Monday, May 26, 2008

Geometry Problem 109

See complete Problem 109
Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. Construct E on AC where AE=BE, then BCE is isosceles. Let AE=BE=CE=1 and AB=x, then by angle bisector theorem DE=1/(x+1), so 1/(x+1)+1=x, giving x=sqrt(2). Thus angle BAC is pi/4, therefore also angle ACB.

  2. If in the construction suggested by Anonymous AE = BE then Tr. AEB is isosceles and BD bisects angle ABE. That's easily understood but why should Tr. BCE be isosceles too?

    1. E is centre of right Tr. ABC, that's why.


  4. The trigonometric solution here is interestingly simple for tan(2α) = BC/AB = BC/CD = sin(3α)/cos(α) applying the Sine Rule in Tr. BCD. Thus sin(2α)/cos(2α) = sin(3α)/cos(α) leading to sin(3α)=sin(5α) which is possible only when α=22.5 deg. ruling out α=67.5 since 2α<90 deg. and thus x = 2α = 45 deg.

  5. The hint about requiring a construction was very helpful. I thought about bisecting angle A to get an isosceles triangle with angles 'alpha' but that seemed to make the figure very complex. I then thought of translating AB such that A maps to D and B to a point E, say, to make another isosceles triangle CDE. It then turns out that CDBE is cyclic and this leads to a proof...

  6. Let O be the midpoint of AC the centre of circle ABC

    OB = b/2. AD = b-c and DO = c-b/2

    Since BD bisects < ABO

    c/b/2 =(b-c)/(c-b/2) and so b^2 = 2c^2

    Hence a= c and x = 45

    Sumith Peiris
    Sri Lanka