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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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0 replies
+1 w
jwelsh
Jul 1, 2025
0 replies
Shortest sequence of coin flips
CyclicISLscelesTrapezoid   13
N 13 minutes ago by dolphinday
Source: USA TSTST 2023/7
The Bank of Pittsburgh issues coins that have a heads side and a tails side. Vera has a row of 2023 such coins alternately tails-up and heads-up, with the leftmost coin tails-up.

In a move, Vera may flip over one of the coins in the row, subject to the following rules:
[list=disc]
[*] On the first move, Vera may flip over any of the $2023$ coins.
[*] On all subsequent moves, Vera may only flip over a coin adjacent to the coin she flipped on the previous move. (We do not consider a coin to be adjacent to itself.)
[/list]
Determine the smallest possible number of moves Vera can make to reach a state in which every coin is heads-up.

Luke Robitaille
13 replies
CyclicISLscelesTrapezoid
Jun 26, 2023
dolphinday
13 minutes ago
"Eulerian" closed walk with of length less than v+e
Miquel-point   1
N 27 minutes ago by AbbyWong
Source: IMAR 2019 P4
Show that a connected graph $G=(V, E)$ has a closed walk of length at most $|V|+|E|-1$ passing through each edge of $G$ at least once.

Proposed by Radu Bumbăcea
1 reply
Miquel-point
May 16, 2025
AbbyWong
27 minutes ago
Find the minimum
sqing   0
33 minutes ago
Source: ZYZ
In triangle $ABC$. Find the minimum value of $cos^2A+cos^2 B+\sqrt{|cosC|} .$
0 replies
sqing
33 minutes ago
0 replies
Shortlist 2017/G4
fastlikearabbit   125
N 40 minutes ago by YaoAOPS
Source: Shortlist 2017, Romanian TST 2018
In triangle $ABC$, let $\omega$ be the excircle opposite to $A$. Let $D, E$ and $F$ be the points where $\omega$ is tangent to $BC, CA$, and $AB$, respectively. The circle $AEF$ intersects line $BC$ at $P$ and $Q$. Let $M$ be the midpoint of $AD$. Prove that the circle $MPQ$ is tangent to $\omega$.
125 replies
fastlikearabbit
Jul 10, 2018
YaoAOPS
40 minutes ago
Trigonometry equation practice
ehz2701   26
N Yesterday at 6:34 PM by vanstraelen
There is a lack of trigonometric bash practice, and I want to see techniques to do these problems. So here are 10 kinds of problems that are usually out in the wild. How do you tackle these problems? (I had ChatGPT write a code for this.). Please give me some general techniques to solve these kinds of problems, especially set 2b. I’ll add more later.

Leaderboard and Solved Problems

problem set 1a (1-10)

problem set 2a (1-20)

problem set 2b (1-20)
answers 2b

General techniques so far:

Trick 1: one thing to keep in mind is

[center] $\frac{1}{2} = \cos 36 - \sin 18$. [/center]

Many of these seem to be reducible to this. The half can be written as $\cos 60 = \sin 30$, and $\cos 36 = \sin 54$, $\sin 18 = \cos 72$. This is proven in solution 1a-1. We will refer to this as Trick 1.
26 replies
ehz2701
Jul 12, 2025
vanstraelen
Yesterday at 6:34 PM
Find the value of angle C
markosa   12
N Yesterday at 6:29 PM by sunken rock
Given a triangle ABC with base BC

angle B = 3x
angle C = x
AP is the bisector of base BC (i.e.) BP = PC
angle APB = 45 degrees

Find x

I know there are multiple methods to solve this problem using cosine law, coord geo
But is there any pure geometrical solution?
12 replies
markosa
Friday at 12:45 PM
sunken rock
Yesterday at 6:29 PM
10 Problems
Sedro   57
N Yesterday at 3:25 PM by Sedro
Title says most of it. I've been meaning to post a problem set on HSM since at least a few months ago, but since I proposed the most recent problems I made to the 2025 SSMO, I had to wait for that happen. (Hence, most of these problems will probably be familiar if you participated in that contest, though numbers and wording may be changed.) The problems are very roughly arranged by difficulty. Enjoy!

