Difference between revisions of "2019 AMC 10B Problems/Problem 16"
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<math>\textbf{(A) }2:3\qquad\textbf{(B) }2:\sqrt{5}\qquad\textbf{(C) }1:1\qquad\textbf{(D) }3:\sqrt{5}\qquad\textbf{(E) }3:2</math> | <math>\textbf{(A) }2:3\qquad\textbf{(B) }2:\sqrt{5}\qquad\textbf{(C) }1:1\qquad\textbf{(D) }3:\sqrt{5}\qquad\textbf{(E) }3:2</math> | ||
+ | |||
+ | ==Diagram== | ||
+ | <asy> | ||
+ | draw((0,0)--(0,8),black); | ||
+ | dot((0,0)); | ||
+ | dot((0,8)); | ||
+ | draw((0,0)--(4,0),black); | ||
+ | dot((4,0)); | ||
+ | draw((0,8)--(4,0),black); | ||
+ | draw((0,0)--(2.5,3),black); | ||
+ | dot((2.5,3)); | ||
+ | draw((0,5)--(2.5,3),black); | ||
+ | dot((0,5)); | ||
+ | </asy> | ||
+ | ~ By Little Mouse | ||
==Solution 1== | ==Solution 1== |
Revision as of 21:30, 21 October 2021
Contents
[hide]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
Diagram
~ By Little Mouse
Solution 1
Without loss of generality, let and
. Let
and
. As
and
are isosceles,
and
. Then
, so
is a
triangle with
.
Then , and
is a
triangle.
In 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
.
Alternatively, once finding the length of one could use the Pythagorean Theorem to find
and consequently
, and then compute the ratio.
Solution 2
Let , and
. (For this solution,
is above
, and
is to the right of
). Also let
, so
, which implies
. Similarly,
, which implies
. This further implies that
.
Now we see that . Thus
is a right triangle, with side lengths of
,
, and
(by the Pythagorean Theorem, or simply the Pythagorean triple
). Therefore
(by definition),
, and
. Hence
(by the double angle formula), giving
.
By the Law of Cosines in , if
, we have
Now
. Thus the answer is
.
Solution 3
WLOG, let , and
.
. Because of this,
is a 3-4-5 right triangle. Draw the altitude
of
.
is
by the base-height triangle area formula.
is similar to
(AA). So
.
is
of
. Therefore,
is
.
~Thegreatboy90
Video Solution 1
~IceMatrix
Video Solution 2
https://youtu.be/4_x1sgcQCp4?t=4245
~ pi_is_3.14
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.