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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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0 replies
jlacosta
May 1, 2025
0 replies
Iran TST Starter
M11100111001Y1R   2
N a minute ago by sami1618
Source: Iran TST 2025 Test 1 Problem 1
Let \( a_n \) be a sequence of positive real numbers such that for every \( n > 2025 \), we have:
\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]Prove that there exists a natural number \( M \) such that for all \( n > M \), the following holds:
\[
a_n < \frac{1}{1404}
\]
2 replies
M11100111001Y1R
Tuesday at 7:36 AM
sami1618
a minute ago
Twin Prime Diophantine
awesomeming327.   23
N an hour ago by HDavisWashu
Source: CMO 2025
Determine all positive integers $a$, $b$, $c$, $p$, where $p$ and $p+2$ are odd primes and
\[2^ap^b=(p+2)^c-1.\]
23 replies
awesomeming327.
Mar 7, 2025
HDavisWashu
an hour ago
Troublesome median in a difficult inequality
JG666   2
N an hour ago by navid
Source: 2022 Spring NSMO Day 2 Problem 3
Determine the minimum value of $\lambda\in\mathbb{R}$, such that for any positive integer $n$ and non-negative reals $x_1, x_2, \cdots, x_n$, the following inequality always holds:
$$\sum_{i=1}^n(m_i-a_i)^2\leqslant \lambda\cdot\sum_{i=1}^nx_i^2,$$Here $m_i$ and $a_i$ denote the median and arithmetic mean of $x_1, x_2, \cdots, x_i$, respectively.

Duanyang ZHANG, High School Affiliated to Renmin University of China
2 replies
JG666
May 22, 2022
navid
an hour ago
Concurrent lines, angle bisectors
legogubbe   0
an hour ago
Source: ???
Hi AoPS!

Let $ABC$ be an isosceles triangle with $AB=AC$, and $M$ an arbitrary point on side $BC$. The internal angle bisector of $\angle MAB$ meets the circumcircle of $\triangle ABC$ again at $P \neq A$, and the internal angle bisector of $\angle CAM$ meets it again at $Q \neq A$. Show that lines $AM$, $BQ$ and $CP$ are concurrent.
0 replies
legogubbe
an hour ago
0 replies
How do I prove this? - Sets and symmetric difference
smadadi1000   1
N 4 hours ago by KSH31415
For sets A, B and C, prove that (A Δ B) Δ C = (A Δ C) Δ (A \ B).

The textbook - Proofs by Jay Cummings - had this definition for symmetric difference:
A Δ B = ( A U B)/(A ∩ B)

this is exercise 3.37 (e).
1 reply
smadadi1000
Yesterday at 3:23 PM
KSH31415
4 hours ago
help me ..
exoticc   2
N Yesterday at 5:00 PM by sodiumaka
Find all pairs of functions (f,g) : R->R that satisfy:
f(1)=2025;
g(f(x+y))+2x+y-1=f(x)+(2x+y)g(y) , ∀x,y∈R
2 replies
exoticc
Yesterday at 9:26 AM
sodiumaka
Yesterday at 5:00 PM
Inequalities
sqing   20
N Yesterday at 3:32 PM by ytChen
Let $ a,b,c\geq 0 ,a+b+c\leq 3. $ Prove that
$$a^2+b^2+c^2+ab +2ca+2bc +  abc \leq \frac{251}{27}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{2}{5}abc  \leq \frac{4861}{540}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{7}{20}abc  \leq \frac{2381411}{26460}$$
20 replies
sqing
May 21, 2025
ytChen
Yesterday at 3:32 PM
21st PMO National Orals #9
yes45   0
Yesterday at 3:22 PM
In square $ABCD$, $P$ and $Q$ are points on sides $CD$ and $BC$, respectively, such that $\angle{APQ} = 90^\circ$. If $AP = 4$ and $PQ = 3$, find the area of $ABCD$.

