Difference between revisions of "2020 AMC 12A Problems/Problem 24"
Awesome1st (talk | contribs) (→Solution) |
Awesome1st (talk | contribs) |
||
Line 26: | Line 26: | ||
draw((3.5,3.5)--(6,4.75)); | draw((3.5,3.5)--(6,4.75)); | ||
label("1", (3.5,3.5)--(6,4.75), SE); | label("1", (3.5,3.5)--(6,4.75), SE); | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
</asy> | </asy> |
Revision as of 12:29, 3 February 2020
Contents
[hide]Problem 24
Suppose that is an equilateral triangle of side length
, with the property that there is a unique point
inside the triangle such that
,
, and
. What is
?
Solution
We begin by rotating by
about
, such that in
,
. We see that
is equilateral with side length
, meaning that
. We also see that
is a
right triangle, meaning that
. Thus, by adding the two together, we see that
. We can now use the law of cosines as following:
giving us that . ~ciceronii
Video Solution
https://www.youtube.com/watch?v=mUW4zcrRL54
See Also
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.