2011 Japan Mathematical Olympiad Finals Problem 1

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2011 Japan Mathematical Olympiad Finals Problem 1

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power of a point

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Source: Japanese MO Finals 2011

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Kunihiko_Chikaya

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Kunihiko_Chikaya

#1 h Feb 11, 2011, 9:13 AM • 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 $AM\cdot PM=BM^2$ .

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#2 h Feb 11, 2011, 9:31 AM • 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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571 posts

yunxiu

#3 h Feb 11, 2011, 12:18 PM • 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

#4 h Feb 11, 2011, 2:51 PM • 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

#5 h Feb 11, 2011, 3:37 PM • 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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#6 h Feb 12, 2011, 12:17 PM • 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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1963 posts

#7 h Feb 13, 2011, 1:22 PM • 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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#8 h Sep 3, 2011, 8:46 AM • 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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7054 posts

#9 h Sep 3, 2011, 9:54 PM • 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. $MB^2=MP\cdot MA$ .

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#10 h Sep 14, 2011, 4:18 PM • 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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2130 posts

Sayan

#11 h Jan 28, 2012, 1:14 PM • 3 Y

Y by mathematiculperson, Adventure10, Mango247

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Attachments:

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#12 h Jan 28, 2012, 10:07 PM • 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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#13 h Jun 6, 2012, 5:17 PM • 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

#14 h Jun 7, 2012, 3:44 PM • 3 Y

Y by mathematiculperson, Adventure10, Mango247

My solution:
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#15 h Jan 14, 2021, 11:28 PM

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$[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

$\color{black}\rule{25cm}{1pt}$
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 $MB^2=MC^2=MP.MA$ .

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#17 h May 8, 2023, 9:49 AM

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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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384 posts

#18 h Jul 14, 2023, 4:37 AM

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Hint: Power, 5 points lie on 1 circle

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#19 h Jul 28, 2023, 6:50 PM

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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 $(I):$
$\Rightarrow \angle PBM=\angle BAP$ $\Rightarrow \boxed{AM\times PM=BM^2}_\blacksquare$ $\color{blue}\rule{24cm}{0.3pt}$

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#20 h Apr 8, 2024, 2:42 PM

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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

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ohiorizzler1434

#22 h Jul 8, 2024, 7:49 AM

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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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#23 h Apr 14, 2025, 2:08 PM

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(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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