2020 CIME II Problems/Problem 7
Let be a triangle with , , and . If is a point on the interior of segment such that the length is an integer, then the average of all distinct possible values of can be expressed in the form , where and are relatively prime positive integers. Find .
Given that the length is an integer and that it lies on the interior of segment , the shortest possible length of is the length of the altitude dropped straight down from vertex . This can be calculated as , which is equal to or . The area of triangle can be found using Heron's formula. It is just The shortest possible length of is can be anything greater than or equal to , but the condition that point lies in the interior of segment limits the values that we can reach. Starting at and heading east (we cannot get because is strictly between and ), we reach the integers , and then as we move further east the length of will start to increase. We then reach . We cannot get because then which is not allowed. The distinct possible values of are The average is . The answer is
|2020 CIME II (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All CIME Problems and Solutions|
The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions.