## Monday, May 26, 2008

### Geometry Problem 109

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

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.

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

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
Moratuwa
Sri Lanka

7. By External Angle Theorem: m∠BDC=3α. Let E be a point on BC such that DE∥AB, we have m∠ABC=m∠DEC=90°, m∠A=m∠EDC=2α, m∠ABD=m∠BDE=α, they are pairs of corresponding angles.
Let F be a point on BC such that DF is the angle bisector ∠EDC, by Leg – Angle theorem: ∆BED≅∆FED, it follows that BD=DF.
Since AB=DC, by SAS: ∆ABD≅∆CDF, therefore: ∠A≅∠C,x=2α, we have x+2α=x+x=90°,x=45°.

8. <DBC=90-a; take E onto BC so that <BDE=2a. Clearly tr. BDE is isosceles and <CDE=<ABD=a, thus triangles BAD and DCE are congruent, s.a.s., wherefrom <DCE=<BAD, i.e. 2x=90, x=45.

9. [For easy typing, I use "a" instead of alpha]
<DBC=90-a & x=90-2a
Consider triangle BCD
sin(90-a)/CD=sinx/BD
CD/BD=sin(90-a)/sinx=cosa/cos2a---------(1)

Consider triangle ABD
sin2a/BD=sin(180-3a)/AB
AB/BD=sin3a/sin2a-------(2)

Since AB=CD, by equating (1) & (2)
cosa/cos2a=sin3a/sin2a
sin3acos2a=sin2acosa
sin5a+sina=sin3a+sina
sin5a=sin3a
5a=3a or 5a=180-3a
a=0 (rej) or 22.5
x=90-2a=45