## Tuesday, March 3, 2015

### Geometry Problem 1092. Equilateral Triangle, Square, Circle, Tangent, 90 Degree, Sangaku

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it. #### 5 comments:

1. Observe that ΔABC is a 30°-60°-90° triangle, and so are ΔAHB, ΔHBD, ΔHMD.
Also observe that HBDM is a kite.

Let BN⊥AC at N which is the mid-point of GE.
Let EP⊥BC at P which is the mid-point of DF.

Let AG=a.
Then
AB = AH + HB = 2 AG + GH = (2+√3)a
BC = 2 ED + DM = 3 AG + 2 GH = (3+2√3)a
AE = AG + GH = (1+√3)a

Hence, AE = BC − AB

2. Join the collinear points H, O, D.
The three 60-30-90 Right Triangles AHG, HDB, HDM are congruent to one another.
Let each side of the equilateral triangle be "a" and each side of the square be "b".
So BD = DM = a - b. Also HB = HM = b, AH = 2AG = 2BD = 2DM = 2a - 2b.
Follows AB = AH + HB = (2a - 2b) + b = 2a - b, AE = AG + GE = (a - b) + b = a
while BC - AB = (BD + DC) - AB = [(a - b) + 2a] - [2a - b] = a.
Hence AE = BC – AB.
(Incidentally AB = BF and AE = FC)

1. Also: angle CAF = 15 degree

3. Let BA meet FE in N. Let the side of the square be a and let AG = b. Easily Tr. EFC is isoceles with FC = a+b.

Now H, O, D are collinear and by congruent Tr.s easily we can prove that HB = a and BD = b. So BC = 2a+3b.

Now since < BNF = 30, Tr. AEN is isoceles and AN = a+b, AH = 2b and HB = a. So BN = 2a+3b = BC.

So AN + AB = BC hence BC - AB = AN = AE.

(Note that a = sqrt3 X b which I have avoided using to maintain clarity)

Sumith Peiris
Moratuwa
Sri Lanka

4. https://1drv.ms/u/s!AuFUZHYD5UUnzWied8htEcHx2GL8
Solved by Fernando Rogério Gonçalves!
Brazil