Difference between revisions of "2019 AIME I Problems/Problem 6"
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− | ==Solution 1 (Trig)== | + | ==Solution 1 (Simple) == |
+ | 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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+ | Note: This solution does not use the condition <math>MN=65</math>. | ||
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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) | ||
− | ==Solution | + | ==Solution 5(Algebraic Bashing)== |
First, let <math>P</math> be the intersection of <math>LO</math> and <math>KN</math>. We can use the right triangles in the problem to create equations. Let <math>a=NP, b=PK, c=NO, d=OM, e=OP, f=PC,</math> and <math>g=NC.</math> We are trying to find <math>d.</math> We can find <math>7</math> equations. They are | First, let <math>P</math> be the intersection of <math>LO</math> and <math>KN</math>. We can use the right triangles in the problem to create equations. Let <math>a=NP, b=PK, c=NO, d=OM, e=OP, f=PC,</math> and <math>g=NC.</math> We are trying to find <math>d.</math> We can find <math>7</math> equations. They are | ||
<cmath>4225+d^2=c^2,</cmath> | <cmath>4225+d^2=c^2,</cmath> | ||
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(Solution by DottedCaculator) | (Solution by DottedCaculator) | ||
− | ==Solution | + | ==Solution 6(5-second PoP)== |
<asy> | <asy> | ||
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(Solution by TheUltimate123) | (Solution by TheUltimate123) | ||
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==Video Solution== | ==Video Solution== | ||
Video Solution: | Video Solution: |
Revision as of 20:11, 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 (Simple)
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(Algebraic Bashing)
First, let be the intersection of
and
. We can use the right triangles in the problem to create equations. Let
and
We are trying to find
We can find
equations. They are
and
We can subtract the fifth equation from the sixth equation to get
We can subtract the fourth equation from the third equation to get
Combining these equations gives
so
Substituting this into the seventh equation gives
Substituting this into the second equation gives
. Subtracting the first equation from this gives
Solving this equation, we find that
(Solution by DottedCaculator)
Solution 6(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.