2025 AMC 10A Problems/Problem 3: Difference between revisions
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A + B > A \implies B > 0 | A + B > A \implies B > 0 | ||
\] | \] | ||
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Since \(A = 2025\) (the longest side): | Since \(A = 2025\) (the longest side): | ||
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giving \(2024\) triangles. | giving \(2024\) triangles. | ||
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Now consider triangles with side lengths \(B, B, 2025\), where \(B < 2025\). | Now consider triangles with side lengths \(B, B, 2025\), where \(B < 2025\). | ||
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which gives \(1012\) triangles. | which gives \(1012\) triangles. | ||
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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. | 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. | ||
\hrule | |||
In total: | In total: | ||
Revision as of 20:59, 6 November 2025
How many isosceles triangles are there with positive area whose side lengths are all positive integers and whose longest side has length 
?
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 the triangle inequalities:
\[ A + A > B \] \[ A + B > A \implies B > 0 \]
\hrule
Since \(A = 2025\) (the longest side):
\[ 2025 + 2025 > B \implies B < 4050 \]
But also, since \(A\) is the longest side, \(B < 2025\). Therefore:
\[ B = 1, 2, 3, \ldots, 2024, \]
giving \(2024\) triangles.
\hrule
Now consider triangles with side lengths \(B, B, 2025\), where \(B < 2025\).
Triangle inequality gives:
\[ B + B > 2025 \implies B > 1012.5 \]
Since \(B\) is an integer and less than \(2025\), the possible values are:
\[ B = 1013, 1014, 1015, \ldots, 2024, \]
which gives \(1012\) triangles.
\hrule
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.
\hrule
In total:
\[ 2024 + 1012 + 1 = \boxed{3037} \]
triangles.
~yuvaG
Video Solution (Fast and Easy)
https://youtu.be/qfvA1uD0--M?si=vhZMvas48oMPyvWx ~ Pi Academy
Video Solution
~MK
Solution 2 (broken down)
Suppose an isosceles triangle has sides 
where 
is the longest side and 
are positive integers.
By triangle inequalities: A + A > B A+A>B A + B > A ⟹ B > 0 A+B>A⟹B>0
Since 
(the longest side):
2025
+
2025
>
B
⟹
B
<
4050
2025+2025>B⟹B<4050
But also 
so 
to 
, giving triangles.
Now consider triangles with side lengths 
, where .
Triangle inequality gives:
B
+
B
>
2025
⟹
B
>
1012.5
B+B>2025⟹B>1012.5
Since 
is integer and , the possible are 
to , or 
triangles.
Finally, don’t forget the case where all sides are (equilateral, which also counts as isosceles by definition in contest math). So add 
more triangle.
In total, 
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
To satisfy restrictions and 
, we can denote the side lengths of a triangle that satisfies these restrictions as 
where and are positive integers. This is possible as the triangle is restricted to be isosceles, meaning at least equal length sides.
In any triangle with side lengths 
, 
, and 
, the following inequalities (Triangle Inequality Theorem) must hold:
![\[a + b > c\]](http://latex.artofproblemsolving.com/0/a/3/0a3d989c0938abd083035fa434aa75b4d3474b65.png)
![\[a + c > b\]](http://latex.artofproblemsolving.com/8/1/f/81faf2590ff435f24a6819ac192e63b6a9a02583.png)
![\[b + c > a\]](http://latex.artofproblemsolving.com/5/1/3/513349439e7e98155a2b8c58beba005da2e65dcd.png)
Therefore, our satisfactory triangles must satisfy
![\[A + A > B\]](http://latex-new.aopstest.com/6/b/b/6bbbe93a7405ce04ea8ded3168b1f5c3d4f9edf9.png)
![\[A + B > A.\]](http://latex-new.aopstest.com/6/e/1/6e1f50372b0f39d7ea7072bfb9cf13baa4238dc9.png)
where 
Now we can focus on restriction 
Let us consider that is equal to ( is the longest side length).
Substituting for we can rewrite our inequalities as follows
![\[2025 + 2025 > B\]](http://latex-new.aopstest.com/e/7/c/e7c2156d7649cde2d8d2ca730faa62d56137f985.png)
![\[2025 + B > 2025\]](http://latex-new.aopstest.com/f/2/1/f219ed66dd0bb01ae13cd364ee491b3a764d4d9b.png)
where 
Furthermore, we can solve this system of linear inequalities like this
![\[4050 > B\]](http://latex-new.aopstest.com/4/3/4/43471c5bb19ce453a99c94c5ee9a75bac8bbd5fd.png)
![\[B > 0\]](http://latex-new.aopstest.com/4/c/f/4cf5c89d447c56fb6578d0171bf50ebd9dc94a9a.png)
![\[0 < B < 4050\]](http://latex-new.aopstest.com/c/a/7/ca79d1a8baf218fc939a31a55ca389be53201053.png)
However, for to be the longest side length, must be less than . Therefore, the solutions are
![\[\text{integers } 0 < B < 2025 \implies 1, 2, 3, 4, 5, \ldots, 2024\]](http://latex-new.aopstest.com/6/1/5/615bae74c8c91b796a6700c73356986bf1fa1e87.png)
This creates cases.
---
Now let’s consider that is the longest side length and it is equal to .
Substituting for , we can rewrite our inequalities as follows
![\[A + A > 2025\]](http://latex-new.aopstest.com/0/d/d/0dd42e96fbc71d6e26c87583f6a819568508f61e.png)
![\[A + 2025 > A\]](http://latex-new.aopstest.com/2/2/5/2252e64426dd8f98f0111adcd88591b8799d479c.png)
where 
Furthermore, we can solve this system of linear inequalities like this
![\[2A > 2025\]](http://latex-new.aopstest.com/f/d/4/fd48fcb088ba9f9799beafb08536b194ea496da9.png)
![\[0 > 2025\]](http://latex-new.aopstest.com/f/6/b/f6b7eb1e50340fd571dd6b7dabb9a1aee5c7d2f5.png)
(which can be ignored as it is always true)
![\[A > 1012.5\]](http://latex-new.aopstest.com/a/f/d/afde7ddebad5429fc21cdd76fa1d16e17188fb6c.png)
However, for to be the longest side length, must be less than . Therefore, the solutions are
![\[\text{integers } 1012.5 < A < 2025 \implies 1013, 1014, 1015, 1016, \ldots, 2024\]](http://latex-new.aopstest.com/8/4/7/847cf0290303aec047aefed1327f48aae43b822f.png)
This creates 
cases.
---
But! here 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 
. We need to assign the longest length to , as it is the only place a length can be applied. So, . We do not need to worry about the Triangle Inequality Theorem as any equilateral shape always satisfies it. For example,
![\[A + A > A, \quad A + A > A, \quad A + A > A\]](http://latex-new.aopstest.com/f/f/8/ff81f0ae2c35ed992341b516cb12f3cd9b5a1686.png)
where 
of the Triangle Inequality Theorem are all equal to and 
is always satisfied for any side length, assuming that is a positive integer.
This creates case.
---
Adding together our cases,
![\[2024 + 1012 + 1 = \boxed{\text{(C) }3037}.\]](http://latex-new.aopstest.com/c/0/3/c0359099465e4686aa4e66713097b77a0b6bd6f8.png)
This is a process called casework.
~ yuvaG (idk why i felt like going this in depth for question 
, but enjoy 😉)
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
See Also
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
