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Feyman's 1/7 Area Triangle: A Beautiful New Proof By Dissection

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Edited by Richard Walker, Monday 30 March 2026 at 20:37

Claim: In any triangle ABC, if we draw a line from each vertex to the point one solidus three along on the side opposite, the inner triangle so formed has area equal to one solidus seven of the area of the orginal triangle.

I posted a proof of this on 28 March but since then I've thought of a simpler and far nicer proof.

Figure 1

In Figure 1 triangle ABC, R, S and T are trisection points of the sides they lie on; and K, L and M the midpoints of the sides. XYZ is the Feynman "one-seventh" triangle formed by lines AT, BS and CR.

In Figure 2 we have rotated triangles XBL, YCM and ZAK through 180 reverse solidus degree about the midpoints L, M and N respectively.

Figure 2

Figure 2 shows triangle ABC can be dissected into a figure consisting of seven congruent triangles, all of equal area, and that the Feynman triangle XYZ, numbered 7 in the diagram, is one of them. It follows immediately that its area is one solidus seven the area of triangle ABC.

Of course, like the proof I gave previously, this must have already been discovered by many before me, but I worked it out for myself and the 'εὕρηκα' moment was very pleasing.

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