Difference between revisions of "2019 AIME I Problems/Problem 6"
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Note that <math>KLMN</math> is cyclic with diameter <math>KN</math> since <math>\angle KLN = \angle KMN = \frac{\pi}{2}</math>. Also, note that we have <math>\triangle KML \sim \triangle KLO</math> by SS similarity. | Note that <math>KLMN</math> is cyclic with diameter <math>KN</math> since <math>\angle KLN = \angle KMN = \frac{\pi}{2}</math>. Also, note that we have <math>\triangle KML \sim \triangle KLO</math> by SS similarity. | ||
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<cmath>98 = x + 8 \Rightarrow x = \boxed{090}.</cmath> | <cmath>98 = x + 8 \Rightarrow x = \boxed{090}.</cmath> | ||
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+ | Note: This solution does not use the condition <math>MN=65</math>. | ||
- gregwwl | - gregwwl |
Revision as of 19:14, 15 March 2019
Contents
[hide]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 1
Note that is cyclic with diameter
since
. Also, note that we have
by SS similarity.
We see this by and
.
The latter equality can be seen if we extend
to point
on
. We know
from which it follows
.
Let . By
we have
Note: This solution does not use the condition .
- gregwwl
Solution 2 (Trig)
Let and
. Note
.
Then, .
Furthermore,
.
Dividing the equations gives
Thus, , so
.
Solution 3 (Similar triangles)
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 4 (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)
Solution 5 (5-second PoP)
Notice that
is inscribed in the circle with diameter
and
is inscribed in the circle with diameter
. Furthermore,
is tangent to
. Then,
and
.
(Solution by TheUltimate123)
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.