Problem 1: An sequence of positive integers $u_1, u_2, \dots, u_8$ has the property for every positive integer $n\le 8$, its $n^\text{th}$ term is greater than the mean of the first $n-1$ terms, and the sum of its first $n$ terms is a multiple of $n$. Let $S$ be the number of such sequences satisfying $u_1+u_2+\cdots + u_8 = 144$. Compute the remainder when $S$ is divided by $1000$.

Problem 2 (solved by fruitmonster97): Rhombus $PQRS$ has side length $3$. Point $X$ lies on segment $PR$ such that line $QX$ is perpendicular to line $PS$. Given that $QX=2$, the area of $PQRS$ can be expressed as $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.

Problem 3 (solved by Math-lover1): Positive integers $a$ and $b$ satisfy $a\mid b^2$, $b\mid a^3$, and $a^3b^2 \mid 2025^{36}$. If the number of possible ordered pairs $(a,b)$ is equal to $N$, compute the remainder when $N$ is divided by $1000$.

Problem 4 (solved by CubeAlgo15): Let $ABC$ be a triangle. Point $P$ lies on side $BC$, point $Q$ lies on side $AB$, and point $R$ lies on side $AC$ such that $PQ=BQ$, $CR=PR$, and $\angle APB<90^\circ$. Let $H$ be the foot of the altitude from $A$ to $BC$. Given that $BP=3$, $CP=5$, and $[AQPR] = \tfrac{3}{7} \cdot [ABC]$, the value of $BH\cdot CH$ can be expressed as $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.

Problem 5 (solved by maromex): Anna has a three-term arithmetic sequence of integers. She divides each term of her sequence by a positive integer $n>1$ and tells Bob that the three resulting remainders are $20$, $52$, and $R$, in some order. For how many values of $R$ is it possible for Bob to uniquely determine $n$?

Problem 6 (solved by Mathsll-enjoy): There is a unique ordered triple of positive reals $(x,y,z)$ satisfying the system of equations \begin{align*} x^2 + 9 &= (y-\sqrt{192})^2 + 4 \\ y^2 + 4 &= (z-\sqrt{192})^2 + 49 \\ z^2 + 49 &= (x-\sqrt{192})^2 + 9. \end{align*}The value of $100x+10y+z$ can be expressed as $p\sqrt{q}$, where $p$ and $q$ are positive integers such that $q$ is square-free. Compute $p+q$.

Problem 7 (solved by sami1618): Let $S$ be the set of all monotonically increasing six-term sequences whose terms are all integers between $0$ and $6$ inclusive. We say a sequence $s=n_1, n_2, \dots, n_6$ in $S$ is symmetric if for every integer $1\le i \le 6$, the number of terms of $s$ that are at least $i$ is $n_{7-i}$. The probability that a randomly chosen element of $S$ is symmetric is $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Compute $p+q$.

Problem 8: For a positive integer $n$, let $r(n)$ denote the value of the binary number obtained by reading the binary representation of $n$ from right to left. Find the smallest positive integer $k$ such that the equation $n+r(n)=2k$ has at least ten positive integer solutions $n$.

Problem 9 (solved by Math-lover1, sami1618): Let $p$ be a quadratic polynomial with a positive leading coefficient. There exists a positive real number $r$ such that $r < 1 < \tfrac{5}{2r} < 5$ and $p(p(x)) = x$ for $x \in \{ r,1,  \tfrac{5}{2r} , 5\}$. Compute $p(20)$.

