Difference between revisions of "1997 JBMO Problems/Problem 3"
Rockmanex3 (talk | contribs) (Solution to Problem 3 (credit to stats11) -- inequality chasing) |
|||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | + | 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 | + | <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, <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> | ||
+ | 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*} | ||
+ | 2(AI + IB + IC) &> AB + AC + BC \\ | ||
+ | AI + IB + IC &> NK + ML + MN \\ | ||
+ | &> NK + (MN + LN) + MN \\ | ||
+ | &> KL + 2 MN \\ | ||
+ | &> KL + BC. | ||
+ | \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: | + | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 19:14, 7 August 2018
Problem
Let be a triangle and let be the incenter. Let , be the midpoints of the sides and respectively. The lines and meet at and respectively. Prove that .
Solution
First, by SAS Similarity, so and That means and since is an isosceles triangle. Similarly, making an isosceles as well. Thus, and
By the Triangle Inequality, and , and That means
See Also
1997 JBMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 | ||
All JBMO Problems and Solutions |