Harmonic Divisions (Lemmas)

by Generic_Username, Aug 12, 2016, 8:07 PM

Let $A,B,C,D$ be four points on a line or circle. The cross-ratio $(A,B;C,D)=\pm\dfrac{CA}{CB}:\dfrac{DA}{DB}$ where the segments are directed and we take the $+$ if $AB\cap CD$ is not in the interior of the circle in the case of a circle and $-$ otherwise.

Let $A,B,C,D$ be four points on a line or circle in this order. Then if $(A,C;B,D)=-1,$ the points $A,B,C,D$ form a harmonic bundle. In the case of a circle, $ABCD$ is called a harmonic quadrilateral.

Lemma 1:
$(A,C;B,D)$ is harmonic iff $MB\cdot MD=MA^2$ where $M$ is the midpoint of segment $AC.$

Proof: Let $A=0,B=b,C=c,D=d$ on the number line. Then $m=\dfrac{c}{2}$ $MB\cdot MD=MA^2\iff (m-b)(m-d)=(m-a)^2\iff (c-2b)(c-2d)=c^2\iff \dfrac{c-0}{c-b}:\dfrac{d-0}{d-b}=-1\iff (A,C;B,D)=-1,$ done.

Lemma 2:
Let $M$ be the midpoint of segment $AB.$ Then $(A,B;M,P)$ is harmonic iff $P$ is the point at infinity on line $AB.$

Proof: By definition, $(A,B;M,P)=\dfrac{MB}{MA}:\dfrac{PB}{PA}=-1$ iff $PA=PB,$ done.

Lemma 3:
In $\triangle ABC,$ let $AX,BY,CZ$ be concurrent cevians. If $YZ\cap BC=X',$ then $(B,C;X,X')$ is harmonic. Converse also holds.

Proof: We only prove one direction, as the proofs are similar.

Assuming the concurrency of cevians, by Ceva we have $\dfrac{XB}{XC}\cdot\dfrac{YC}{YA}\cdot\dfrac{ZA}{ZB}=1.$ By Menelaus WRT $Z,Y,Z'$ we have $\dfrac{X'B}{X'C}\cdot\dfrac{YC}{YA}\cdot\dfrac{ZA}{ZB}=-1,$ and dividing yields the desired result.

Lemma 4: Let $A,B,C$ be three points lying on circle $\omega.$ Let the tangents at $A$ and $C$ intersect at a point $P$ and let the line $PB$ intersect $\omega$ again at $D$. Then $ABCD$ is harmonic. Converse also holds.

Proof: Again, we only prove one direction. Since $BD$ is a symmedian in $\triangle ABC,\triangle ADC$ we have by Steiner's $\dfrac{BA^2}{BC^2}=\dfrac{AX}{CX}=\dfrac{DA^2}{DC^2}$ so $(A,C;B,D)=-1.$

Lemma 5: Let $A,B,C,D$ be points lying in this order on line $d,$ and let $P$ be a point not lying on this line. Take another line $d'$ and consider the intersections $A',B',C',D'$ of the lines $PA,PB,PC,PD$ with another line $d'.$ Then $(A,C;B,D)=(A',C';B',D').$

Proof: Let $\angle APB=x,\angle BPC=y,\angle CPD=z.$ By Ratio Lemma, $\left(\dfrac{BA}{BC}\right)\left(\dfrac{DC}{DA}\right)=\left(\dfrac{PA\sin x}{PC\sin y}\right)\left(\dfrac{PC\sin z}{PA\sin{(x+y+z)}}\right)$ which is a ratio solely dependent on angles, so we are done.

Lemma 6: Let $A,B,C,D$ be four points lying on a line in this order. If $X$ does not lie on this line, two of these statements imply the last;

i) $(A,C;B,D)=-1.$
ii) $XB$ is the internal bisector of $\angle AXC.$
iii) $XB\perp XD.$

Proof:
$\text{i}+\text{ii}\implies \text{iii}$:
$XB,XD$ are the internal and external bisectors of $\angle AXC$ so we are immediately done.