Answer Confirmation
Solution
0 replies
yes45
Yesterday at 3:22 PM
0 replies
Inequalities
sqing   0
Yesterday at 2:31 PM
Let $ a,b> 0 $ and $2a+2b+ab=5. $ Prove that
$$ \frac{a^4}{b^4}+\frac{1}{a^4}+42ab-a^4\geq  43$$$$ \frac{a^5}{b^5}+\frac{1}{a^5}+64ab-a^5\geq  65$$$$ \frac{a^6}{b^6}+\frac{1}{a^6}+90ab-a^6\geq  91$$$$ \frac{a^7}{b^7}+\frac{1}{a^7}+121ab-a^7\geq  122$$
0 replies
sqing
Yesterday at 2:31 PM
0 replies
Original Problem (qrxz17)
qrxz17   0
Yesterday at 1:32 PM
Problem. Suppose that
\begin{align*}
    (a^2+b^2)^2 + (b^2+c^2)^2 +(c^2+a^2)^2 &= 138 \\
    (a^2+b^2+c^2)^2 &=100.
\end{align*}Find \(a^2(a^2-1) + b^2(b^2-1)+c^2(c^2-1)\).
Answer: Click to reveal hidden text
Solution. Subtracting the two given equations, we get
\begin{align*}
        a^4+b^4+c^4=38.
    \end{align*}Taking the square root of the second equation, we get

\begin{align*}
        a^2+b^2+c^2 = 10.
    \end{align*}
Then,
\begin{align*}
        a^4+b^4+c^4-(a^2+b^2+c^2) &= a^4-a^2+b^4-b^2+c^4-c^2 \\
        &= a^2(a^2-1)+b^2(b^2-1)+c^2(c^2-1) = \boxed{28}.
    \end{align*}
0 replies
qrxz17
Yesterday at 1:32 PM
0 replies
17th PMO Area Stage #5
yes45   0
Yesterday at 1:31 PM
Triangle \(ABC\) has a right angle at \(B\), with \(AB = 3\) and \(BC = 4\). If \(D\) and \(E\) are points
on \(AC\) and \(BC\), respectively, such that \(CD = DE = \frac{5}{3}\), find the perimeter of quadrilateral
\(ABED\).
Answer Confirmation
Solution
0 replies
yes45
Yesterday at 1:31 PM
0 replies
Sipnayan 2019 SHS Orals Semifinal Round B Difficult \#3
qrxz17   0
Yesterday at 1:30 PM
Problem. Suppose that \(a\), \(b\), \(c\) are positive integers such that \((a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) = 63\). Find all possible values of \(a+b+c\).
Answer. Click to reveal hidden text
Solution. Expanding and cancelling "like" terms, we get

\begin{align*}
        (a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) &= (a^4 +b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2) - 2(a^4+b^4+c^4) \\
        &= 2a^2b^2+2a^2c^2+2b^2c^2 - a^4 -b^4-c^4
    \end{align*}
We multiply the entire equation by \(-1\) to rearrange the terms and place the higher-degree terms at the front of the expression in order to have a more recognizable form.

\begin{align*}
        a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2 = -63.
    \end{align*}
Simplifying the left-hand expression, we get

\begin{align*}
        a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2 &=a^4-2a^2(b^2+c^2)+b^4-2b^2c^2+c^4 \\
        &= a^4-2a^2(b^2+c^2)+(b^2+c^2)^2 - 4b^2c^2 \\
        &= [a^2 - (b^2+c^2)]^2-(2bc)^2 \\
        &= (a^2-b^2-c^2+2bc)(a^2-b^2-c^2-2bc)\\
        &=[a^2-(b-c)^2][a^2-(b+c)^2] \\
        &= (a+b-c)(a-b+c)(a+b+c)(a-b-c) = -63.
    \end{align*}
If \(a=b=c\), we get
\begin{align*}
     -3a^4 &=-63 \\
     a^4 &= 21
    \end{align*}

in which there are no real solutions for \(a\).

If we either have \(a=b\), \(a=c\), or \(b=c\), we get something of the form

\[
    a^2(a+2b)(a-2b)=-63.
    \]
Since \(1\) and \(9\) are the only square factors of \(63\), we have \(a = \{1, 3\}\).

In this case, the solutions for \((a, b, c)\) are \((1, 4, 4)\) and \((3, 2, 2)\), including all their permutations.

If \(a, b, c\) are all distinct positive integers, there are no solutions.

