Isogonal Line Lemma (updated)

by utkarshgupta, Jan 5, 2016, 7:14 AM

So I happened to stumble upon this lemma while solving a problem.
I know I should have used a diagram to prevent configuration thing but I don't really know how to use geogebra.
if someone can provide the diagram.

Lemma :
Let $AP,AS$ and $AQ,AR$ be two pairs of isogonal lines with respect to $\angle BAC$.
Let $PR \cap  QS = X$ and $PQ \cap RS = Y$.
Then $AX,AY$ are isogonal line with respect to $\angle BAC$

Proof


Now let us solve some problems using this lemma.

Problem 1 (India Postals 2015 Set 2) :
Let $ABCD$ be a convex quadrilateral. In the triangle $ABC$ let $I$ and $J$ be the incenter and the excenter opposite the vertex $A$, respectively. In the triangle $ACD$ let $K$ and $L$ be the incenter and the excenter opposite the vertex $A$, respectively. Show that the lines $IL$ and $JK$, and the bisector of the angle $BCD$ are concurrent.

Solution

Problem 2 (USAMO 2008} :
Let $ ABC$ be an acute, scalene triangle, and let $ M$, $ N$, and $ P$ be the midpoints of $ \overline{BC}$, $ \overline{CA}$, and $ \overline{AB}$, respectively. Let the perpendicular bisectors of $ \overline{AB}$ and $ \overline{AC}$ intersect ray $ AM$ in points $ D$ and $ E$ respectively, and let lines $ BD$ and $ CE$ intersect in point $ F$, inside of triangle $ ABC$. Prove that points $ A$, $ N$, $ F$, and $ P$ all lie on one circle.

Solution

Problem $3$ (China TST 2002) (Generalization of INMO 2003) :
Let $E$ and $F$ be the intersections of opposite sides of a convex quadrilateral $ABCD$. The two diagonals meet at $P$. Let $O$ be the foot of the perpendicular from $O$ to $EF$. Show that $\angle BOC=\angle AOD$.

Solution


Problem $4$ (Iran TST 2015)
$AH$ is the altitude of triangle $ABC$ and $H^\prime$ is the reflection of $H$ trough the midpoint of $BC$. If the tangent lines to the circumcircle of $ABC$ at $B$ and $C$, intersect each other at $X$ and the perpendicular line to $XH^\prime$ at $H^\prime$, intersects $AB$ and $AC$ at $Y$ and $Z$ respectively, prove that $\angle ZXC=\angle YXB$.

Solution

Problem $5$ (Tournament of Towns Spring Senior A 2006)(given by Anant Mudgal)
In triangle $ABC$, let $AA'$ be the bisector, and let $X$ be any point on $AA'$ and let $BX,CX$ meet $AC,AB$ again at $B',C'$. Let $BX$ meet $A'C'$ at $P$ and $CX$ meet $A'B'$ at $Q$. Prove that $\angle PAC=\angle QAB$.

Solution
This post has been edited 3 times. Last edited by utkarshgupta, Mar 4, 2016, 12:38 PM

Comment

22 Comments

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Another proof is by root BC + angle chase + isogonal conjugates.

Synthetic.

by anantmudgal09, Jan 7, 2016, 9:48 PM

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You inverted so that isn't exactly synthetic :P
This post has been edited 1 time. Last edited by utkarshgupta, Jan 8, 2016, 3:06 AM

by utkarshgupta, Jan 8, 2016, 3:05 AM

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Lol :)

This is kind of interesting as many people in different countries have different meanings when they say "synthetic".

by anantmudgal09, Jan 8, 2016, 6:12 AM

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This is useful :) Nice application of a well-known and technical lemma of $a+b=x+y<180$, $\frac{\sin a}{\sin b} = \frac{\sin x}{\sin y} \implies a=x, b=y$

by rkm0959, Feb 24, 2016, 4:08 PM

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Thanks :D

by utkarshgupta, Feb 26, 2016, 10:04 AM

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You should add RMM 2016 P1 in this list!

by rkm0959, Feb 27, 2016, 3:10 PM

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@above, shhh, you actually spoiled it for him, lol :P

Just kidding! :)

by anantmudgal09, Feb 27, 2016, 10:52 PM

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Another nice problem, which is reduced to triviality's oblivion, by our friend, is

Tournament of Towns Spring Senior A 2006

In triangle $ABC$, let $AA'$ be the bisector, and let $X$ be any point on $AA'$ and let $BX,CX$ meet $AC,AB$ again at $B',C'$. Let $BX$ meet $A'C'$ at $P$ and $CX$ meet $A'B'$ at $Q$. Prove that $\angle PAC=\angle QAB$.

