Difference between revisions of "2019 AIME I Problems/Problem 6"
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In convex quadrilateral <math>KLMN</math> side <math>\overline{MN}</math> is perpendicular to diagonal <math>\overline{KM}</math>, side <math>\overline{KL}</math> is perpendicular to diagonal <math>\overline{LN}</math>, <math>MN = 65</math>, and <math>KL = 28</math>. The line through <math>L</math> perpendicular to side <math>\overline{KN}</math> intersects diagonal <math>\overline{KM}</math> at <math>O</math> with <math>KO = 8</math>. Find <math>MO</math>. | In convex quadrilateral <math>KLMN</math> side <math>\overline{MN}</math> is perpendicular to diagonal <math>\overline{KM}</math>, side <math>\overline{KL}</math> is perpendicular to diagonal <math>\overline{LN}</math>, <math>MN = 65</math>, and <math>KL = 28</math>. The line through <math>L</math> perpendicular to side <math>\overline{KN}</math> intersects diagonal <math>\overline{KM}</math> at <math>O</math> with <math>KO = 8</math>. Find <math>MO</math>. | ||
| − | ==Solution== | + | ==Solution (Similar triangles)== |
| + | (writing this, don't edit) | ||
| + | <asy> | ||
| + | size(100); | ||
| + | draw((0,0)--(1,0)); | ||
| + | </asy> | ||
| + | |||
| + | ==Video Solution== | ||
Video Solution: | Video Solution: | ||
https://www.youtube.com/watch?v=0AXF-5SsLc8 | https://www.youtube.com/watch?v=0AXF-5SsLc8 | ||
Revision as of 07:23, 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)
![[asy] size(100); draw((0,0)--(1,0)); [/asy]](http://latex.artofproblemsolving.com/5/c/a/5cab5fb8c3bb6b434fa9fbf91d4253bb3b0e695f.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
