2025 AMC 10A Problems/Problem 3: Difference between revisions
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==Solution 1: Casework== | ==Solution 1: Casework== | ||
Split the problem into two cases: | Split the problem into two cases:<br> | ||
Case 1: 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. | 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 2: 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. | 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.<br> | ||
Therefore, the total number of cases is <imath>1012 + 2025 = \boxed{3037}</imath> | Therefore, the total number of cases is <imath>1012 + 2025 = \boxed{3037}</imath><br> | ||
-cw | -cw | ||
Revision as of 15:08, 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
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.
Therefore, the total number of cases is 
-cw
