+ \frac{1}{k-\frac{1}{a}}=0 \implies p=\frac{1}{2} (a+\frac{1}{a})$ Consider the transformation $x \rightarrow 2x-a$ . Then it suffices to show $b$ , $b+1- \frac{b}{a}$ , $\frac{1}{a}$ are collinear. $\frac{\frac{b}{a}-1}{b-\frac{1}{a}}=\frac{\frac{a}{b}-1}{a-\frac{1}{b}} \implies \frac{b-a}{ab-1}=\frac{a-b}{1-ab}$ which is true.

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#6 h Aug 5, 2024, 9:31 PM

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Yet another complex bash sol, that i somehow quickly despite having zero practice.
Let $(ABC)$ be unit circle then $o=0$ , $h=\frac{1}{2} \left( a+b+c-\frac{bc}{a} \right)$ , $p=\frac{a+\frac{c^2}{a}}{2}$ and $m=\frac{a+b}{2}$ , so we need $h,p,m$ colinear but by the composed shifting $X \to (2X-a)a$ and colinearity lemma all we need is $\frac{c^2-ab}{c^2+bc-ab-ac} \in \mathbb R$ .
But this is easy to prove by taking the conjugates and multiplying numerator and denominator by $abc^2$ thus $HP$ bisects $AB$ as desired, and done :cool:

:cool:

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This post has been edited 3 times. Last edited by MathLuis, Aug 5, 2024, 9:33 PM

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#7 h Dec 5, 2024, 3:52 PM

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Direct
Sol

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#8 h Feb 18, 2025, 1:36 AM

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Solution

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#9 h Mar 17, 2025, 6:02 PM

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来个学习意志
WLOG let $(ABC)$ be unit circle on Argand Plane. By complex foot formula, $h=\frac{1}{2}(a+b+c-\frac{bc}{a}), p=\frac{1}{2}(a+\frac{c^2}{a})$ . Only need to prove that:
$\begin{vmatrix} 1&h&\overline{h}\\ 1&p&\overline{p}\\ 1&\frac{a+b}{2}&\overline{(\frac{a+b}{2})}\\ \end{vmatrix}= \begin{vmatrix} 2&a+b+c-\frac{bc}{a}&\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{a}{BC}\\ 2&a+\frac{c^2}{a}&\frac{1}{a}+\frac{a}{c^2}\\ 2&a+b&\frac{1}{a}+\frac{1}{b}\\ \end{vmatrix}=\begin{vmatrix} 0&c-\frac{bc}{a}&\frac{1}{c}-\frac{a}{bc}\\ 0&\frac{c^2}{a}-b&\frac{a}{c^2}-\frac{1}{b}\\ 2&a+b&\frac{1}{a}+\frac{1}{b}\\ \end{vmatrix}=0$ which is equivalent to
$(c-\frac{bc}{a})(\frac{a}{c^2}-\frac{1}{b})-(\frac{c^2}{a}-b)(\frac{1}{c}-\frac{a}{bc})=0.$ It is obviously true.

This post has been edited 5 times. Last edited by daixiahu, Mar 17, 2025, 6:04 PM
Reason: .

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3194 posts

#10 h Apr 1, 2025, 3:49 AM

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Let $M$ be the intersection of $HP$ and $AB$ . It suffices to show that $M$ is the circumcenter of $\triangle{ABH}$ , or that $\angle{MBH}=\angle{BHM}$ . If $\angle{ACO}=\theta$ , then $\angle{AOC}=180-2\theta\implies \angle{ABH}=\angle{MBH}=90-\theta$ . Additionally, $APHC$ is cyclic so $\angle{AHP}=\theta\implies \angle{BHM}=90-\theta$ .

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#11 h Apr 25, 2025, 7:05 AM

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Let $A, B, C$ be $a, b, c.$ We have that $H$ is $\frac12(a + b + c - \overline a bc),$ and since $P \in \overline{OC},$ $P$ lies on the diameter, and hence $p = \frac12(a + c - c - \overline a c(-c)) = \frac{1}{2}(a + \overline a c^2).$ Now, by applying the function $\tau\colon \alpha = 2\alpha - a - b,$ we have that the midpoint of $\overline{AB}$ maps to $0,$ and $\tau(H) = c - \overline{a}bc, \tau(P) = -b + \overline a c^2.$ However, since $\overline{\left(\frac{c - \overline abc}{-b + \overline ac^2}\right)} = \overline{\left(\frac{ac - bc}{-ab + c^2}\right)} = \frac{ac - bc}{-ab + c^2}$ we have that they are collinear.

