IMO Shortlist 2017 Geometry

by liberator, Oct 4, 2018, 9:40 PM

Here are my solutions to some of the geometry problems from this year's IMO Shortlist.

G1. Let $ABCDE$ be a convex pentagon such that $AB=BC=CD$ , $\angle{EAB}=\angle{BCD}$ , and $\angle{EDC}=\angle{CBA}$ . Prove that the perpendicular line from $E$ to $BC$ and the line segments $AC$ and $BD$ are concurrent.

My solution

G3. Let $O$ be the circumcenter of an acute triangle $ABC$ . Line $OA$ intersects the altitudes of $ABC$ through $B$ and $C$ at $P$ and $Q$ , respectively. The altitudes meet at $H$ . Prove that the circumcenter of triangle $PQH$ lies on a median of triangle $ABC$ .

My solution

G4. 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 $\omega$ .

My solution

geometry

ISL

Proving collinearities with the Newton-Gauss line

by liberator, Apr 11, 2017, 10:29 PM

So apparently I'm being peer pressured into posting

Here is the desired blog entry, a whole 364 days after the last! I hope that this satisfies your requests/demands or whatever you want to call it.

This is just to illustrate a nice little trick that I've used to solve several (read: three) problems.

To show that three points are collinear, we can use the converse of the Newton-Gauss line property, by proving that circles on appropriate diameters are coaxal (orthocentres often come in handy here), which can give us the fourth side of a complete quadrilateral. This is perhaps best demonstrated by three examples:

Problem 1: Let $ABC$ be a triangle. Let $\triangle P_aP_bP_c$ and $\triangle Q_aQ_bQ_c$ be the circumcevian triangles of points $P$ and $Q$ respectively. Prove that $\triangle ABC$ and the triangle determined by lines $P_aQ_a,P_bQ_b,P_cQ_c$ are in perspective.

[Jerabek's theorem]

$[asy] unitsize(2cm); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10); path carc(pair A, pair B, pair C, real d=0, bool dir) { pair O=circumcenter(A,B,C); return arc(O,circumradius(A,B,C),degrees(A-O)+d,degrees(C-O)-d,dir); } pair A=dir(110), B=dir(200), C=dir(-20), Q=(-0.1,-0.05), P_a=-A, P_b=-B, P_c=-C, Q_a=IP(unitcircle,L(A,Q,-0.1,10)), Q_b=IP(unitcircle,L(B,Q,-0.1,10)), Q_c=IP(unitcircle,L(C,Q,-0.1,10)), X=extension(P_a,Q_a,B,C), Y=extension(P_b,Q_b,C,A), Z=extension(P_c,Q_c,A,B); D(X--B--A--C); D(unitcircle,heavygreen); D(carc(A,foot(A,B,C),X,CCW),red); D(X--Q_a--A--foot(A,B,C)); D("A",A,NW); D("B",B,SW); D("C",C,SE); D("Q",Q,E); D("Q_a",Q_a); D("P_a",P_a,N); D("X",X,E); D("H",A+B+C,SSE); [/asy]$
Solution. Let $X=P_aQ_a\cap BC$ , and define $Y,Z$ analogously. Take a homography fixing $(ABC)$ and sending $P$ to the centre. It is easy to see that the circles on diameters $\overline{AX},\overline{BY},\overline{CZ}$ are coaxal, with common radical axis $QH$ . Then by the converse of the Newton-Gauss line, it follows that $X,Y,Z$ are collinear. $\square$

[Sorry for the lack of complete diagrams for the next two problems. Some of the points are very far apart, meaning that the diagrams would probably not be very clear anyway]

Problem 2: A circle $\omega$ with centre $P$ intersects the sides $BC,CA,AB$ of triangle $ABC$ at $A_1$ and $A_2$ , $B_1$ and $B_2$ , $C_1$ and $C_2$ respectively. Let $A'$ be the circumcentre of triangle $A_1A_2P$ , $B'$ the circumcentre of triangle $B_1B_2P$ , $C'$ the circumcentre of triangle $C_1C_2P$ . Prove that $AA'$ , $BB'$ and $CC'$ are concurrent.

