Difference between revisions of "2024 AMC 12B Problems/Problem 21"
m (Minor grammar edits) |
(Added Solution 7 to 2024 AMC 12B Problem 21.) |
||
(One intermediate revision by one other user not shown) | |||
Line 82: | Line 82: | ||
==Solution 6 (Law of Cosines)== | ==Solution 6 (Law of Cosines)== | ||
− | We can start by scaling the <math>3-4-5</math> right triangle up and "gluing" it to the <math>5-12-13</math> triangle's longer leg. Let <math>\alpha</math>, <math>\beta</math>, and <math>\theta</math> be the smallest angles in the <math>3-4-5</math>, <math>5-12-13</math> and unknown triangle respectively. We can construct the following diagram of the two known triangles. | + | We can start by scaling the <math>3-4-5</math> right triangle up by a factor of <math>3</math> and "gluing" it to the <math>5-12-13</math> triangle's longer leg. Let <math>\alpha</math>, <math>\beta</math>, and <math>\theta</math> be the smallest angles in the <math>3-4-5</math>, <math>5-12-13</math> and unknown triangle respectively. We can construct the following diagram of the two known triangles. |
<asy> | <asy> | ||
pair A = (0,0); | pair A = (0,0); | ||
Line 125: | Line 125: | ||
~Phinetium (again) | ~Phinetium (again) | ||
+ | |||
+ | ==Solution 7 (No Trig, Coord Bash)== | ||
+ | [[File:AMC 12B 2024 Problem 21.png|None]] | ||
+ | |||
+ | Set up a coordinate system. Let <math>(0, 0)</math>, <math>(4, 0)</math>, and <math>(0, 3)</math> be the vertices of the base <math>3-4-5</math> right triangle. In this case, the three smallest angles will all be at <math>(4, 0)</math>, and one of the coordinates of the unknown triangle has to lie on the line <math>x = 4</math>. Now, scale down the <math>5-12-13</math> triangle by <math>\frac{5}{12}</math> so that the new sides are <math>\frac{25}{12}-5-\frac{65}{12}</math>, and place the side with length 5 at the coordinates <math>(0, 3)</math> and <math>(4, 0)</math>. The line passing through these two points can be written as <math>y = -\frac{3}{4}x + 3</math>, so the perpendicular of this line at <math>(0, 3)</math> can be written as <math>y = \frac{4}{3}x + 3</math>. Since the length of the other side is <math>\frac{25}{12}</math>, after drawing smaller <math>3-4-5</math> right triangles, we find that the third coordinate of the <math>\frac{25}{12}-5-\frac{65}{12}</math> is at <math>(\frac{5}{4}, \frac{14}{3})</math>. This coordinate will be one of the coordinates for our unknown triangle. We can place the other coordinate of the unknown triangle at <math>(4, \frac{14}{3})</math> and the third is, by definition, at <math>(4, 0)</math>. The distance from <math>(\frac{5}{4}, \frac{14}{3})</math> to <math>(4, \frac{14}{3})</math> is <math>\frac{11}{4}</math>, and the distance from <math>(4, \frac{14}{3})</math> to <math>(4, 0)</math> is <math>\frac{14}{3}</math>, and from before, the distance from <math>(\frac{5}{4}, \frac{14}{3})</math> to <math>(4, 0)</math> is <math>\frac{65}{12}</math>. Scaling up the sides so that they are integers, we see that the side lengths make a <math>33-56-65</math> right triangle. The perimeter is then <math>33+56+65=\boxed{\textbf{(C)\ 154}}</math>. | ||
+ | |||
+ | ~mathwizard123123 | ||
==Video Solution by Innovative Minds== | ==Video Solution by Innovative Minds== |
Latest revision as of 13:39, 16 July 2025
Contents
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)
Solution 6 (Law of Cosines)
We can start by scaling the right triangle up by a factor of
and "gluing" it to the
triangle's longer leg. Let
,
, and
be the smallest angles in the
,
and unknown triangle respectively. We can construct the following diagram of the two known triangles.
We can also construct a diagram for the third, unknown triangle like so. We know that , and therefore that
. We also know that the other acute angle in this third triangle will have a measure of
by the triangle angle sum theorem.
We can use the law of cosines on the triangle in the first diagram to get the equation
. Isolating
, we get
, which further simplifies to
. Since the third triangle has to be a primitive Pythagorean triple, we must take this trig ratio into its most simple form, namely
. Using this information in our second diagram, we know that the smaller, adjacent leg must have length
, and the hypotenuse must have length
. We can then use the Pythagorean theorem to find the other, unknown leg, which has length
. Adding these three lengths together, we find that the perimeter of this right triangle is
.
~Phinetium
Solution 6.1 (Faster Ending)
Instead of computing to find the second leg in the unknown triangle by hand, we can use process of elimination.
and
are the only answers within the realm of possibility, because
would entitle a triangle with a negative side length, and
and
would require legs greater than the length of the hypotenuse. The answer choice
would force the second leg to have a length of
, which is smaller than the smallest leg in the triangle. (We know that the leg with length
is the smallest leg in the triangle by the side-angle relationship theorem, because it is opposite the smallest angle in the triangle.) Therefore, the only valid answer choice remaining is
.
~Phinetium (again)
Solution 7 (No Trig, Coord Bash)
Set up a coordinate system. Let ,
, and
be the vertices of the base
right triangle. In this case, the three smallest angles will all be at
, and one of the coordinates of the unknown triangle has to lie on the line
. Now, scale down the
triangle by
so that the new sides are
, and place the side with length 5 at the coordinates
and
. The line passing through these two points can be written as
, so the perpendicular of this line at
can be written as
. Since the length of the other side is
, after drawing smaller
right triangles, we find that the third coordinate of the
is at
. This coordinate will be one of the coordinates for our unknown triangle. We can place the other coordinate of the unknown triangle at
and the third is, by definition, at
. The distance from
to
is
, and the distance from
to
is
, and from before, the distance from
to
is
. Scaling up the sides so that they are integers, we see that the side lengths make a
right triangle. The perimeter is then
.
~mathwizard123123
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.