Challenging Geometry Puzzle: Problem 1558. Share your solution by posting it in the comment box provided.

Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.

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Perpendiculars from C1, C and C2 to B1B2 (extended to R) meet B1B2 at P, Q and R respectively.

ReplyDeleteLet C1B1= a and CB2 = b.

Let B1P = t1, PB = t2, BQ = t3, QB2 = t4 and B2R = t5.

Let C1P = u and C2R = v

C1PRC2 is a trapezoid with C1P//CQ//C2R, so by connecting C2P and applying the midpoint theorem,

CQ = (u+v)/2 ..........(1) and

t2 + t3 = t4 + t5 ......(2)

Now it is given that t1 + t2 = t3 + t4....... (3)

From (2) and (3), t3 = (t1 + t5) / 2..........(4)

Now consider Right Triangles B1C1P, BCQ and B2C2R

CQ = (u + v)/2

BQ = t3 = (t1+ t5)/2

and the included angle is 90

So BC = (a + b)/2

Similarly we can show that CA = (a+b)/2 and AB = (a+b)/2

Therefore Triangle ABC has BC = CA = AB and is thus equilateral

Sumith Peiris

Moratuwa

Sri Lanka