Difference between revisions of "2011 AIME I Problems/Problem 4"
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(→Solution: Nitpicking. Also previous post was done as it is clear that the assertion of OQ being a midline and MQC, NOC being isosceles was not addressed as some have asked why the facts were true) |
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Since <math>{OQ}</math> is a midline, it then follows that <math>{OC} = 58.5</math> and <math>{QC} = 60</math>. Since <math>\triangle MQC</math> and <math>\triangle NOC</math> are both isosceles, we have that <math>ON = OC = 58.5</math> and <math>MQ = QC = 60</math>. Since <math>OQ</math> is a midline, <math>OQ = 62.5</math>. We want to find <math>MN</math>, which is just <math>ON + MQ - OQ</math>. | Since <math>{OQ}</math> is a midline, it then follows that <math>{OC} = 58.5</math> and <math>{QC} = 60</math>. Since <math>\triangle MQC</math> and <math>\triangle NOC</math> are both isosceles, we have that <math>ON = OC = 58.5</math> and <math>MQ = QC = 60</math>. Since <math>OQ</math> is a midline, <math>OQ = 62.5</math>. We want to find <math>MN</math>, which is just <math>ON + MQ - OQ</math>. | ||
− | Substituting, the answer is <math>58.5 + 60 - 62.5 = \boxed {56}</math>. | + | Substituting the values of <math>ON, MQ, OQ</math>, we have that the answer is <math>58.5 + 60 - 62.5 = \boxed {56}</math>. |
== See also == | == See also == | ||
{{AIME box|year=2011|n=I|num-b=3|num-a=5}} | {{AIME box|year=2011|n=I|num-b=3|num-a=5}} |
Revision as of 17:38, 14 March 2012
Problem 4
In triangle , , and . The angle bisector of angle intersects at point , and the angle bisector of angle intersects at point . Let and be the feet of the perpendiculars from to and , respectively. Find .
Solution
Extend such that it intersects lines and at points and , respectively.
Lemma 1: are midpoints of and
Proof: Consider the reflection of the vertex over the line , and let this point be . Since , we have that is the image of after reflection over , and from the definition of reflection . Then it is easily seen that since is an angle bisector, that , so lies on . Similarly, if we define to be the reflection of over , then we find that lies on . Then we can now see that , with a homothety of ratio taking the first triangle to the second. Then this same homothety takes everything on the line to everything on the line . So since lie on , this homothety also takes to so they are midpoints, as desired.
Lemma 2: are isosceles triangles
Proof: To show that is isosceles, note that , with similarity ratio of . So it suffices to show that triangle is isosceles. But this follows quickly from Lemma 1, since is both an altitude and an angle bisector of . is isosceles by the same reasoning.
Since is a midline, it then follows that and . Since and are both isosceles, we have that and . Since is a midline, . We want to find , which is just .
Substituting the values of , we have that the answer is .
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |