Difference between revisions of "2011 AIME I Problems/Problem 4"
(→Problem 4) |
(→Problem 4) |
||
Line 1: | Line 1: | ||
== Problem 4 == | == Problem 4 == | ||
In triangle <math>ABC</math>, <math>AB=125</math>, <math>AC=117</math> and <math>BC=120</math>. The angle bisector of angle <math>A</math> intersects <math> \overline{BC} </math> at point <math>L</math>, and the angle bisector of angle <math>B</math> intersects <math> \overline{AC} </math> at point <math>K</math>. Let <math>M</math> and <math>N</math> be the feet of the perpendiculars from <math>C</math> to <math> \overline{BK}</math> and <math> \overline{AL}</math>, respectively. Find <math>MN</math>. | In triangle <math>ABC</math>, <math>AB=125</math>, <math>AC=117</math> and <math>BC=120</math>. The angle bisector of angle <math>A</math> intersects <math> \overline{BC} </math> at point <math>L</math>, and the angle bisector of angle <math>B</math> intersects <math> \overline{AC} </math> at point <math>K</math>. Let <math>M</math> and <math>N</math> be the feet of the perpendiculars from <math>C</math> to <math> \overline{BK}</math> and <math> \overline{AL}</math>, respectively. Find <math>MN</math>. | ||
+ | |||
+ | |||
+ | == Solution == | ||
+ | |||
+ | 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. Notice that <math>{OQ}</math> is a midline; it then follows that <math>{OC} = 58.5</math> and <math>{QC} = 60</math>. Now notice that <math>\triangle MQC</math> and <math>\triangle NOC</math> are both isosceles. Thus, <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>. | ||
== 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 14:22, 6 April 2011
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. Notice that is a midline; it then follows that and . Now notice that and are both isosceles. Thus, 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 |