Therefore, all possible values of \(a+b+c\) are \(\boxed{7 \text{ and }9}\).
0 replies
qrxz17
Yesterday at 1:30 PM
0 replies
MATHirang MATHibay 2011 Orals Survival Round Average \#1
qrxz17   0
Yesterday at 1:27 PM
Problem. Solve for all possible values of y: $\sqrt{y^2 + 4y + 8} + \sqrt{y^2+4y+4}=\sqrt{2(y^2+4y+6)}$
Answer. Click to reveal hidden text
Solution. Let \( N = y^2+4y+6\). We have
\begin{align*}
        \sqrt{N+2} + \sqrt{N-2} = \sqrt{2N}
    \end{align*}
Taking the square of the equation, we get

\begin{align*}
        (N+2)+(N-2)+2\sqrt{N^2-4}&=2N \\
        \sqrt{N^2-4} &=0
    \end{align*}
Taking the square of the equation again, we get

\begin{align*}
        N^2-4&=0 \\
        N^2&=4
    \end{align*}
If N is -2, then \(\sqrt{2N}\) is an imaginary number. Thus, N must be equal to +2. We then have

\begin{align*}
        y^2 +4y+6&=2 \\
        y^2+4y+4&=0 \\
        (y+2)^2&=0 \\
        y+2&=0 \\
        y &= \boxed{-2}.
    \end{align*}
0 replies
qrxz17
Yesterday at 1:27 PM
0 replies
Original Problem (qrxz17)
qrxz17   0
Yesterday at 1:12 PM
Problem. Given that \( 2^{\log_{16} 71} = a \) and \( x^{\log_{256} 71} = 71a \), find the value of \(x\).
Answer. Click to reveal hidden text
Solution. Let \(k=\log_{16}71\). We have
\begin{align*}
    2^k &= a \\   
    16^k &= 71
    \end{align*}
So,
\begin{align*}
        2^k\cdot16^k&=a\cdot71 \\
        32^k &= 71a = x^{\log_{256} 71}
    \end{align*}
Substituting the value of k, we get

\begin{align*}
       32^{\log_{16}71} &= x^{\log_{256} 71}
   \end{align*}
By using the change-of-base formula, we get

\begin{align*}
       32^{\frac{\log_{256}71}{\log_{256}16}}=32^{\frac{\log_{256}71}{\frac{1}{2}}} = 32^{2 \cdot \log_{256}71} = 1024^{\log_{256}71}
       \end{align*}
Therefore, \(x=\boxed{1024}\).
0 replies
qrxz17
Yesterday at 1:12 PM
0 replies
2011 Japan Mathematical Olympiad Finals Problem 1
Kunihiko_Chikaya   20
N Apr 14, 2025 by zhoujef000
Source: Japanese MO Finals 2011
Given an acute triangle $ABC$ with the midpoint $M$ of $BC$. Draw the perpendicular $HP$ from the orthocenter $H$ of $ABC$ to $AM$.
Show that $AM\cdot PM=BM^2$.
20 replies
Kunihiko_Chikaya
Feb 11, 2011
zhoujef000
Apr 14, 2025
2011 Japan Mathematical Olympiad Finals Problem 1
G H J
Source: Japanese MO Finals 2011
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Kunihiko_Chikaya
14514 posts
#1 • 4 Y
Y by ShahinH, MathLuis, Adventure10, Mango247
Given an acute triangle $ABC$ with the midpoint $M$ of $BC$. Draw the perpendicular $HP$ from the orthocenter $H$ of $ABC$ to $AM$.
Show that $AM\cdot PM=BM^2$.
This post has been edited 1 time. Last edited by Kunihiko_Chikaya, Feb 12, 2011, 1:34 AM
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vntbqpqh234
286 posts
#2 • 3 Y
Y by mathematiculperson, Adventure10, Mango247
Let BE,CF are altitudes of triangle $ABC$. EF meets BC at T.
Easy for we have T,H, P is collinear.
have A,F,H,P,E lie on a circle. hence $\angle MAB=\angle CHP$(1)
P,E,M,C lie on a circle then $\angle MPC=\angle ACB$
hence HPCB is inscribed in a circle or $\angle PHC=\angle MBP$(2)
With (1) and (2)
We have QED.
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yunxiu
571 posts
#3 • 3 Y
Y by mathematiculperson, Adventure10, Mango247
Let $BE,CF$ are altitudes of triangle,than $A,F,H,P,E$ lies on a circle.So $\angle APE=\angle AFE=\angle ACB$,hence $H,P,C,B$ lies on a circle.
Since $ME=MC$,we have $\angle MPC=\angle MEC=\angle MCE$,so $\triangle MPC\sim \triangle MCA$.So $BM^2=CM^2=MA\times{MP}$.
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jgnr
1343 posts
#4 • 3 Y
Y by mathematiculperson, Adventure10, and 1 other user
Let $AD,BE,CF$ be altitudes of $ABC$. Then $AEPHF$, $HPMD$, and $EFDM$ are cyclic, and the radical axis $EF, PH, MD$ concur at $T$. Note that $APDT$ is cyclic because $\angle APT=\angle ADT=90^{\circ}$, so $AM\cdot PM=TM\cdot DM=BM^2$. The last equality is because $(C,B;D,T)$ is harmonic.
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WakeUp
1347 posts
#5 • 3 Y
Y by mathematiculperson, Adventure10, Mango247
Let $AD,BE,CF$ be the altitudes. It is easy to see $A,F,P,H,E$ are concyclic and $P,H,D,M$ are concyclic.
Then $\angle AMD=\angle AHP=\angle AEP$ so $P,E,C,M$ are conyclic. Then $P$ must be the Miquel point of $F,E,M$ which implies $M,B,F,P$ are concyclic. Then $\angle BPM=\angle BFM=\angle BMF$ ($\triangle BMF$ is isosceles) and so $\triangle BPM$ is similar to $\triangle ABM$ which implies $AM\cdot PM=BM^2$.
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tonotsukasa
12 posts
#6 • 3 Y
Y by mathematiculperson, Adventure10, Mango247
Brutal-force solution

