Difference between revisions of "2019 AIME I Problems/Problem 6"
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+ | If you're wondering why <math>KX\cdot KN=KL^2,</math> it's because <math>KX\cdotKN=KX\cdot(KX+XN)=KX^2+KX\cdotKN=KX^2+LX^2=KL^2</math> (last part by similarity). | ||
==Solution 6 (Alternative PoP)== | ==Solution 6 (Alternative PoP)== |
Revision as of 04:20, 27 July 2020
Contents
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 (Trig)
Let and
. Note
.
Then, .
Furthermore,
.
Dividing the equations gives
Thus, , so
.
Solution 2 (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 3 (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 4 (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 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)
If you're wondering why it's because $KX\cdotKN=KX\cdot(KX+XN)=KX^2+KX\cdotKN=KX^2+LX^2=KL^2$ (Error compiling LaTeX. Unknown error_msg) (last part by similarity).
Solution 6 (Alternative PoP)
(Diagram by vedadehhc)
Call the base of the altitude from to
point
. Let
. Now, we have that
by the Pythagorean Theorem. Once again by Pythagorean,
. Using Power of a Point, we have
(
is the intersection of
with the circle
)
.
(Solution by RootThreeOverTwo)
Solution 7 (just one pair of similar triangles)
Note that since
, quadrilateral
is cyclic. Therefore, we have
so
, giving
Therefore,
.
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.