## Saturday, October 2, 2010

### Problem 524: Circle, Equilateral Triangles, Midpoint, Side, Measurement

Geometry Problem
Level: High School, SAT Prep, College geometry.
Click the figure below to see the complete problem 524 about Circle, Equilateral Triangles, Midpoint, Side, Measurement.

1. Let O be the centre of C1. Then B is (0,a/√3) and C is (a/2,-a/2√3) which makes D:a/4,a/4√3). If F is (x1,y1) then x1^2+y1^2=a^2/3 but y1=a/4√3 since DF//AC. Thus x1^2 + a^2/48=a^3/3 or x1^2=15a^2/48 or x1=√5*a/4 and thus x=√5*a/4-a/4 or x =(√5-1)a/4
Ajit

2. G midpoint of EF,R the radius of C1
OGF is a rigth triangle
OF=R,OD=R/2,R=a.sqr(3)/3,DG=x.sqr(3)/2,GF=x/2
and the th.of Pythagoras gives the solution

3. Question to Anonymous: Pythagoras Th. applied to which triangle gives the desired result?
Vihaan

4. Anonymous you need not reply. When applied to Rt Tr. OGF, Pythagoras gives us (a/(2√3)+x√3/2)^2+(x/2)^2=(a/√3)^2 from which the necessary result follows. I understand now.
Vihaan

5. prolongando FD hasta cortar la circunferencia C1 en Q y a AB en T (2TD = AC = a ).
Tomando las cuerdas AB y CQ se cumple: FD.QD=BD.DC

x(x+a/2)=(a/2).(a/2) x = a/4(sqr(5)-1)

RUDDY CRUZ MENDEZ LIMA - PERU

6. To Ruddy, why are the products FD.QD and BD.DC equal? Nice solution.