$\text{ii}+\text{iii}\implies \text{i}$:
By the angle bisector theorem we may immediately conclude.

$\text{iii}+\text{i}\implies \text{ii}$:
Using notation from the previous lemma, $(A,C;B,D)=-\dfrac{\sin x\sin z}{\sin y\sin(x+y+z)}=-\dfrac{\cos y\sin x}{\sin y\cos x}=-\dfrac{\tan x}{\tan y}=-1$ iff $x=y.$
1. Let $ABC$ be a triangle with orthocenter $H$ and let $D,E,F$ be the feet of the altitudes lying on the sides $BC,CA,AB.$ Let $T=EF\cap BC.$ Prove $TH$ is perpendicular to the $A$-median of $\triangle ABC.$

Solution

2. In $\triangle ABC$ let $M$ be the midpoint of $AB,$ let $X$ be the second intersection of $BC$ with the circumcircle of $\triangle AMC$ and let $\omega$ be the circle which is tangent to $AC$ at $C$ and passes through $X.$ Furthermore, let $XM\cap \omega=Z,XM\cdot AC=Y.$ Prove that $AX,BY,CZ$ are concurrent.

Solution (incomplete)

3. (IMO 2012) Given triangle ABC the point $J$ is the centre of the excircle opposite the vertex $A$. This excircle is tangent to the side $BC$ at $M$, and to the lines $AB$ and $AC$ at $K$ and $L$, respectively. The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G$. Let $S$ be the point of intersection of the lines $AF$ and $BC$, and let $T$ be the point of intersection of the lines $AG$ and $BC$. Prove that $M$ is the midpoint of $ST$. (The excircle of $ABC$ opposite the vertex $A$ is the circle that is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond $C$.)

Solution

4. (IMO 2003) Let $ABCD$ be a cyclic quadrilateral. Let $P,Q,R$ be the feet of the perpendiculars from $D$ to the lines $BC,CA,AB,$ respectively. Show that $PQ=QR$ if and only if $ABCD$ is harmonic.

Solution

5. (Sharygin 2013) Let $D$ be the foot of the $B-$internal angle bisector of $\triangle ABC.$ Points $I_a,I_c$ are the incenters of $\triangle ABD,\triangle CBD$ respectively. The line $I_aI_c$ meets $AC$ at point $Q.$ Prove that $\angle DBQ=90^{\circ}.$

Solution

6. (USA TST 2011) In an acute scalene triangle $ABC,$ points $D,E,F$ lie on sides $BC,CA,AB,$ respectively, such that $AD\perp BC,BE\perp CA,CF\perp AB.$ Altitudes $AD,BE,CF$ meet at orthocenter $H.$ Points $P$ and $Q$ lie on segment $EF$ such that $AP\perp EF$ and $HQ\perp EF.$ Lines $DP$ and $QH$ intersect at point $R.$ Compute $\dfrac{HQ}{HR}.$

Solution (projective)
Solution (Yufei)

7. Let $\omega$ be a circle with center $O$ and $A$ a point outside it. Denote by $B,C$ the points where the tangents from $A$ to $\omega$ meet the circle, $D$ the point on $\omega$ for which $O$ lies on line $AD,$ $X$ the foot of the perpendicular from $B$ to $CD,$ $Y$ the midpoint of segment $BX,$ and $Z$ the second intersection of $DY$ with $\omega.$ Prove that $ZA\perp ZC.$

Solution

8. (APMO 2013) Let $ABCD$ be a quadrilateral inscribed in a circle $\omega$, and let $P$ be a point on the extension of $AC$ such that $PB$ and $PD$ are tangent to $\omega$. The tangent at $C$ intersects $PD$ at $Q$ and the line $AD$ at $R$. Let $E$ be the second point of intersection between $AQ$ and $\omega$. Prove that $B$, $E$, $R$ are collinear.

