## Saturday, August 21, 2010

### Problem 509: Triangle, 120 Degrees, Angles, Congruence, Mind Map

Geometry Problem
Click the figure below to see the complete problem 509 about Triangle, 120 Degrees, Angles, Congruence.

Complete Problem 509

Level: High School, SAT Prep, College geometry

1. En espaĆ±ol: Sea P punto tal que el PBAD sea paralelogramo, y sea Q un punto tal que BQC sea equilatero y no se superponga a la figura inicial. Notemos las siguientes relaciones angulares DBC=x,BAD=DPB=PBQ=60-2x, BDP=120-x, luego como BP=AD=BC=QB, se tiene que el BQP es isosceles en B lo que implica que BPQ=60+x, DPQ=120-x=180-QAD, luego el cuadrilatero BDPQ es ciclico, entonces DQB=60-2x, DQC=2x, pero entonces DQA=DAQ, DQ=AD=QC, luego el triangulo QDC es isosceles en Q, pero QDC=60+2x y DQC=2x, luego 2(60+2x)+2x=180 y por lo tanto x=10.

2. The solution is indeed x=10. Can someone provide the diagram of the above solution?

3. This is my solution, enjoy it:
https://mega.nz/#!Fo4F3aLT!kFAEM_LxdAlsAq_50h4Z5_J0HWIXYpM8REcnCyKeRkw

Pedro Miranda

4. I'm pretty sure x= 12 as I have done and proved this question around 3 months ago,but I have no proof so i don't know what to say.

5. Extend AB to E such that AD = AE

Considering Triangle ADE, < BDE = 60+x (180 - 2(60-2x) -3x) = < DBE
Hence AD = BC = DE = BE = CE (since Triangle BCE is thus equilateral)
So E is the circumcentre of Triangle BCD and
< BCE = <BED / 2
2x = (60-2x)/2 from which
x = 10

Sumith Peiris
Moratuwa
Sri Lanka