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Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 from Geometry, based on Area of Equilateral Triangle (Question 4).

Point B lies on line segment AC with AB =16 and BC =4. Points D and E lie on the same side of line AC forming equilateral triangle ABD and traingle BCE. Let M be the midpoint of AE, and N be the midpoint of CD. The area of triangle BMN is x. Find \(x^{2}\).

- is 107
- is 507
- is 840
- cannot be determined from the given information

Algebra

Theory of Equations

Geometry

But try the problem first...

Answer: is 507.

Source

Suggested Reading

AIME, 2015, Question 4

Geometry Revisited by Coxeter

First hint

Let A(0,0), B(16,0),C(20,0). let D and E be in first quadrant. then D =\((8,8\sqrt3)\), E=\((18,2\sqrt3\)).

Second Hint

M=\((9,\sqrt3)\), N=(\(14,4\sqrt3\)), where M and N are midpoints

Final Step

since BM, BN, MN are all distance, BM=BN=MN=\(2\sqrt13\). Then, by area of equilateral triangle, x=\(13\sqrt3\) then\(x^{2}\)=507.

- https://www.cheenta.com/number-and-series-aime-2015/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s

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