Problem 10 (solved by aaravdodhia, sami1618): Find the number of ordered triples of positive integers $(a,b,c)$ such that $a+b+c=995$ and $ab+bc+ca$ is a multiple of $995$.
57 replies
Sedro
Jul 10, 2025
Sedro
Yesterday at 3:25 PM
Trigonometry prove
Studying_geometry   2
N Yesterday at 2:39 PM by CuriousMathBoy72
Prove that $ sin36^\circ - cos18^\circ = \frac{1}{2} $
2 replies
Studying_geometry
Yesterday at 2:18 PM
CuriousMathBoy72
Yesterday at 2:39 PM
PROVINCIAL MATHEMATICS 9 MATH QUESTIONS FOR REFERENCE
tuananh_vvvbb   0
Yesterday at 1:33 PM
Hello friends, I would like to share with you a reference to an HSG question for me to get a score of 70% or more, which is quite difficult. Readers, please refer to me for an explanation. Thank you all very much. Good health.
0 replies
tuananh_vvvbb
Yesterday at 1:33 PM
0 replies
Inequalitis
sqing   11
N Yesterday at 10:07 AM by sqing
Let $ a,b,c\geq  0 , a^2+b^2+c^2 =3.$ Prove that
$$a^3 +b^3 +c^3 +\frac{11}{5}abc  \leq \frac{26}{5}$$
11 replies
sqing
May 31, 2025
sqing
Yesterday at 10:07 AM
Inequalities of integers
nhathhuyyp5c   3
N Yesterday at 9:07 AM by Pal702004
Let $m,n$ be positive integers, $m$ is even such that $\sqrt{2}<\dfrac{m}{n}<\sqrt{2}+\dfrac{1}{2}$. Prove that there exist positive integers $k,l$ satisfying $$\left|\frac{k}{l}-\sqrt{2}\right|<\frac{m}{n}-\sqrt{2}.$$
3 replies
nhathhuyyp5c
Jun 14, 2025
Pal702004
Yesterday at 9:07 AM
Inequalities
sqing   5
N Yesterday at 8:05 AM by sqing
Let $ a,b,c\geq 0, \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=\frac{3}{2}.$ Prove that
$$ \left(a+b+c-\frac{17}{6}\right)^2+9abc   \geq\frac{325}{36}$$$$   \left(a+b+c-\frac{5}{2}\right)^2+12abc \geq\frac{49}{4}$$$$\left(a+b+c-\frac{14}{5}\right)^2+\frac{49}{5}abc \geq\frac{49}{5}$$
5 replies
sqing
Jun 30, 2025
sqing
Yesterday at 8:05 AM
x+y+z+1/x+1/y+1/z=0
nhathhuyyp5c   1
N Yesterday at 7:48 AM by sqing
Let $x,y,z$ be reals such that $|x|,|y|,|z|\geq1$ and $x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0$. Find $\max P=x+y+z$.
1 reply
nhathhuyyp5c
Yesterday at 6:38 AM
sqing
Yesterday at 7:48 AM
On a generalization of a^3+b^3+c^3-3abc
Sivin   1
N Yesterday at 4:05 AM by Sivin
We note that $x_1^2+x_2^2-2x_1x_2=(x_1-x_2)^2$ and
$${x_1^3+x_2^3+x_3^3-3x_1x_2x_3=(x_1+x_2+x_3)(x_1^2+x_2^2+x_3^2-x_1x_2-x_2x_3-x_3x_1)}$$are reducible polynomials in $\mathbb{C}.$ However, $x_1^4+x_2^4+x_3^4+x_4^4-4x_1x_2x_3x_4$ is an irreducible polynomial in $\mathbb{C}.$ So what are all the $n$ such that the polynomial:
$$f_n(x_1,x_2,\dots,x_n)=x_1^n+x_2^n+\dots+x_n^n-nx_1x_2\dots x_n$$is reducible in $\mathbb{C}$?
1 reply
Sivin
Dec 3, 2024
Sivin
Yesterday at 4:05 AM
Rectangle EFGH in incircle, prove that QIM = 90
v_Enhance   70
N Jul 21, 2025 by hectorleo123
Source: Taiwan 2014 TST1, Problem 3
Let $ABC$ be a triangle with incenter $I$, and suppose the incircle is tangent to $CA$ and $AB$ at $E$ and $F$. Denote by $G$ and $H$ the reflections of $E$ and $F$ over $I$. Let $Q$ be the intersection of $BC$ with $GH$, and let $M$ be the midpoint of $BC$. Prove that $IQ$ and $IM$ are perpendicular.
70 replies
v_Enhance
Jul 18, 2014
hectorleo123
Jul 21, 2025
Rectangle EFGH in incircle, prove that QIM = 90
G H J
Source: Taiwan 2014 TST1, Problem 3
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dolphinday
1339 posts
#62 • 1 Y
Y by fearsum_fyz
Let the altitude of $I$ onto $BC$ be $D$, and the antipode of $D$ wrt the incircle be $D'$. Then let the reflection of $Q$ over $I$ be $Q'$. Notice that the pole of $EF$ is $A$ and the pole of $\overline{D'D'}$ is $Q$. By reflection, we have $\overline{EF} \cap \overline{D'D'} = Q'$ and by La Hire's, $Q'$ is the pole of $AD'$. This means that the pole of $Q$ is parallel to $AD'$, and $AD' \parallel IM$ finishes.
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HamstPan38825
8904 posts
#63
Y by
Here's a fairly quick complex bash. Let the incircle be the unit circle, so $g=-e$ and $h=-f$, and in particular
\begin{align*}
q &= \frac{2def+d^2e+d^2f}{ef-d^2} \\
m &= \frac{df}{d+f} + \frac{de}{d+e}.
\end{align*}Hence $$\frac qm = \frac{(d+e)(d+f)}{ef-d^2}$$is the negative of its conjugate and thus imaginary, as needed.
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joshualiu315
2535 posts
#64 • 1 Y
Y by dolphinday
Let $D$ be the tangency of the incircle with $\overline{BC}$ and denote $D'$ as the antipode of $D$ with respect to the incircle. Then, reflect $Q$ over $I$ to point $Q'$, noting that $Q'$ is obviously tangent to the incircle and it lies on $\overline{EF}$. Since $A$ is the pole of $\overline{EF}$, we must have $\overline{AD'}$ be the polar of $Q'$. The well-known property that $\overline{AD'} \parallel \overline{IM}$ finishes. $\square$
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OronSH
1842 posts
#65
Y by
let $T$ be antipode of $D$, let $P=TT\cap EF$ clearly $I$ is the midpoint of $PQ$. now $AT\parallel IM$ by homothety at $D$ and $AT$ is the polar of $P$ implies the conclusion
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ErTeeEs06
89 posts
#66
Y by
Let $D'$ be the reflection of $D$ in $I$ and $P$ be the reflection of $Q$ in $I$. We wish to show $MI\perp IQ$. It is well known that $MI\parallel AD'$ because $AD'$ passes through the point where the $A$-excircle touches $BC$. So it satisfies to show $AD'\perp IP$. Because of reflections we know that $PD'$ is tangent to the incircle. Now since $P$ is on $EF$, which is the polar of $A$ we know by La Hire that $A$ is on the polar of $P$. Since $PD'$ is tangent we see that $AD'$ is the polar of $P$ which is ofcourse perpendicular to $PI$ and therefore we are done.
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daixiahu
7 posts
#67
Y by
来个学习意志
WLOG let $\odot I$ be unit circle on Argand plane and let $BC$ tangent to $\odot I$ at $z=1$, then $g=-e, h=-f$. By ice cream cone formula, $a=\frac{2ef}{e+f},b=\frac{2f}{f+1},c=\frac{2e}{e+1}$. By mid point formula, $m=\frac{f}{f+1}+\frac{2e}{e+1}=\frac{2ef+e+f}{ef+e+f+1}$. By complex interception formula, $q=\frac{2ef+e+f}{ef-1}$, want to show $q\overline{m}=-\overline{q}m$:
$$
LHS=\frac{2ef+e+f}{ef-1}\cdot\frac{\frac{e+f+2}{ef}}{\frac{e+f+1}{ef}+1}=\frac{2ef+e+f}{ef-1}\cdot\frac{e+f+2}{ef+e+f+1},
$$$$
RHS=\frac{\frac{e+f+2}{ef}}{1-\frac{1}{ef}}\cdot\frac{2ef+e+f}{ef+e+f+1}=\frac{e+f+2}{ef-1}\cdot\frac{2ef+e+f}{ef+e+f+1}.
$$
This post has been edited 5 times. Last edited by daixiahu, Mar 18, 2025, 7:01 AM
Reason: .
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Ilikeminecraft
734 posts
#68
Y by
Reflect $D, Q$ across $I$ to get $D’, Q’.$
Clearly, $D’Q’$ is tangent to the incircle. Note that $AD’\parallel IM$ since $AD’$ hits the $A$-extouch point, and take homothety centered at $D$. Clearly, $A$ lies on the polar of $Q’$ since $Q’\in EF.$ Thus, $AD’$ is the polar of $Q’.$ Thus, $AD’\perp IQ’.$ This finishes.
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Ilikeminecraft
734 posts
#69
Y by
well heres a complex

