https://artofproblemsolving.com/wiki/index.php?title=2013_IMO_Problems/Problem_3&feed=atom&action=history2013 IMO Problems/Problem 3 - Revision history2024-03-28T23:50:42ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2013_IMO_Problems/Problem_3&diff=74600&oldid=prevLiberator: Create page2016-01-17T13:41:56Z<p>Create page</p>
<p><b>New page</b></p><div>==Problem==<br />
Let the excircle of triangle <math>ABC</math> opposite the vertex <math>A</math> be tangent to the side <math>BC</math> at the point <math>A_1</math>. Define the points <math>B_1</math> on <math>CA</math> and <math>C_1</math> on <math>AB</math> analogously, using the excircles opposite <math>B</math> and <math>C</math>, respectively. Suppose that the circumcentre of triangle <math>A_1B_1C_1</math> lies on the circumcircle of triangle <math>ABC</math>. Prove that triangle <math>ABC</math> is right-angled.<br />
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<i>Proposed by Alexander A. Polyansky, Russia</i><br />
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==Solution==<br />
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<asy><br />
unitsize(2.5cm);<br />
void b() {<br />
pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10); pen dd=linetype("4 8");<br />
/* Define the excenter */<br />
pair excenter(pair A=(0,0), pair B=(0,0), pair C=(0,0)) { <br />
return extension(A,bisectorpoint(C,A,B),B,rotate(90,B)*bisectorpoint(A,B,C));<br />
}<br />
/* Draw points */<br />
pair A=D("A",dir(115),N),<br />
B=D("B",(-1,0),W),<br />
C=D("C",(1,0),E),<br />
I=D(incenter(A,B,C)),<br />
Ia=D("I_a",excenter(A,B,C),S),<br />
Ib=D("I_b",excenter(B,C,A),E),<br />
Ic=D("I_c",excenter(C,A,B),W), <br />
A1=D("A_1",foot(Ia,B,C),SSW),<br />
B1=D("B_1",foot(Ib,C,A),E),<br />
C1=D("C_1",foot(Ic,A,B),N),<br />
Ma=D("M_a",midpoint(Ib--Ic),N),<br />
Mb=D("M_b",midpoint(Ic--Ia),SW),<br />
Mc=D("M_c",midpoint(Ia--Ib),SE),<br />
V=D("V",circumcenter(Ia,Ib,Ic),SE);<br />
/* Draw paths */<br />
D(unitcircle,heavyblue);<br />
D(circumcircle(B,C,V),linetype("2 2")+rgb(0.6,0,1));<br />
D(circumcircle(B,C1,V),linetype("2 2")+rgb(0.6,0,1));<br />
D(A--B--C--cycle);<br />
D(Ia--Ib--Ic--cycle,gray(0.3)+linewidth(1));<br />
D(A1--B1--C1--cycle);<br />
D(I--Ia,dd+red); D(I--Ib,dd+red); D(I--Ic,dd+red);<br />
D(V--Ia,dd+heavygreen); D(V--Ib,dd+heavygreen); D(V--Ic,dd+heavygreen);<br />
}<br />
b(); pathflag=false; b();<br />
</asy><br />
Let the excenters opposite <math>A,B,C</math> be <math>I_a,I_b,I_c</math>. Let the midpoint of <math>\overline{I_bI_c}</math> be <math>M_a</math>, which lies on <math>(ABC)</math>, the nine-point circle of <math>\triangle I_aI_bI_c</math>; analogously define <math>M_b,M_c</math>.<br />
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<math>M_aB=M_aC</math> and <math>BC_1=s-a=B_1C</math>, so <math>\triangle M_aBC_1\cong\triangle M_aCB_1</math> (SAS), thus <math>M_a</math> is equidistant from <math>B_1,C_1</math>, with analogous results for <math>M_b,M_c</math>. It follows that the circumcentre of <math>\triangle A_1B_1C_1</math> is one of <math>M_a,M_b,M_c</math>; WLOG, suppose it is <math>M_a</math>.<br />
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By isogonal conjugacy, <math>I_aA_1,I_bB_1I_cC_1</math> concur at the Bevan point <math>V</math> of <math>\triangle ABC</math>. <math>M_aM_b</math> is the common perpendicular bisector of <math>\overline{C_1A_1}</math> and <math>\overline{I_cC}</math>, so <math>C_1A_1\parallel I_cC</math>. <math>(A_1C_1M_b)</math> is the circle on diameter <math>\overline{VB}</math>, so by Reim's theorem, <math>V \in (I_bI_cBC)</math>.<br />
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Hence <math>\angle I_cI_aI_b=\tfrac{1}{2}\angle I_cVI_b=45^{\circ}\implies\angle CAB=180^{\circ}-2\angle I_cI_aI_b=90^{\circ}</math>, as required.<br />
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{{alternate solutions}}<br />
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==See Also==<br />
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{{IMO box|year=2013|num-b=2|num-a=4}}<br />
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[[Category:Olympiad Geometry Problems]]</div>Liberator