Let A,B,C be A$(2a,2b)$B$(0,0)$C$(2c,0)$ (where $b,c>0$). Then M is $(c,0)$ and we should prove $AM \times PM=BM^2=c^2$.
$AM=\sqrt{(2a-c)^2+(2b)^2}$, and H is $\left( 2a,\frac{2ac-2a^2}{b} \right)$.
Line AC is $2bx+(c-2a)y-2bc=0$, and line HP is $y=-(x-2a) \frac{2a-c}{2b}+\frac{2ac-2a^2}{b}$.
Thus P's x coordinate is $x_{p}=\frac{4a^2c - 2ac^2 + 4b^2c}{(2a-c)^2+(2b)^2}$.
We thus obtain $PM=\sqrt{1+\frac{(2b)^2}{(2a-c)^2}} |x_{p}-c|=\frac{c^2}{AM}$ and hence $AM \times PM=BM^2$.
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Headhunter
1963 posts
#7 • 2 Y
Y by Adventure10, Mango247
Let $AD$,$BE$,$CF$ be the feet of three altitudes of the triangle $ABC$. Then, $\square BCEF$ is cyclic.
It's well known that $O$ is the orthocenter of the triangle made by $BF\cap CE$, $BC\cap FE$, $BE\cap CF$
Thus, $GP$ is the polar of $A$ w.r.t. ($O$) and then $MP\cdot MA=MB^{2}$, which is the property of harmonic division.
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sunken rock
4401 posts
#8 • 2 Y
Y by Adventure10, Mango247
Take $A'$ the antipode of $A$ on the circle $\odot(ABC)$, $\{D\}=\odot(ABC)\cap \odot(HAP)$; obviously, $A', M, D$ are collinear, hence by power of $M$ w.r.t. $\odot (HAP)$: $MA\cdot MP=MH\cdot MD$, but $MA'=MH$, so $MA\cdot MP=MD\cdot MA'$; from power of $M$ w.r.t. $\odot (ABC)$ we get $MB\cdot MC=MD\cdot MA'$, so it's done.

