2019 AMC 10B Problems/Problem 16
Problem
In with a right angle at
, point
lies in the interior of
and point
lies in the interior of
so that
and the ratio
. What is the ratio
Solution
Without loss of generality, let and
. Let
and
. As
and
are isosceles,
and
. Then
, so
is a 3-4-5 triangle with
.
Then , and
is a 1-2-
triangle.
On isosceles triangles and
, drop altitudes from
and
onto
; denote the feet of these altitudes by
and
respectively. Then
by AAA similarity, so we get that
, and
. Similarly we get
, and
.
Solution 2
, and
. (For this solution, A is above C, and B is to the right of C). Denote the angle of point A as "t". Then
is
degrees, which implies that
is
degrees. Similarly, the angle of point B is
degrees, which implies that
is
degrees. This further implies that
is
degrees.
This may seem strange, but if you draw the diagram, the solution will work itself out like this.
Now we see that . Thus triangle CDE is a right triangle, with side lengths of 3x, 4x, and by the pythaogrean theorem, 5x. Now we see that AC is 4x (by definition), BC is 5x+3x = 8x, and AB is
x. Now, we find the cosine of 2y - this is
. which is
Using law of cosines on triangle BED, and denoting the length of BD as "d", we get
Since this is DB, and we know AB, to find the ratio we find AD, which is
, which is
. Thus the answer is
~IronicNinja
Solution 3
Draw a nice big diagram and measure. Only use as last resort.
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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All AMC 10 Problems and Solutions |
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