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March 22, 2020

Challenge Problem 13: What Is The Area of Triangle 1984 AIME?

The American Invitational Mathematics Exam (AIME) is the second qualifying test for selection in the US Mathematical Olympiad team.

The test is 3 hours long with 15 questions (an average of 12 min/problem), and no calculators are allowed. It’s a challenging exam, and even the talented students who take it only average (mean and median) a score of 3–a mere 20 percent.

Today’s puzzle is problem 3 from the 1984 test. I also want to thank a viewer in Sweden who told me a very similar problem was asked in a competition there for 14 year olds. It is not an outdated problem!

Point P is in the interior of triangle ABC, and the lines through P are parallel to the sides of ABC. The three triangles shown in the diagram have areas of 4, 9, and 49. What is the area of triangle ABC?

Label the diagram as follows.

Because the lines through P are parallel to the sides of ABC, each of the three small triangles HIP, DEP, FGP is similar to ABC and so they are all similar to each other.

A triangle’s area is proportional to its side length squared, so the square root of its area is proportional to its side length. We are given the information:

areas HIP:DEP:FGP = 4:9:49

Because the triangles are similar, their sides will have a ratio equal to the square root of their areas.

sides HIP:DEP:FGP = 2:3:7

Suppose HI = 2x. Because the triangles are similar, DP = 3x and PG = 7x.

Now notice ADPI is a parallelogram. This is because lines through P are parallel to the sides of ABC, so IP and AD are parallel and so are DP and AI. Opposite sides in a parallelogram have equal length, so AI = DP = 3x.

Similarly HPGC is a parallelogram, and HC = PG = 7x.

Consequently we can calculate AC has a length:

= AI + IH + HC
= 3x + 2x + 7x
= 12x

Triangles ABC and HIP are similar, and their sides are in a ratio:

= (12x):(2x)
= 6:1

Consequently their areas will be in a ratio of the square of their sides:

area AC:IH
= 6^2:1^2
= 36:1

Thus triangle ABC has an area 36 times as large as HIP, so its area is:

36(4) = 144


I was emailed a similar problem from a student in Sweden.

Art of Problem Solving 1984 AIME Problem 3

Web 2.0 Calc forum similar question

Math 7210 solution at the University of Georgia

Mind Your Decision

Yêu Tiếng Anh



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