[UK Training for RMM 2017]

$[asy] unitsize(2cm); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10); pair P=origin, A_1=dir(230), A_2=dir(-50), B_1=dir(-20), B_2=dir(95), C_1=dir(120), C_2=dir(185), A=extension(B_1,B_2,C_1,C_2), B=extension(C_1,C_2,A_1,A_2), C=extension(A_1,A_2,B_1,B_2), Ap=circumcenter(A_1,A_2,P), Bp=circumcenter(B_1,B_2,P), Cp=circumcenter(C_1,C_2,P), X=2/(conj(IP(A--Ap,unitcircle))+conj(OP(A--Ap,unitcircle))); D(A--B--C--cycle); DPA(A--Ap^^B--Bp^^C--Cp); D(unitcircle,heavygreen); D("A",A,NW); D("B",B,SW); D("C",C,SE); D("A'",Ap,E); D("B'",Bp); D("C'",Cp,SW); D("P",P,E); [/asy]$
Solution. Let the polar of $A$ in $\omega$ meet the polar of $A'$ at $X$ , and define $Y, Z$ analogously. It suffices to show that $X, Y, Z$ are collinear. Let $\triangle DEF$ be the triangle formed by the polars of $A', B', C'$ in $\omega$ , which is homothetic to $\triangle ABC$ with centre $P$ . Consider the circles on diameters $\overline{DX},\overline{EY},\overline{FZ}$ . Easily, $P$ and the orthocentre of $\triangle DEF$ lie on a common radical axis of these circles. Thus the midpoints of $\overline{DX},\overline{EY},\overline{FZ}$ are collinear, so the result follows by the converse of the Newton-Gauss line. $\square$

Problem 3: A circle $\omega$ with centre $P$ intersects the sides $BC,CA,AB$ of triangle $ABC$ at $A_1$ and $A_2$ , $B_1$ and $B_2$ , $C_1$ and $C_2$ respectively. Let the lines tangent to $\omega$ at $A_1$ and $A_2$ meet at $A'$ and define similarly $B'$ and $C'$ . Prove that the lines $AA'$ , $BB'$ , and $CC'$ are concurrent.

[Silouanas Brazitikos, UK IMO Training and Selection Camp April 2017]
$[asy] unitsize(2cm); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10); pair P=origin, A_1=dir(230), A_2=dir(-50), B_1=dir(-20), B_2=dir(95), C_1=dir(120), C_2=dir(185), A=extension(B_1,B_2,C_1,C_2), B=extension(C_1,C_2,A_1,A_2), C=extension(A_1,A_2,B_1,B_2), Ap=2/(conj(A_1)+conj(A_2)), Bp=2/(conj(B_1)+conj(B_2)), Cp=2/(conj(C_1)+conj(C_2)), X=extension(B,C,tangent(A,P,1),tangent(A,P,1,2)); D(A--B--C--cycle); DPA(A--Ap^^B--Bp^^C--Cp); D(unitcircle,heavygreen); D("A",A,NW); D("B",B,SW); D("C",C,SE); D("A'",Ap); D("B'",Bp,NE); D("C'",Cp,NW); D("P",P,E); [/asy]$
Solution. Let the polar of $A$ in $\omega$ meet $BC$ at $X$ , and define $Y,Z$ analogously. It suffices to show that $X,Y,Z$ are collinear. Consider the circles on diameters $\overline{AX},\overline{BY},\overline{CZ}$ . Easily, $P$ and the orthocentre of $\triangle ABC$ lie on a common radical axis of these circles. Thus the midpoints of $\overline{AX},\overline{BY},\overline{CZ}$ are collinear, so the result follows by the converse of the Newton-Gauss line. $\square$

geometry

Parallelogram Isogonality Lemma

by liberator, Apr 12, 2016, 1:43 PM

Well I thought that I better make a blog post within the year so here it is.

This nice lemma allows us to solve a variety of olympiad geometry problems. I first encountered it as BrMO2 2013/2. It is a generalisation of A Simple Lemma.

Lemma. The point $P$ lies inside triangle $ABC$ so that $\angle ABP = \angle PCA$ . The point $Q$ is such that $PBQC$ is a parallelogram. Prove that $\angle QAB = \angle CAP$ .
$[asy] unitsize(2.8cm); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10pt); real x=0.54; pair A=dir(110), B=dir(205), C=dir(-25), X=WP(C--A,x), Y=IP(circumcircle(B,C,X),A--B), P=IP(B--X,C--Y), Q=B+C-P, Ap=A+B-P; D(unitcircle,heavygreen); D(A--B--C--cycle,heavygrey+linewidth(1)); D(Q--A--P,purple); DPA(B--P--C--Q--cycle^^Ap--A^^Ap--B^^Ap--Q); /* Angle marks */ DPA(anglemark(P,B,A,5)^^anglemark(A,C,P,5),orange); /* Dots and labels */ D("A",A,dir(A)); D("B",B,dir(B)); D("C",C,dir(C)); D("P",P,dir(130)); D("Q",Q,dir(Q)); D("A'",Ap,dir(Ap)); [/asy]$
Proof. Let $A'$ be such that $APBA',ACQA'$ are parallelograms. Then $\angle A'AB=\angle PBA=\angle PCA=\angle A'QB$ , so $AA'BQ$ is cyclic. But then $\angle BAQ=\angle BA'Q=\angle PAC$ , as required. $\square$

