2019 AIME II Problems/Problem 7
Triangle has side lengths , and . Lines , and are drawn parallel to , and , respectively, such that the intersections of , and with the interior of are segments of lengths , and , respectively. Find the perimeter of the triangle whose sides lie on lines , and .
Let the points of intersection of with divide the sides into consecutive segments . Furthermore, let the desired triangle be , with closest to side , closest to side , and closest to side . Hence, the desired perimeter is since , , and .
Note that , so using similar triangle ratios, we find that , , , and .
We also notice that and . Using similar triangles, we get that Hence, the desired perimeter is -ktong
Let the diagram be set up like that in Solution 1.
By similar triangles we have Thus
Since and , the altitude of from is half the altitude of from , say . Also since , the distance from to is . Therefore the altitude of from is .
By triangle scaling, the perimeter of is of that of , or
Notation shown on diagram. By similar triangles we have So, email@example.com, vvsss
|2019 AIME II (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|