Difference between revisions of "2019 AIME I Problems/Problem 6"
(→Solution (Similar triangles)) |
(→Solution (Similar triangles)) |
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Line 17: | Line 17: | ||
draw(K--M--N--cycle); | draw(K--M--N--cycle); | ||
draw(L--M); | draw(L--M); | ||
− | label("K", K, | + | label("K", K, SW); |
label("L", L, NW); | label("L", L, NW); | ||
label("M", M, NE); | label("M", M, NE); | ||
− | label("N", N, | + | label("N", N, SE); |
draw(L--P); | draw(L--P); | ||
label("P", P, S); | label("P", P, S); | ||
Line 29: | Line 29: | ||
</asy> | </asy> | ||
− | 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. | + | 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>. |
==Video Solution== | ==Video Solution== |
Revision as of 10:54, 15 March 2019
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, .
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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