2025 AMC 10A Problems/Problem 3: Difference between revisions
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Case <imath>1</imath>: The two sides are both smaller than <imath>2025</imath>, which means that they range from <imath>1013</imath> to <imath>2024</imath>. There are <imath>1012</imath> such cases.<br> | Case <imath>1</imath>: The two sides are both smaller than <imath>2025</imath>, which means that they range from <imath>1013</imath> to <imath>2024</imath>. There are <imath>1012</imath> such cases.<br> | ||
Case <imath>2</imath>: There are two sides of length <imath>2025</imath>, so the last side must be in the range <imath>1</imath> to <imath>2025</imath>. There are <imath>2025</imath> such cases. Keep in mind, an equilateral triangle also counts as an isosceles triangle, since it has <b>at least</b> 2 sides.<br> | Case <imath>2</imath>: There are two sides of length <imath>2025</imath>, so the last side must be in the range <imath>1</imath> to <imath>2025</imath>. There are <imath>2025</imath> such cases. Keep in mind, an equilateral triangle also counts as an isosceles triangle, since it has <b>at least</b> 2 sides.<br> | ||
Therefore, the total number of cases is <imath>1012 + 2025 = \boxed{3037}</imath><br> | Therefore, the total number of cases is <imath>1012 + 2025 = \boxed{(D) 3037}</imath><br> | ||
~cw, minor edit by sd | ~cw, minor edit by sd | ||
Revision as of 20:09, 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 1: Casework
You can split the problem into two cases:
Case 
: The two sides are both smaller than , which means that they range from 
to 
. There are 
such cases.
Case 
: There are two sides of length , so the last side must be in the range to . There are 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 
~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
~MK
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 | ||
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