2019 AIME II Problems/Problem 7
Contents
Problem
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 .
Diagram
~MRENTHUSIASM
Solution 1
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
Solution 2
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
~ Nafer
Solution 3
Notation shown on diagram. By similar triangles we have So, vladimir.shelomovskii@gmail.com, vvsss
Solution 4 (Way too short, just keep track of which side is which)
Let's squish a triangle with side lengths 15, 22.5, and 27.5 into a equilateral triangle with side length 1. Then, the original triangle gets turned into a equilateral triangle with side length 8. Since 15 is one eighth of 120, it has a length of one. Since 45 and 55 are one fourth of 180 and 220 respectively, they are both two long. We extend the three segments to form a big equilateral triangle shown in black. Notice it has a side length of 11. Now that our task is done, let's undo the distortion. We get 11(15+22.5+27.5)=11(65)=715.
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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