Difference between revisions of "2019 AIME I Problems/Problem 6"
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First, let <math>P</math> be the intersection of <math>LO</math> and <math>KN</math>. Note that <math>m\angle KPL = 90^{\circ}</math> as given in the problem. Since <math>\angle KPL \cong \angle KLN</math> and <math>\angle PKL \cong \angle LKN</math>, <math>\triangle PKL \sim \triangle LKN</math> by AA similarity. Similarly, <math>\triangle KMN \sim \triangle KPO</math>. | First, let <math>P</math> be the intersection of <math>LO</math> and <math>KN</math>. Note that <math>m\angle KPL = 90^{\circ}</math> as given in the problem. Since <math>\angle KPL \cong \angle KLN</math> and <math>\angle PKL \cong \angle LKN</math>, <math>\triangle PKL \sim \triangle LKN</math> by AA similarity. Similarly, <math>\triangle KMN \sim \triangle KPO</math>. | ||
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+ | ==Solution 2 (Similar triangles, orthocenters)== | ||
+ | Extend <math>KL</math> and <math>NM</math> past <math>L</math> and <math>M</math> respectively to meet at <math>P</math>. Let <math>H</math> be the intersection of diagonals <math>KM</math> and <math>LN</math> (this is the orthocenter of <math>\triangle KNP</math>). | ||
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+ | As <math>\triangle KOL \sim \triangle KHP</math> (as <math>LO \parallel PH</math>, using the fact that <math>H</math> is the orthocenter), we may let <math>OH = 8k</math> and <math>LP = 28k</math>. | ||
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+ | Then using similarity with triangles <math>\triangle KLH</math> and <math>\triangle KMP</math> we have | ||
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+ | <cmath>\frac{28}{8+8k} = \frac{8+8k+HM}{28+28k}</cmath> | ||
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+ | Cross-multiplying and dividing by <math>4+4k</math> gives <math>2(8+8k+HM) = 28 \cdot 7 = 196</math> so <math>MO = 8k + HM = \frac{196}{2} - 8 = \boxed{090}</math>. (Solution by scrabbler94) | ||
==Video Solution== | ==Video Solution== |
Revision as of 13:57, 15 March 2019
Contents
Problem 6
In convex quadrilateral side is perpendicular to diagonal , side is perpendicular to diagonal , , and . The line through perpendicular to side intersects diagonal at with . Find .
Solution (Similar triangles)
(writing this, don't edit)
First, let be the intersection of and . Note that as given in the problem. Since and , by AA similarity. Similarly, .
Solution 2 (Similar triangles, orthocenters)
Extend and past and respectively to meet at . Let be the intersection of diagonals and (this is the orthocenter of ).
As (as , using the fact that is the orthocenter), we may let and .
Then using similarity with triangles and we have
Cross-multiplying and dividing by gives so . (Solution by scrabbler94)
Video Solution
Video Solution: https://www.youtube.com/watch?v=0AXF-5SsLc8
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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