## Monday, July 26, 2010

### Problem 480: Triangle, Circle, Center, Altitude, Angle, Circumcenter

Geometry Problem
Click the figure below to see the complete problem 480 about Triangle, Circle, Center, Altitude, Angle, Circumcenter.

Complete Problem 480

Level: High School, SAT Prep, College geometry

1. join A to O, draw OT perpendicular to AB ( T on AB )
ang BCD = 90° - ß = 1/2 arc ACB (1)
ang AOB = arc ACB =>
ang AOT = 1/2 arc ACB (2)
from (1) & (2)
ang BCD = ang AOT =90°-α =90°-ß (3)( AT=TB,OT perpen )
from (3) =>

α = ß

2. I take the E point of maximum arc AB (not arc ACB).Is <BCD=<AEB=γ,then β+γ=90.
But <AOB=2.<AEB=2γ,so 2α+2γ=180 or α+γ=90. Therefore α=β.

3. 1. Draw in AO.
2. Note triangles ABO ACO and BCO are all isosceles.
3. Let angle OBA = x and angle ABC = y (greek letters are hard in comments)
4. Then angle OAB = OBA = x from the isosceles triangle.
5. Angle OCB = OBC must be x + y from its isosceles triangle
6. Angle AOC is 2y since its the central angle and ABC is the inscribed one.
6. Angle OAC = ACO = 90 - y since this is also isosceles.
7. Angle BCD = 180 - (angle ACO + angle OCB) = 90 - x
8. Therefore CBD = 180 - 90 - BCD = x.

4. In general, ortocenter and circuncenter are isogonal conjugates.

5. < AOC = 2B & < COB = 2A
So 2A + 2B + 2@ = 180
Hence A + B + @ = 90 ...(1)
Now < BCD = A + B
So A + B + Beta = 90......(2)

Comparing (1) & (2) the result follows

Sumith Peiris
Moratuwa
Sri Lanka

6. [For easy typing, I use a for alpha & b for beta]
a=b

7. Simple Solution

< AOB = 180 - 2@
So Reflex < AOB = 180 + 2@

Hence < ACB = Reflex <AOB /2 = 90 + @
So < BCD = 90 - @

And Beta = 90- (90-@) = @

Sumith Peiris
Moratuwa
Sri Lanka