Some example problems:

Problem 1: Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ meet at $E$ . The extensions of the sides $AD$ and $BC$ beyond $A$ and $B$ meet at $F$ . Let $G$ be the point such that $ECGD$ is a parallelogram, and let $H$ be the image of $E$ under reflection in $AD$ . Prove that $D,H,F,G$ are concyclic.

[ISL 2012 G2]

My solution

Problem 2: Let $P$ be a point inside triangle $ABC$ , and suppose lines $AP$ , $BP$ , $CP$ meet the circumcircle again at $T$ , $S$ , $R$ (here $T \neq A$ , $S \neq B$ , $R \neq C$ ). Let $U$ be any point in the interior of $PT$ . A line through $U$ parallel to $AB$ meets $CR$ at $W$ , and the line through $U$ parallel to $AC$ meets $BS$ again at $V$ . Finally, the line through $B$ parallel to $CP$ and the line through $C$ parallel to $BP$ intersect at point $Q$ . Given that $RS$ and $VW$ are parallel, prove that $\angle CAP = \angle BAQ$ .

[Taiwan TST2 2014 Problem 6]

My solution

Problem 3: Let $ABCD$ be a cyclic quadrilateral, and let diagonals $AC$ and $BD$ intersect at $X$ .Let $C_1,D_1$ and $M$ be the midpoints of segments $CX,DX$ and $CD$ , respecctively. Lines $AD_1$ and $BC_1$ intersect at $Y$ , and line $MY$ intersects diagonals $AC$ and $BD$ at different points $E$ and $F$ , respectively. Prove that line $XY$ is tangent to the circle through $E,F$ and $X$ .

[EGMO 2016 Problem 2]

My solution

This post has been edited 1 time. Last edited by liberator, Apr 15, 2016, 8:21 PM
Reason: added EGMO 2016/2

geometry

isogonality lemma

IMO 2011 Problem 6

by liberator, Jul 21, 2015, 4:57 PM

Problem: Let $ABC$ be an acute triangle with circumcircle $\Gamma$ . Let $\ell$ be a tangent line to $\Gamma$ , and let $\ell_a, \ell_b$ and $\ell_c$ be the lines obtained by reflecting $\ell$ in the lines $BC$ , $CA$ and $AB$ , respectively. Show that the circumcircle of the triangle determined by the lines $\ell_a, \ell_b$ and $\ell_c$ is tangent to the circle $\Gamma$ .

Proposed by Japan

My solution

This post has been edited 2 times. Last edited by liberator, Jul 22, 2015, 1:35 PM

IMO 2015 Problem 3

by liberator, Jul 14, 2015, 2:48 PM

Problem: Let $ABC$ be an acute triangle with $AB > AC$ . Let $\Gamma$ be its cirumcircle, $H$ its orthocenter, and $F$ the foot of the altitude from $A$ . Let $M$ be the midpoint of $BC$ . Let $Q$ be the point on $\Gamma$ such that $\angle HQA = 90^{\circ}$ and let $K$ be the point on $\Gamma$ such that $\angle HKQ = 90^{\circ}$ . Assume that the points $A$ , $B$ , $C$ , $K$ and $Q$ are all different and lie on $\Gamma$ in this order.

Prove that the circumcircles of triangles $KQH$ and $FKM$ are tangent to each other.

Proposed by Ukraine

My solution

This post has been edited 2 times. Last edited by liberator, Jul 14, 2015, 8:58 PM

Coaxal circles in incenter/excenter configuration

by liberator, Apr 15, 2015, 7:18 PM

Problem: Let $ABC$ be a triangle, whose excircle (opposite $A$ ) touches $BC,CA,AB$ at $P,Q,R$ respectively. Denote $D$ as the intersection of the lines $PQ$ and $AB$ , and $E$ as the intersection of the lines $RP$ and $CA$ . If $I_a$ is the excenter of $\triangle ABC$ , opposite $A$ , prove that the circumcircles of triangles $PQE, PRD$ and $PI_aA$ are coaxal.

Commentary: We may replace "excircle" with "incircle", "excenter" with "incenter", and the result still holds.

See interactive diagram here.

My solution

Attachments:

geometry

Reim s theorem

power of a point

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