Solution

9. (Sharygin 2013) Let $AD$ be a bisector of triangle $ABC.$ Points $M$ and $N$ are projections of $B$ and $C$ respectively on $AD.$ The circle with diameter $MN$ intersects $BC$ at points $X$ and $Y.$ Prove that $\angle BAX=\angle CAY.$

Solution

10. (ELMO SL 2014) Let $AB=AC$ in $\triangle ABC,$ and let $D$ be a point on segment $AB.$ The tangent at $D$ to the circumcircle $\omega$ of triangle $BCD$ hits $AC$ at $E$ The other tangent from $E$ to $\omega$ touches it at $F,$ and $G=BF\cap CD,H=AH\cap BC.$ Prove that $BH=2HC.$

Solution
This post has been edited 4 times. Last edited by Generic_Username, Aug 23, 2016, 5:55 PM

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6 Comments

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Lemma 7 (not included in lemmas): Let $\triangle DEF$ be the orthic triangle of $\triangle ABC$ and let $H$ be the orthocenter of $\triangle ABC,$ $X=EF\cap AD.$ Then $(A,H;X,D)$ is harmonic.

Proof: Let $EF\cap BC=T.$ By Lemma 3, $-1=(T,D;B,C)\stackrel{E}{=}(A,H;X,D)$ so we are done.

by Generic_Username, Aug 13, 2016, 2:15 AM

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Useless lemma: Let $\triangle DEF$ be the contact triangle of $\triangle ABC$ and let $I$ be the incenter. Then $(A,I;E,F)$ is harmonic.

Let $T$ be a point distinct from $A$ such that $AT$ is tangent to $(AEIF)$. Since $AI$ is a diameter of $(AEIF)$, we merely need to show that $EF || AT$. But as $\angle TAF = \angle AEF$ due to tangency and $\angle AEF = \angle AFE$ as $AEF$ is isosceles, we are done.

by cjquines0, Aug 14, 2016, 10:54 AM

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hello generic i gave you admin

by wu2481632, Aug 19, 2016, 10:26 PM

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I'll edit soon

by Generic_Username, Aug 22, 2016, 1:43 AM

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k done with all bar 2 and 10 but I'll do those later

by Generic_Username, Aug 23, 2016, 1:16 AM

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Greetings Wu, I am Wu.

by JustinW, Jun 14, 2019, 10:14 AM

DARN

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  • fr /1888 char

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  • :omighty: harmonic divisions lol

    by gracemoon124, May 3, 2023, 4:11 AM

  • I got here after searching Midpoint of Harmonic Bundles Lemma on 2019 AIME I #15

    by the_mathmagician, Jan 23, 2022, 6:39 PM

  • wait help how do i prepare for jmo

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  • cool problem, wu!

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  • I knew there was a reason I named myself First.

    by First, May 6, 2017, 2:38 AM

  • he was the First to predict such an occurrence

    by PiDude314, May 6, 2017, 12:56 AM

  • first was right all along

    by wu2481632, May 6, 2017, 12:52 AM

  • :rotfl: I am more than sure you can solve a JMO #3, but sometimes Olympians regret spending time on a #3 in their best subject and getting crushed by it vs a #1 in one of their worse subjects

    by First, Sep 27, 2016, 2:02 AM

  • What makes you think I can never solve a JMO3-level question?

    by wu2481632, Sep 23, 2016, 2:01 PM

  • There has been consistently two geo's on the JMO, but they aren't always just #1 or #2.

    by First, Sep 16, 2016, 1:18 AM

  • Wu told me to shout for him again
    He says "I'm screwed" (this is a real quote)
    Lmao I think he should do NT and combo but apparently he just gets bored by that

    by Generic_Username, Sep 11, 2016, 8:59 PM

  • Wu what happens if there is only one geo on the JMO or they are too hard for you. Shouldn't you do a lot of problems in another subject as well

    by First, Sep 11, 2016, 7:29 PM

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