2024 AMC 12B Problems/Problem 21
Contents
[hide]Problem
The measures of the smallest angles of three different right triangles sum to . All three triangles have side lengths that are primitive Pythagorean triples. Two of them are
and
. What is the perimeter of the third triangle?
Solution 1
Let and
be the smallest angles of the
and
triangles respectively. We have
Then
Let
be the smallest angle of the third triangle. Consider
In order for this to be undefined, we need
so
Hence the base side lengths of the third triangle are
and
. By the Pythagorean Theorem, the hypotenuse of the third triangle is
, so the perimeter is
.
Solution 2 (Complex Number)
The smallest angle of triangle can be viewed as the arguement of
, and the smallest angle of
triangle can be viewed as the arguement of
.
Hence, if we assume the ratio of the two shortest length of the last triangle is (
being some rational number), then we can derive the following formula of the sum of their arguement.
Since their arguement adds up to
, it's the arguement of
. Hence,
where
is some real number.
Solving the equation, we get Hence
Since the sidelength of the theird triangle are co-prime integers, two of its sides are and
. And the last side is
, hence, the parameter of the third triangle if
.
~Prof. Joker
Solution 3 (Another Trig)
Denote the smallest angle of the triangle as
, the smallest angle of the
triangle as
, and the smallest angle of the triangle we are trying to solve for as
. We then have
Taking the hypotenuse to be
and one of the legs to be
, we compute the last leg to be
Giving us a final answer of .
~tkl
Solution 3.1 (Different Flavor of the same thing)
Consider using the cosine addition identity. Instead of using the Pythagorean theorem, we can use Euclid's formula since we're dealing with primitive triples.
Combining that, we get and
. Using this, we can get that the other leg must be
. We add the lengths and get that the perimeter is
.
~ sxbuto
Solution 4 (Similarity)
Let's arrange the triangles and
as shown in the diagram.
vladimir.shelomovskii@gmail.com, vvsss
Solution 5 (Complex)
Suppose the triangle has legs . We want
This is equivalent to
Since the argument of this complex number is
its real part must be
. Matching real and imaginary parts yields
or
. The smallest pair
that works is
which yields a hypotenuse of
The perimeter of this triangle is
-Benedict T (countmath1)
Video Solution by Innovative Minds
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=cyiF8_5fEsM
See also
2024 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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 |
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