Solution:- Trivial by our Lemma!

by anantmudgal09, Mar 2, 2016, 11:15 PM

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Wow, this even kills India TST 2014 that circumcenters etc problem. Just wow, dude, this is so awesome.... Hats off! :)

by anantmudgal09, Mar 5, 2016, 12:08 AM

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Add 2007 G3 we did on the ride as well. This Lemma is cool!

by anantmudgal09, Mar 9, 2016, 4:19 AM

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Another victim is USA TST 2012, Problem 2.

by Ankoganit, Mar 25, 2016, 10:29 AM

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Who else thinks there should be an AoPS Wiki page on this? :)

by Ankoganit, Apr 17, 2016, 2:43 PM

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Another one is ELMO 2016 P6a

by navi_09220114, Jan 3, 2018, 8:04 AM

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Don't forget ISL 2006/G3!

Again trivialized by this lemma.

by Vrangr, May 2, 2018, 5:23 AM

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Add Sharygin 2018 Problem 9.7 also to this list. Another problem reduced to the oblivion of triviality by this lemma.

by math_pi_rate, Aug 2, 2018, 8:27 AM

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not sure if someone has mentioned this but i think the proof of this lemma can also be done using DDIT?

by JasperL, Mar 17, 2019, 6:28 PM

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Talking of synthetic, my KVPY interviewers didn't know what synthetic geometry is

by MathPassionForever, May 6, 2019, 9:59 AM

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Why in the problem 2, the proof is completed if $AF$ is simedianns (more details please) ;) :read:

by Mathsy123, Nov 28, 2019, 10:39 PM

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Here's another short proof (I use this notation)

Fix $ABC$ as reference and let $P=(p_1,p_2,p_3)$ and $Q=(q_1,q_2,q_3).$ Then $BP: (p_1:t:p_3)$ and $CQ:(q_1:q_2:t),$ which gives $L_1=BP \cap CQ=\left( p_1q_1:q_2p_1:p_3q_1 \right).$ Similarly $L_2=BQ \cap CP=\left( p_1q_1:p_2q_1:q_3p_1 \right).$

Then since $$\frac{q_2p_1}{p_3q_1} \cdot \frac{p_2q_1}{q_3p_1}=\frac{p_2}{p_3} \cdot \frac{q_2}{q_3},$$hence $AL_1,AL_2$ are isogonal if and only if $AP,AQ$ are isogonal in $\angle BAC.$

by Wizard_32, Feb 8, 2020, 5:17 AM

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My favourite proof of this lemma is using polar duality. As if you know the background behind duality angle chasing, then this problem is dual to simple inscribed angle theorem :-).

As @JasperL mentioned DDIT. For those who understand DIT(Desargues Involution Theorem) and DDIT(Dual Desargues Involution Theorem), they really don't need this as en lemma, it's a direct consequence of DDIT, but DDIT is more powerful, as it can do similar tricks with all involutions and not just reflection. So If you like this lemma, I really recommend you to try and understand Desargues Involutions, it's just like this lemma, but better :-)

Note: It really looks like this problem is better view as it's dual as polar duality kills it and Dual Desargues does as well.

by riadok, Mar 19, 2020, 10:41 AM

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As I do not see a proof using DDIT in the comment section, so here it goes. :P

Firstly, we have that $(AP,AS)$ and $(AQ,AR)$ swap under the Involution of reflection over the angle-bisector of $\angle BAC$.

Now using DDIT on $PQSR$ (we use DDIT on the concave quadrangle) w.r.t. $A$, we get that $(AP,AS)$, $(AQ,AR)$ and $(AX,AY)$ swap under this Involution. This means that $AX$ and $AY$ are isogonal and we are done!

by kamatadu, Jul 7, 2023, 10:19 AM

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USAMO 2012 P5 as well!

by Math_legendno12, Jan 27, 2025, 2:07 PM

Stay insane,Coz it's your will, labour and pain,which takes you to the top of the mountain.

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  • Here goes first post of 2025! Great blog.

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  • First post of 2024

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  • Nice blog ! Your isogonality lemma is really powerful !

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  • Post plss....

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  • alas,this is ded

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  • Thanks for the nice blog.

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  • I think this might be silly but ... when should we expect to have another post ?? I am very keen to see it :D

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  • Let's all echo what's written in the blog description - Stay Insane / 'Cause it's your labor, will and pain/ That takes you to the top of soda fountain :D

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  • hey utkarsh jee is over now ... continue your elementary blog pleaseeeeeee!

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  • Congrats on becoming a contest moderator!

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  • INTERSTING BLOG

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  • I have no plans for this blog right now....
    No time here people !
    But lets see....
    I may try some combinatorics :P

    by utkarshgupta, Feb 15, 2017, 12:47 PM

  • Thanks for the nice blog!

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  • Revive it!!!
    Best blog out there, for sure!

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