Difference between revisions of "2011 AIME I Problems/Problem 4"
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+ | Extend <math>{MN}</math> such that it intersects lines <math>{AC}</math> and <math>{BC}</math> at points <math>O</math> and <math>Q</math>, respectively. | ||
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+ | '''Lemma 1: <math>O, Q</math> are midpoints of <math>AC</math> and <math>BC</math>''' | ||
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+ | '''Proof:''' Consider the reflection of the vertex <math>C</math> over the line <math>BM</math>, and let this point be <math>C_1</math>. Since <math>\angle{BMC} = 90^{\circ}</math>, we have that <math>C_1</math> is the image of <math>C</math> after reflection over <math>M</math>, and from the definition of reflection <math>\angle{MBC} = \angle{MBC_1}</math>. Then it is easily seen that since <math>BM</math> is an angle bisector, that <math>\angle{MBC_1} = \angle{MBA}</math>, so <math>C_1</math> lies on <math>AB</math>. Similarly, if we define <math>C_2</math> to be the reflection of <math>C</math> over <math>N</math>, then we find that <math>C_2</math> lies on <math>AB</math>. Then we can now see that <math>\triangle{CMN} \sim \triangle{CC_1C_2}</math>, with a homothety of ratio <math>2</math> taking the first triangle to the second. Then this same homothety takes everything on the line <math>MN</math> to everything on the line <math>AB</math>. So since <math>O, Q</math> lie on <math>MN</math>, this homothety also takes <math>O, Q</math> to <math>A, B</math> so they are midpoints, as desired. <math>\Box</math> | ||
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+ | '''Lemma 2: <math>\triangle{MQC}, \triangle{NOC}</math> are isosceles triangles''' | ||
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+ | '''Proof:''' To show that <math>\triangle{MQC}</math> is isosceles, note that <math>\triangle{MQC} \sim \triangle{C_1BC}</math>, with similarity ratio of <math>\frac{1}{2}</math>. So it suffices to show that triangle <math>\triangle{C_1BC}</math> is isosceles. But this follows quickly from Lemma 1, since <math>BM</math> is both an altitude and an angle bisector of <math>\angle{C_1BC}</math>. <math>\triangle{NOC}</math> is isosceles by the same reasoning. <math>\Box</math> | ||
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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>. | ||
Substituting, the answer is <math>58.5 + 60 - 62.5 = \boxed {56}</math>. | Substituting, the answer is <math>58.5 + 60 - 62.5 = \boxed {56}</math>. |
Revision as of 18:36, 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 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 |