Difference between revisions of "1997 JBMO Problems/Problem 3"

(Solution to Problem 3 (credit to stats11) -- inequality chasing)
 
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== Problem ==
 
== Problem ==
  
''(Greece)'' Let <math>ABC</math> be a triangle and let <math>I</math> be the incenter. Let <math>N</math>, <math>M</math> be the midpoints of the sides <math>AB</math> and <math>CA</math> respectively. The lines <math>BI</math> and <math>CI</math> meet <math>MN</math> at <math>K</math> and <math>L</math> respectively. Prove that <math>AI+BI+CI>BC+KL</math>.
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Let <math>ABC</math> be a triangle and let <math>I</math> be the incenter. Let <math>N</math>, <math>M</math> be the midpoints of the sides <math>AB</math> and <math>CA</math> respectively. The lines <math>BI</math> and <math>CI</math> meet <math>MN</math> at <math>K</math> and <math>L</math> respectively. Prove that <math>AI+BI+CI>BC+KL</math>.
  
 
== Solution ==
 
== Solution ==
  
== See also ==
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<asy>
 +
size(9.22 cm);
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pair B=(0,0), A=(50,120), C=(140,0), N=(25,60), M=(95,60), I=(60,40), K=(90,60), L=(20,60);
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draw(B--A--C--B);
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draw(circle(I,40));
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dot(A);
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label("A",(50,125));
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dot(B);
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label("B",B,SW);
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dot(C);
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label("C",C,SE);
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dot(N);
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label("N",(24,65));
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dot(M);
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label("M",M,NE);
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dot(I);
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label("I",I,S);
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dot(K);
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label("K",K,NW);
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dot(L);
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label("L",L,NW);
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draw(L--C,dotted);
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draw(K--B,dotted);
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draw(L--M);
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draw(anglemark(C,B,A,200));
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draw(anglemark(N,K,I,200));
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draw(anglemark(A,C,B,200));
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draw(anglemark(A,C,B,150));
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draw(anglemark(C,L,K,400));
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draw(anglemark(C,L,K,450));
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 +
</asy>
 +
 
 +
First, by SAS Similarity, <math>\triangle ANM \sim \triangle ABC,</math> so <math>NM \parallel BC</math> and <math>MN = \tfrac12 BC.</math>  That means <math>\angle IBC = \angle IKN,</math> and since <math>\angle IBN = \angle IBC,</math> <math>\triangle NBK</math> is an [[isosceles triangle]].  Similarly, <math>\angle MLC = \angle LCB = \angle LCM,</math> making <math>\triangle MLC</math> an isosceles as well.  Thus, <math>ML = MC</math> and <math>NB = NK.</math>
 +
 
 +
<br>
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By the [[Triangle Inequality]], <math>AI + IB > AB,</math> and <math>AI + IC > AC</math>, and <math>BI + IC > BC.</math>  That means
 +
<cmath>\begin{align*}
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2(AI + IB + IC) &> AB + AC + BC \\
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AI + IB + IC &> NK + ML + MN \\
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&> NK + (MN + LN) + MN \\
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&> KL + 2 MN \\
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&> KL + BC.
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\end{align*}</cmath>
 +
 
 +
== See Also ==
  
 
{{JBMO box|year=1997|num-b=2|num-a=4}}
 
{{JBMO box|year=1997|num-b=2|num-a=4}}
  
[[Category:Intermediate Geometry Problems]]
+
[[Category:Olympiad Geometry Problems]]

Latest revision as of 19:14, 7 August 2018

Problem

Let $ABC$ be a triangle and let $I$ be the incenter. Let $N$, $M$ be the midpoints of the sides $AB$ and $CA$ respectively. The lines $BI$ and $CI$ meet $MN$ at $K$ and $L$ respectively. Prove that $AI+BI+CI>BC+KL$.

Solution

[asy] size(9.22 cm);  pair B=(0,0), A=(50,120), C=(140,0), N=(25,60), M=(95,60), I=(60,40), K=(90,60), L=(20,60); draw(B--A--C--B); draw(circle(I,40));  dot(A); label("A",(50,125)); dot(B); label("B",B,SW); dot(C); label("C",C,SE); dot(N); label("N",(24,65)); dot(M); label("M",M,NE); dot(I); label("I",I,S); dot(K); label("K",K,NW); dot(L); label("L",L,NW); draw(L--C,dotted); draw(K--B,dotted); draw(L--M);  draw(anglemark(C,B,A,200)); draw(anglemark(N,K,I,200)); draw(anglemark(A,C,B,200)); draw(anglemark(A,C,B,150)); draw(anglemark(C,L,K,400)); draw(anglemark(C,L,K,450));  [/asy]

First, by SAS Similarity, $\triangle ANM \sim \triangle ABC,$ so $NM \parallel BC$ and $MN = \tfrac12 BC.$ That means $\angle IBC = \angle IKN,$ and since $\angle IBN = \angle IBC,$ $\triangle NBK$ is an isosceles triangle. Similarly, $\angle MLC = \angle LCB = \angle LCM,$ making $\triangle MLC$ an isosceles as well. Thus, $ML = MC$ and $NB = NK.$


By the Triangle Inequality, $AI + IB > AB,$ and $AI + IC > AC$, and $BI + IC > BC.$ That means \begin{align*} 2(AI + IB + IC) &> AB + AC + BC \\ AI + IB + IC &> NK + ML + MN \\ &> NK + (MN + LN) + MN \\ &> KL + 2 MN \\ &> KL + BC. \end{align*}

See Also

1997 JBMO (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5
All JBMO Problems and Solutions
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