Let us shift our triangle such that the incircle is the unit circle. Define $D, E, F$ as $d, e, f.$ Thus, we have that $B, C, G, E$ are $\frac{2df}{d + f}, \frac{2de}{d + e}, -e, -f.$ By taking $Q = \overline{GH} \cap \overline{DD},$ we get that $Q = \frac{-2efd-d^2(e + f)}{ef - d^2} = d\cdot\frac{2ef + d(e + f)}{d^2 - ef}.$ Now, we prove that $\frac{IQ}{IM} \in i\mathbb R$:
\begin{align*}
	\frac{IQ}{IM} & = \frac{d\cdot\frac{2ef + d(e + f)}{d^2 - ef}}{\frac{df(d + e) + de(d + f)}{(d + f)(d + e)}} \\
	& = \frac{(d + f)(d + e)}{d^2 - ef} \\
	\overline{\left(\frac{(d + f)(d + e)}{d^2 - ef}\right)} & = \frac{(\frac1d + \frac1f)(\frac1d + \frac1e)}{\frac1{d^2} - \frac1e\frac1f} = \frac{(d +f)(d + e)}{ef - d^2}
\end{align*}and hence we are done.
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lpieleanu
3145 posts
#70
Y by
Solution
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zuat.e
96 posts
#71
Y by
Let $Q'$ be the reflection of $Q$ over $I$, then $Q=QI\cap EF$ and $D'Q'$ is tangent to the incircle, where $D'$ is the antipode of the tangency point $D$ with $BC$.
Let $E=AE\cap BC$. AS $IM\parallel AE$, it suffices to show that $AE\cap QI$.
Note that $Q'$ lies both on the tangent to the incircle at $D'$ and on $EF$, hence the polar of $Q'$ is precisely $AD'$, as desired.
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Learning11
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#72
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We complex bash.

