## Friday, June 3, 2016

### Geometry Problem 1222: Doubling the Cube, Triangle, 90, 120 degrees, Perpendicular, Congruence

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

Click the figure below to view more details of problem 1222.

1. Drop a perpendicular BH to AC. With the usual notation BC = a, and AC = b = DB. CH = a/2 and BH = sqrt3 a/2

From similar triangles c/(c+b) = b/(b+a/2) which yields ac = 2b^2.....(1)

Applying Pythagoras,

(b+c)^2 -(b+a/2)^2 = 3a^2/4

Substituting for a from (1)

3b^4/c^2 = (b+c)^2 - (b+b^2/c)^2

Simplifying the algebra and factorizing

(c^3-2b^3)(c+2b) = 0

c+2b = 0 is not a permissible solution

Hence c^3 = 2b^3

Sumith Peiris
Moratuwa
Sri Lanka

2. let AC=BD= x , AD=y and DC=h

Applying sine rule to triangle BDC
=> sin(30)/x = sin(B)/h
=> sin(B) = h/2x -------------- (1)

Applying sine rule to triangle ABC
=> sin(B)/x = sin(120)/(x+y)
=> sin(B) = sqrt(3).(x)/2.(x+y) ----------------- (2)

(1) = (2)
=> h/2x = sqrt(3).(x)/2.(x+y)
=> sqrt(y2-x2)/2.x = sqrt(3).(x)/2.(x+y) (Applying pythogrous to tri ADC => h2 = y2-x2)
=> sqrt(3).x2 = (sqrt(y2-x2)).(x+y)

Squaring on both sides
=> 3.x4 = (y2-x2).(x2+y2+2xy)
=> 3.x4 = y4-x4+2xy3-2x3y
=> 3 = y4/x4 - 1 + 2y3/x3 - 2y/x

let p = y/x

=> p4-2p+2p3-4 = 0
=> p(p3-2)+2(p3-2)=0
=> (p+2)(p3-2) = 0

Since p!= -2 => p3 = 2 => y3/x3 = 2