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 1 (Trig)== | + | |
+ | ==Solution 1 (Simplicity) == | ||
+ | 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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+ | We see this by <math>\angle LKM = \angle OKL</math> and <math>\angle KLO = \angle KML</math>. | ||
+ | The latter equality can be seen if we extend <math>LP</math> to point <math>L'</math> on <math>(KLMN)</math>. We know <math>LK = KL'</math> from which it follows <math>\angle KLO = \angle KML</math>. | ||
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+ | Let <math>MO = x</math>. By <math>\triangle KML \sim \triangle KLO</math> we have | ||
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+ | <cmath>\frac{KL}{KO} = \frac{KM}{KL} \Rightarrow \frac{28}{8} = \frac{x+8}{28}.</cmath> | ||
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+ | <cmath>98 = x + 8 \Rightarrow x = \boxed{090}.</cmath> | ||
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+ | - gregwwl | ||
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+ | ==Solution 2 (Trig)== | ||
Let <math>\angle MKN=\alpha</math> and <math>\angle LNK=\beta</math>. Note <math>\angle KLP=\beta</math>. | Let <math>\angle MKN=\alpha</math> and <math>\angle LNK=\beta</math>. Note <math>\angle KLP=\beta</math>. | ||
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Thus, <math>MK=\frac{MN}{\tan\alpha}=98</math>, so <math>MO=MK-KO=\boxed{090}</math>. | Thus, <math>MK=\frac{MN}{\tan\alpha}=98</math>, so <math>MO=MK-KO=\boxed{090}</math>. | ||
− | ==Solution | + | ==Solution 3 (Similar triangles)== |
<asy> | <asy> | ||
size(250); | size(250); | ||
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Solution by vedadehhc | Solution by vedadehhc | ||
− | ==Solution | + | ==Solution 4 (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>). | 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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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) | 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) | ||
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==Solution 5 (5-second PoP)== | ==Solution 5 (5-second PoP)== |
Revision as of 19:12, 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 (Simplicity)
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
- 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.