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seriousPossibilist

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seriousPossibilist

#12 h May 14, 2025, 5:40 PM

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Cartesian coordinate bash: Shift the triangle so that $B,C$ lie on the $x-$ axis and $A$ lies
on the $y-$ axis.

The line $AB$ has equation $\frac{y}{x-b} = \frac{-a}{b} \implies y=\frac{-a}{b}x+a$ .
Hence the perpendicular bisector of segment $AB$ has the form
$y = \frac{b}ax + c$ for an arbitrary constant $c$ . Since the perpendicular bisector
of the line passes through $(\frac{b}2, \frac{a}2)$ , $c$ turns out to be $\frac{a^2-b^2}{2a}$ .
Hence the perpendicular bisector of segment $AB$ is $y = \frac{b}ax + \frac{a^2-b^2}{2a}$ .
Note that $x = \frac{b+c}{2}$ is the perpendicular bisector of $BC$ .
The point of intersection of both the perpendicular bisectors, i.e the circumcenter,
turns out to be: $O = \left(\frac{b+c}{2}, \frac{a^2+bc}{2a}\right).$ Now, the equation of line $CO =$ $\frac{y}{x-c} = \frac{a^2+bc}{a(b-c)} \implies y = x \left(\frac{a^2+bc}{a(b-c)} \right) + c \left(\frac{a^2 + bc}{a(c-b)} \right)$ Let this line be $\ell_1$ .
We aready to solve for the coordinate of $P$ : e
Any line perpendicular to line $CO$ has the form: $y = x\left( \frac{a(c-b)}{a^2 + bc} \right) + p \text{ for an arbitrary constant $p$ and since $(0,a)$ lies on the line through $P$, $p=a$. }$ Let this line be $\ell_2$ . Solving $\ell_1,\ell_2$ gives us the coordinates of $P$ . Hence:
$\frac{ax(c-b)}{a^2+bc}-\frac{a^2x-bcx}{a(b-c)} = \frac{a^2c+bc^2}{a(c-b)} - a \implies x = \frac{b(a^2+c^2)(a^2+bc)}{(ab-ac)^2+(a^2+bc)^2}$ $\implies \text{(from $\ell_2$) } y = \frac{b(a^2+c^2)(ac - ab)}{(ab-ac)^2+(a^2+bc)^2} + a \implies \frac yx = \frac{ac-ab}{a^2 + bc} + \frac ax$ Note that $\frac ax$ = $\frac{a(ab-ac)^2+a(a^2+bc)^2}{b(a^2+c^2)(a^2+bc)} = \frac{a^5+a^3b^2+a^3c^2+ab^2c^2}{a^4b+a^2b^2c+a^2bc^2+b^2c^3} =\frac{a\left(a^2+b^2\right)\left(a^2+c^2\right)}{b\left(a^2+bc\right)\left(a^2+c^2\right)} =\frac{a\left(a^2+b^2\right)}{b\left(a^2+bc\right)}$ $\implies \frac yx = \frac{a\left(a^2+b^2\right)}{b\left(a^2+bc\right)} + \frac{ac-ab}{a^2+bc} =\frac{a\left(a^2+b^2\right)+\left(ac-ab\right)b}{b\left(a^2+bc\right)} =\frac{a^3+abc}{b\left(a^2+bc\right)} = \frac ab.$ Hence if the coordinates of $P = (h,k)$ , then $\frac hk = \frac ba$ . Note that $H = (0,0) \implies$ the equation
of line $HP = \frac yx=\frac hk$ if $P = (h,k) \implies$ equation of $HP=\frac yx = \frac ba$ . But then the midpoint of
$AB = \left(\frac b2, \frac a2 \right)$ clearly lies on this line. Done. $\hfill \blacksquare$

This post has been edited 1 time. Last edited by seriousPossibilist, May 14, 2025, 5:42 PM
Reason: Minor formatting

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