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> as shown above. 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>. Using these similarities we see that |
+ | <cmath>\frac{KP}{KL} = \frac{KL}{KN}</cmath> | ||
+ | <cmath>KP = \frac{KL^2}{KN} = \frac{28^2}{KN} = \frac{784}{KN}</cmath> | ||
+ | and | ||
+ | <cmath>\frac{KP}{KO} = \frac{KM}{KN}</cmath> | ||
+ | <cmath>KP = \frac{KO \cdot KM}{KN} = \frac{8\cdot KM}{KN}</cmath> | ||
+ | Combining the two equations, we get | ||
+ | <cmath>\frac{8\cdot KM}{KN} = \frac{784}{KN}</cmath> | ||
+ | <cmath>8 \cdot KM = 28^2</cmath> | ||
+ | <cmath>KM = 98</cmath> | ||
+ | Since <math>KM = KO + MO</math>, we get <math>MO = 98 -8 = \boxed{090}</math>. | ||
+ | |||
+ | Solution by vedadehhc | ||
==Solution 2 (Similar triangles, orthocenters)== | ==Solution 2 (Similar triangles, orthocenters)== |
Revision as of 19:06, 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
Let and . Note .
Then, . Furthermore, .
Dividing the equations gives
Thus, , so .
Solution (Similar triangles)
(writing this, don't edit)
First, let be the intersection of and as shown above. Note that as given in the problem. Since and , by AA similarity. Similarly, . Using these similarities we see that and Combining the two equations, we get Since , we get .
Solution by vedadehhc
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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