## Wednesday, December 23, 2009

### Problem 406. Right triangle, 15 degrees, Midpoints, Congruence

Proposed Problem
Click the figure below to see the complete problem 406 about Right triangle, 15 degrees, Midpoints, Congruence.

Complete Problem 406
Level: High School, SAT Prep, College geometry

1. Join G to B
|GB|=1
Join F to B
|FB|=1
see m(FBG)=60 and FBG is equilateral triangle
|FG|=1

2. draw BM perpendicular to FG, GN perp to AC

Right tr BMG and GNC are congruent ( BG=GC)=> GN = x/2
but GN = 1/2 (ang 30) => x = 1

3. Connect BF and BG, AD=CE=2; AD and CE both hypotenuses. BF and BG are medians for them, so BF=BG=1. The median reveals Angle ABF=Angle CBG=15 Degrees. Angles (ABF+ABG+CBG)=90 Degrees, so Angle ABG=60 Degrees, which means FBG is equilateral.
Therefore, FG=x=1.

4. Solved in a very complex manner but Newzad's solution is so simple.

5. See the drawing

∠BAD= π/2 => B, A and D are concyclic, circle with center F middle of BD
=> FB=FA=FD and ∠BAF=15°
In the same way, GC=GA=GE and ∠CAG=15°
=>∠FAG=90-15-15=60°
ΔFAG is isosceles in A (AF=AG) and ∠FAG=60° => ΔFAG is equilateral
Therefore x=FG=BD/2=CE/2

triangle BAD congruent to triangle triangle BCE (AAS)
=>BA=BC & BD=BE
AE=BA-BE
CD=BC-BD
=>AE=CD

Let intersecting point of AD & EC be H
AE=CE (proved)
triangle EAH congruent to triangle DCH (AAS)
=>AH=CH
<HAC=<HCA
<EAC=<DCA=45

sin105/AC=sin45/2
AC=2sin105/sin45
AC=2sqrt2*sin105
AC=sqrt3+1

In triangle AHC
<AHC=120
sin120/AC=sin30/AH
AH=ACsin30/sin120
AH=AC/sqrt3=(sqrt3+1)/sqrt3

Triangle FHG ~ Triangle AHC
x/AC=FH/AH
x/AC=(AH-1)/AH
x/AC=1/(sqrt3+1)
x=1