Difference between revisions of "2019 AMC 12A Problems/Problem 19"
Soyamyboya (talk | contribs) (→Solution 2) |
Sevenoptimus (talk | contribs) (Improved clarity and formatting) |
||
Line 8: | Line 8: | ||
==Solution 1== | ==Solution 1== | ||
− | + | Notice that by the Law of Sines, <math>a:b:c = \sin{A}:\sin{B}:\sin{C}</math>, so let's flip all the cosines using <math>\sin^{2}{x} + \cos^{2}{x} = 1</math> (sine is positive for <math>0^{\circ}\le x \le 180^{\circ}</math>, so we're good there). | |
<math>\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\sin C=\frac{\sqrt{15}}{4}</math> | <math>\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\sin C=\frac{\sqrt{15}}{4}</math> |
Revision as of 20:25, 10 February 2019
Contents
Problem
In with integer side lengths,
What is the least possible perimeter for
?
Solution 1
Notice that by the Law of Sines, , so let's flip all the cosines using
(sine is positive for
, so we're good there).
These are in the ratio , so our minimal triangle has side lengths
,
, and
.
is our answer.
-Rowechen Zhong
Solution 2
is obtuse since its cosine is negative, so we let the foot of the altitude from
to
be
. Let
,
,
, and
. By the Pythagorean Theorem,
and
. Thus,
. The sides of the triangle are then
,
, and
, so for some integers
,
and
, where
and
are minimal. Hence,
, or
. Thus smallest possible positive integers
and
that satisfy this are
and
, so
. The sides of the triangle are
,
, and
, so
is our answer.
-soyamyboya
See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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.