Best regards,
sunken rock
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Virgil Nicula
7054 posts
#9 • 3 Y
Y by Adventure10, Mango247, and 1 other user
kunny wrote:
Let an acute $\triangle ABC$ with the midpoint $M$ of $[BC]$ . Draw the perpendicular $HP$ from the orthocenter $H$ to $AM$ . Show that $MA\cdot MP=BM^2$ .
Proof. Let $D\in BC$ , $E\in CA$ , $F\in AB$ be the projections of $H$ to the sides of $\triangle ABC$ and $T\in EF\cap BC$ . Are well-known that
$T\in HP$ and $(T,B,D,C)$ is a harmonical division. In conclusion, $MB^2=MD\cdot MT=MP\cdot MA$ , i.e. $MB^2=MP\cdot MA$ .
This post has been edited 1 time. Last edited by Virgil Nicula, Jan 28, 2012, 10:36 PM
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Vikernes
77 posts
#10 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Let the altitudes of the triangle be $AD,BE$ and $CF$, thus the points $A,F,P,H,E$ are concyclic. It is well known that $MF$ is tangent to the circle $\omega$ that passes throught the points $A,F,H,E$ (just denote $J$ the midpoint of $AH$, thus $J$ is the center of $\omega$ and show that $JF\perp MF$, it is an easy angle chase).
So by power of point $PM\cdot AM=MF^2=BM^2$.
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Sayan
2130 posts
#11 • 3 Y
Y by mathematiculperson, Adventure10, Mango247
Click to reveal hidden text
Attachments:
This post has been edited 1 time. Last edited by Sayan, Dec 10, 2012, 3:07 AM
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Virgil Nicula
7054 posts
#12 • 2 Y
Y by Adventure10, Mango247
PP. Given an acute triangle $ABC$ with the orthocenter $H$ and the midpoint $M$ of

$[BC]$. Denote the projection $P$ of $H$ on $AM$ . Show that $MA\cdot MP=MB^2$ .


Another proof (metric). Denote $D\in AH\cap BC$ . Using the power of $A$ w.r.t. the circumcircle of the quadrilateral $HDMP$

obtain that $AH\cdot AD=AP\cdot AM\iff$ $2Rh_a\cos A=$ $m_a\cdot AP\iff$ $AP\stackrel{(2Rh_a=bc)}{\ \ =\ \ }\frac {bc\cdot \cos A}{m_a}$ $\implies$

$MA\cdot MP=m_a\left(m_a-\frac {bc\cdot \cos A}{m_a}\right)=$ $m_a^2-bc\cdot \cos A=\frac 14\cdot\left[2\left(b^2+c^2\right)-a^2-2\cdot \left(b^2+c^2-a^2\right)\right]=\frac {a^2}{4}=MB^2$ .
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LarrySnake
42 posts
#13 • 1 Y
Y by Adventure10
Let $A'$ and $B'$ be the foot of altitudes from $A$ and $B$,respectively. Let $\angle PAB' = \alpha$. It is obviously that $\angle ACB = \angle AHB' = \angle APB' $ where $H $is the orthocenter of $ \triangle ABC $and $BM = B'M= MC$ ,because $\triangle BB'C$ is right-angled. From sine law for $\triangle AB'M$ and $\triangle AB'P $, we get:
$\frac{B'M}{\sin\ \alpha}= \frac{AM}{\sin(180- \angle ACB)}= \frac{AM}{\sin C}$ $\Longrightarrow$ $\frac{B'M}{AM} = \frac{\sin\ \alpha}{\sin C}$

$\frac{B'P}{\sin\ \alpha} = \frac{AB'}{\sin C} \Longrightarrow \frac{B'P}{AB'} = \frac{\sin\ \alpha}{\sin C}
=\frac{B'M}{AM}$

From these result we claim that $B'M$ is tangent in $B'$ to the circumcircle of $\triangle APB'$. From here follows that : $BM^{2} = B'M^{2} = AM\cdot PM $
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Mahi
52 posts
#14 • 3 Y
Y by mathematiculperson, Adventure10, Mango247
My solution:
Click to reveal hidden text
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EulersTurban
386 posts
#15
Y by
[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(12cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -13.451618314675148, xmax = 21.36861995749222, ymin = -12.20607728014102, ymax = 9.550882058546675;  /* image dimensions */

