## Sunday, October 29, 2023

### Geometry Problem 1558: Three Equilateral Triangles

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.

Gain comprehensive insights! Click below to reveal the complete details.

Share your solution by clicking 'Comment' below the post or entering your solution or comment in the 'Enter Comment' field and pressing 'Publish'.

#### 1 comment:

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