2025 AMC 10A Problems/Problem 3

How many isosceles triangles are there with positive area whose side lengths are all positive integers and whose longest side has length $2025$ ?

$\textbf{(A)}~2025\qquad\textbf{(B)}~2026\qquad\textbf{(C)}~3012\qquad\textbf{(D)}~3037\qquad\textbf{(E)}~4050$

Solution 1: Casework

You can split the problem into two cases:
Case $1$ : The two sides are both smaller than $2025$ , which means that they range from $1013$ to $2024$ . There are $1012$ such cases.
Case $2$ : There are two sides of length $2025$ , so the last side must be in the range $1$ to $2025$ . There are $2025$ such cases. Keep in mind, an equilateral triangle also counts as an isosceles triangle, since it has at least 2 sides.
Therefore, the total number of cases is $1012 + 2025 = \boxed{(D) 3037}$
~cw, minor edit by sd

Side note not by the author: If you're unsure whether the equilateral triangle is a valid case, notice that the other answers are much farther away, as compared to only a one-off error from 3037, which would make 3037 still the best answer.

Video Solution (Fast and Easy)

https://youtu.be/qfvA1uD0--M?si=vhZMvas48oMPyvWx ~ Pi Academy

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

Solution 2 (broken down)

Suppose an isosceles triangle has sides $A, A, B$ where $A$ is the longest side and $A, B$ are positive integers.

By triangle inequalities: A + A > B A+A>B A + B > A ⟹ B > 0 A+B>A⟹B>0

Since $A = 2025$ (the longest side): 2025 + 2025 > B ⟹ B < 4050 2025+2025>B⟹B<4050 But also $B < 2025$ so $B = 1$ to $2024$ , giving $2024$ triangles.

Now consider triangles with side lengths $B, B, 2025$ , where $B < 2025$ . Triangle inequality gives: B + B > 2025 ⟹ B > 1012.5 B+B>2025⟹B>1012.5 Since $B$ is integer and $B < 2025$ , the possible $B$ are $1013$ to $2024$ , or $1012$ triangles.

Finally, don’t forget the case where all sides are $2025$ (equilateral, which also counts as isosceles by definition in contest math). So add $1$ more triangle.

In total, $2024 + 1012 + 1 = \boxed{3037}$ triangles.

Solution 3 (more broken down)

To solve this question, we first must observe the following restrictions given by the problem - 1.) Be an isosceles triangle 2.) Positive integer side lengths (causing a positive area) 3.) Longest length is $2025$

To satisfy restrictions $1$ and $2$ , we can denote the side lengths of a triangle that satisfies these restrictions as $A, A, B,$ where $A$ and $B$ are positive integers. This is possible as the triangle is restricted to be isosceles, meaning at least $2$ equal length sides.

In any triangle with side lengths $a$ , $b$ , and $c$ , the following inequalities (Triangle Inequality Theorem) must hold:

$a + b > c$ $a + c > b$ $b + c > a$

Therefore, our satisfactory triangles must satisfy

$A + A > B$ $A + B > A.$

Where $A > 0, B > 0.$

Now we can focus on restriction $3.$ Let us consider that $A$ is equal to $2025$ ( $A$ is the longest side length).

Substituting for $A$ we can rewrite our inequalities as follows

$2025 + 2025 > B$ $2025 + B > 2025$

where $B > 0.$

Furthermore, we can solve this system of linear inequalities like this

$4050 > B$ $B > 0$

$0 < B < 4050$

However, for $A$ to be the longest side length, $B$ must be less than $2025$ . Therefore, the solutions are

$\text{integers } 0 < B < 2025 \implies 1, 2, 3, 4, 5, \ldots, 2024$

This creates $2024$ cases.

---

Now let’s consider that $B$ is the longest side length and it is equal to $2025$ .

Substituting for $B$ , we can rewrite our inequalities as follows

$A + A > 2025$ $A + 2025 > A$

where $A > 0.$

Furthermore, we can solve this system of linear inequalities like this

$2A > 2025$ $0 > 2025$ (which can be ignored as it is always true)

$A > 1012.5$

However, for $B$ to be the longest side length, $A$ must be less than $2025$ . Therefore, the solutions are

$\text{integers } 1012.5 < A < 2025 \implies 1013, 1014, 1015, 1016, \ldots, 2024$

This creates $2024 - 1012 = 1012$ cases.

---

But there is also a case of an isosceles triangle where all side lengths of the triangle are equal in length and congruent - this special isosceles is referred to as an equilateral triangle. We can denote the lengths of such a triangle as $A, A, A$ . We need to assign the longest length to $A$ , as it is the only place a length can be applied. So, $A = 2025$ . We do not need to worry about the Triangle Inequality Theorem as any equilateral shape always satisfies it. For example,

$A + A > A, \quad A + A > A, \quad A + A > A$

where $a,b,c$ of the Triangle Inequality Theorem are all equal to $A$ and $A > 0$

is always satisfied for any side length, assuming that $A$ is a positive integer.

This creates $1$ case.

---

Adding together our cases,

$2024 + 1012 + 1 = \boxed{\text{(C) }3037}.$

This is a process called casework.

~ yuvaG (idk why i felt like going this in depth for question $3$ , but enjoy 😉)

Video Solution by Daily Dose of Math

https://youtu.be/LN5ofIcs1kY

~Thesmartgreekmathdude

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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

Page

Toolbox

Search