 /* draw figures */
draw((1.2217680901856096,5.147978491948061)--(-2.468863291036393,-4.078599961106974), linewidth(0.8) + blue); 
draw((-2.468863291036393,-4.078599961106974)--(12.802714838158101,-4.587652565413459), linewidth(0.8) + blue); 
draw((12.802714838158101,-4.587652565413459)--(1.2217680901856096,5.147978491948061), linewidth(0.8) + blue); 
draw((1.053897946824669,0.1118741911198956)--(2.983266984406037,0.9146988913215361), linewidth(2)); 
draw((1.2217680901856096,5.147978491948061)--(5.166925773560854,-4.333126263260216), linewidth(0.8) + blue); 
draw(circle((-2.292829886339947,1.2024021797863635), 5.2839352164545925), linewidth(0.8) + linetype("4 4") + red); 
draw((2.983266984406037,0.9146988913215361)--(-2.468863291036393,-4.078599961106974), linewidth(0.8) + blue); 
draw((2.983266984406037,0.9146988913215361)--(12.802714838158101,-4.587652565413459), linewidth(0.8) + blue); 
draw(circle((13.210755203392592,7.653558391621359), 12.248009741761717), linewidth(0.8) + linetype("4 4") + red); 
draw(circle((1.137833018505139,2.629926341533978), 2.5194506799028615), linewidth(0.8) + linetype("4 4") + red); 
draw((-0.5379741022876767,0.7486230107648315)--(1.053897946824669,0.1118741911198956), linewidth(0.8) + blue); 
draw((1.053897946824669,0.1118741911198956)--(12.802714838158101,-4.587652565413459), linewidth(0.8) + blue); 
draw((1.2217680901856096,5.147978491948061)--(1.053897946824669,0.1118741911198956), linewidth(0.8) + blue); 
draw((2.983266984406037,0.9146988913215361)--(-0.5379741022876767,0.7486230107648315), linewidth(0.8) + blue); 
draw((1.053897946824669,0.1118741911198956)--(-2.468863291036393,-4.078599961106974), linewidth(0.8) + blue); 
draw(circle((5.0829907018803855,-6.851178413674303), 8.044730573858503), linewidth(0.8) + linetype("4 4") + red); 
 /* dots and labels */
dot((1.2217680901856096,5.147978491948061),dotstyle); 
label("$A$", (1.3185350439108823,5.386108461287419), NE * labelscalefactor); 
dot((-2.468863291036393,-4.078599961106974),dotstyle); 
label("$B$", (-2.3683137143186044,-3.853771759954426), NE * labelscalefactor); 
dot((12.802714838158101,-4.587652565413459),dotstyle); 
label("$C$", (12.902522808965257,-4.35445492465226), NE * labelscalefactor); 
dot((5.166925773560854,-4.333126263260216),linewidth(4pt) + dotstyle); 
label("$M$", (5.255725384489285,-4.14962999363951), NE * labelscalefactor); 
dot((1.053897946824669,0.1118741911198956),linewidth(4pt) + dotstyle); 
label("$H$", (1.1364684385662163,0.2882435116367458), NE * labelscalefactor); 
dot((2.983266984406037,0.9146988913215361),linewidth(4pt) + dotstyle); 
label("$P$", (3.0709261203532927,1.1075432356877468), NE * labelscalefactor); 
dot((-0.5379741022876767,0.7486230107648315),linewidth(4pt) + dotstyle); 
label("$H'$", (-0.45661435819961127,0.9254766303430799), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]
Nice exercise for Humpty points :D
$\color{black}\rule{25cm}{1pt}$
Obviously, by definition we have that $P$ is the Humpty point of $ABC$.
Thus we have that $BHPC$ is a cyclic quadrilateral.
Now assume, without loss of generality, that $AB < AC$ and let $H'$ be the intersection of $HC$ with $AB$.
Obviously we have that $AH'HP$ is cyclic, since $\angle HPA = \angle HH'A= 90$.
Then we have that:
$$\angle PBC = \angle PHC = \angle H'AP = \angle BAP$$Thus we have that $MB$ is tangent to $(ABP)$.
Now we easily see that $\angle PCB= \angle PAC$, which implies that $MC$ is tangent to $(PAC)$
Thus we must have that $MB^2=MC^2=MP.MA$.
This post has been edited 1 time. Last edited by EulersTurban, Jan 14, 2021, 11:28 PM
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allaith.sh
26 posts