Let the incircle be the unit circle. Then $a=\frac{2ef}{e+f}$, $b=\frac{2df}{d+e}$, $h=-e$, $g=-f$, and $m=\frac{df}{d+f}+\frac{de}{d+e}$. By complex intersection, $$q=\frac{hg(d+d)-d^2(h+g)}{hg-d^2}=\frac{2def+d^2(e+f)}{ef-d^2}.$$To show perpendicularity, it suffices to show $\frac{0-q}{0-m}+\overline{\left(\frac{0-q}{0-m}\right)}=0$. We have $$\frac{q}{m}=\frac{2def+d^2(e+f)}{ef-d^2}\cdot \frac{(d+f)(d+e)}{2def+d^2(e+f)}=\frac{(d+f)(d+e)}{ef-d^2},$$$$\overline{\left(\frac{q}{m}\right)}=\frac{\frac{2}{def}+\frac{1}{d^2}\left(\frac{1}{e}+\frac{1}{f}\right)}{\frac{1}{ef}-\frac{1}{d^2}} \cdot \frac{\left(\frac{1}{d}+\frac{1}{e}\right)\left(\frac{1}{d}+\frac{1}{e}\right)}{\frac{1}{d^2f}+\frac{1}{d^2e}+\frac{2}{def}}=\frac{(d+f)(d+e)}{d^2-ef}$$so we are done.
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SimplisticFormulas
162 posts
#73
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??
sol
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Jndd
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#74
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Let $D'$ be the antipode of $D$ with respect to the incircle of $ABC$, which we will call $\omega$. Then, let $Q'=EF\cap QI$. Since we can rotate triangle $GDQ$ $180$ degrees about $I$ to get triangle $FD'Q'$, we can say that the tangent to $\omega$ at $D$ intersects $Q'$. Since $Q'$ lies on the polar of $A$ with respect to $\omega$, we have that $A$ lies on the polar of $Q'$, but since $Q'D'$ is tangent to $\omega$, line $AD'$ must be precisely the polar of $Q'$. Let $AD'$ intersect $BC$ at $X$, which is the tangency point of the $A$-excircle to line $BC$. Then, it is well known that $BD=CX$ (and not hard to prove), so $BM=MX$ and $DX=2DM$. Since we have that $D'D=2ID$, we get $D'X\parallel IM$. Since we know that $AX$ is the polar of $Q'$ and the center of $(ID'Q')$ lies on $IQ'$, we have that $IQ'\perp AX$, giving that $IQ'\perp IM$. Since $Q$, $I$, and $Q'$ are collinear, we have that $QI\perp IM$, as desired.
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fearsum_fyz
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#75
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Let $Q'$ be the reflection of $Q$ over $I$ and let $D'$ be the antipode of $D$ in the incircle (where $D$ denotes the $A$-intouch point).

