See complete Problem 109

Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 26, 2008

### Geometry Problem 109

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## Monday, May 26, 2008

###
Geometry Problem 109

Online Geometry theorems, problems, solutions, and related topics.

See complete Problem 109

Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Subscribe to:
Post Comments (Atom)

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.

ReplyDeleteIf 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?

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

Deletehttp://geometri-problemleri.blogspot.com/2009/05/problem9-solution-on-10-may.html

ReplyDeleteThe 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.

ReplyDeleteThe 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...

ReplyDeleteLet O be the midpoint of AC the centre of circle ABC

ReplyDeleteOB = 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

Moratuwa

Sri Lanka