#17
Y by
Just reflect about $M$ then we have $H'$ is the antipode of $A$ on the circle $\odot(ABC)$ and $P'$ is $AM \cap \odot(ABC)$
$\implies ABCP'$ is cyclic $\implies AM \cdot MP=AM \cdot MP' = MB\cdot MC =BM^2$
This post has been edited 1 time. Last edited by allaith.sh, May 8, 2023, 9:52 AM
Reason: .
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David_Kim_0202
384 posts
#18
Y by
Hint: Power, 5 points lie on 1 circle
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hectorleo123
347 posts
#19
Y by
Kunihiko_Chikaya wrote:
Given an acute triangle $ABC$ with the midpoint $M$ of $BC$. Draw the perpendicular $HP$ from the orthocenter $H$ of $ABC$ to $AM$.
Show that $AM\cdot PM=BM^2$.
$\color{blue}\boxed{\textbf{Proof:}}$
$\color{blue}\rule{24cm}{0.3pt}$
Let $F\equiv CH\cap AB$
$P$ is the A-Humpty point of $\triangle ABC$
$$\Rightarrow BHPC \text{ is cyclic}$$$$\Rightarrow \angle PBM=\angle PHC...(I)$$
$$\angle HFA=\angle HPA=90$$$$\Rightarrow AFHP \text{ is cyclic}$$$$\Rightarrow \angle PHC=\angle PAF$$By $(I):$
$$\Rightarrow \angle PBM=\angle BAP$$$$\Rightarrow \boxed{AM\times PM=BM^2}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$
This post has been edited 1 time. Last edited by hectorleo123, Jul 28, 2023, 6:51 PM
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AshAuktober
1011 posts
#20
Y by
Simple enough.
Let $E, F$ be the feet of the $B-$ and $C-$altitudes. Then as $\angle{APH} = \angle{AEH} = \angle{AFH} = \frac{\pi}{2}$, we get $A, E, P, F, H$ to be concyclic. Now from the Three Tangents Lemma in EGMO Chapter 1 (Alternatively a simple angle chase), combined with Power of a Point, we get $MA \cdot MP = MF^2$. Furthermore, since $\Delta BFC$ is right-angled at $F$, we get $MB = MF$, so $MA \cdot MP = MB^2$. $\square$
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ohiorizzler1434
804 posts
#22
Y by
Skibiditastic problem! P is the humpty point by definition! Then it is well known that BM is a tangent to APB, and then the relation is true by power of a point!
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zhoujef000
322 posts
#23
Y by
(We will use the standard conventions of $BC=a,$ $AC=b,$ $AB=c,$ $\angle BAC=A,$ $\angle ABC=B,$ and $\angle ACB=C.$)
[asy]
import olympiad;
pair B=(0,0), C=(21,0), A=(5,12), D=(5,0), M=(21/2,0);
pair H=orthocenter(A, B, C), E=foot(B, A, C), F=foot(C, B, A);
pair P=foot(H, A, M);
draw(A--B--C--cycle);
draw(M--A--D);
draw(C--F);
draw(H--P);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$M$", M, S);
label("$F$", F, NW);
label("$P$", P-(0.5, 0), NE);
label("$H$", H, SW);
draw(circumcircle(H, D, M));
draw(rightanglemark(H, P, M, 20));
draw(rightanglemark(H, D, M, 20));
draw(rightanglemark(A, F, C, 20));
[/asy]
Let $D$ be the foot of the altitude from $A$ to $BC$ and let $F$ be the foot of the altitude from $C$ to $AB.$ Observe that $\angle HDM=90^{\circ}=\angle HPM,$ so $HPMD$ is cyclic. Now, by power of a point, $AH\cdot AD=AP\cdot AM.$

Observe that $AF=b\cos(A).$ As such, since $\angle HAF=90^{\circ}-B,$ $AH=\dfrac{b\cos(A)}{\sin(B)}.$ Also, $AD=c\sin(B).$ Thus, $AH\cdot AD=\dfrac{b\cos(A)}{\sin(B)}\cdot c\sin(B)=bc\cos(A).$

Also, by Apollonius's theorem, we have $AM^2=\dfrac{b^2+c^2}{2}-\dfrac{a^2}{4},$ so $AM\cdot PM=AM(AM-AP)=AM^2-AP\cdot AM=AM^2-AH\cdot AD=\left(\dfrac{b^2+c^2}{2}-\dfrac{a^2}{4}\right)-bc\cos(A)=\dfrac{b^2+c^2-2bc\cos(A)}{2}-\dfrac{a^2}{4}.$

By the law of cosines, $b^2+c^2-2bc\cos(A)=a^2,$ so $AM\cdot PM=\dfrac{b^2+c^2-2bc\cos(A)}{2}-\dfrac{a^2}{4}=\dfrac{a^2}{2}-\dfrac{a^2}{4}=\dfrac{a^2}{4}=\left(\dfrac{a}{2}\right)^2=BM^2,$ as desired. $\Box$
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