Claim: $AD'$ is the polar of $Q'$.
Proof. Flipping everything over $I$, we see that $Q'D'$ is tangent to the incircle, so $D'$ lies on the polar of $Q'$.
Also, $Q'$ lies on $\overleftrightarrow{EF}$, the polar of $A$, so by La Hire's theorem $A$ lies on the polar of $Q'$.
This implies that the line $\overleftrightarrow{AD'}$ is the polar of $Q'$.

Hence $AD' \perp \overleftrightarrow{Q'IQ}$.
But it is well known that $AD' \parallel IM$ (for a proof, let $X$ denote the intersection of $AD'$ with $BC$ and apply midpoint theorem to $\Delta{XDD'}$).
Therefore $IQ \perp IM$ as desired.
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This post has been edited 1 time. Last edited by fearsum_fyz, Jul 21, 2025, 4:45 PM
Reason: improved flow
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hectorleo123
368 posts
#76 • 1 Y
Y by Yiyj1
Posted only because apparently no one noticed the Newton line that shortens two lines
Let \( A' \) be the reflection of \( A \) over \( I \), and \( D' \) the midpoint of \( AD \), then \( A' \) is the intersection of the tangents from \( G \) and \( H \) to \( (I) \).
The polar of \( Q \) with respect to \( (I) \) passes through \( A' \) and \( D \Rightarrow IQ \perp A'D \), by homothety at \( A \) with ratio \( 2:1 \).
\( A'D \parallel ID' \), which is the Newton line of the degenerate quadrilateral \( ABDC \Rightarrow I, D', M \) are collinear
\(\Rightarrow A'D \parallel IM \Rightarrow IM \perp IQ  _\blacksquare\)
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