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four equilateral triangles solution

Solution to the Four Equilateral Triangles Puzzle

Posted on Aug 13, 2018

Four Equilateral Triangles

Here are 44 equilateral triangles. Find the missing area.

Solution by Properties of Equilateral Triangles

Four equilateral triangles labelled

With the points labelled as above, the equilateral triangle CEGC E G is formed by taking four copies of CDHC D H and assembling them into a single triangle. Its area is therefore 4×5=204 \times 5 = 20, so it is congruent to triangle AIJA I J. Since CHC H and HGH G have the same length, this means that the height of HH above ACA C is half the height of II.

The point KK is so that HKH K is parallel to ACA C. Since the height of HH above ACA C is half that of II, KK is also the midpoint of IBI B. As triangle BHIB H I is equilateral, angle BKHB K H is a right-angle and so IBI B is perpendicular to ACA C. The length of IBI B is thus the height of the triangle AIJA I J. From lengths in an equilateral triangle, the length of IBI B is 32\frac{\sqrt{3}}{2} times the length of AIA I, so the area of BHIB H I is 34\frac{3}{4} that of triangle AIJA I J, namely 1515.

Solution by Dissection and Symmetry

Four equilateral triangles dissected

In the above diagram, the original diagram has been extended so that the outer shape is an equilateral triangle. The outer triangle consists of four copies of the middle triangle of the original diagram. Looking at the upper right triangle, the area of one of these four is 20+5×5=4520 + 5\times 5 = 45.

In the central triangle, the three right-angled triangles are all half of the pink triangle, so have area 1010. This leaves 453×10=1545 - 3 \times 10 = 15 for the red triangle.