## Wednesday, February 18, 2009

### Problem 14: Triangle, Cevian, Midpoint, Angles

Proposed Problem See complete Problem 14 at:
http://gogeometry.com/problem/problem014.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

1. WLOG one may assume D:(0,0), A:(-1,0), C:(1,0). Let angle BCA = m and angle BAC=angle CBD = 2m
Thus angle BDA = 3m. Further, let tan(m)=t
Now, DB: y =(-tan3m)x =-x(3t-t^3)/(1-3t^2)
CB:y=(1-x)t which give B as((3t^2-1)/2(t2+1),(3t-t^3)/2(t2+1)) while A is (-1,0). hence slope BA=
(3t-t^3)/(5t^2+1) which as we know is tan(2m)=2t/(1-t^2). The equation obtained: (3t-t^3)/(5t^2+1)=2t/(1-t^2) has five solutions of which the only admissible solution is t=0.2679491924 which gives m=15 deg. Thus, angle BCA = 15 deg.
Alternatively, in triangle ABD, AD/BD =sin(5m)/sin(2m) while triangle BDC gives, CD/BD=sin(2m)/sin(m). Since AD=CD, we can say that sin(5m)/sin(2m)=sin(2m)/sin(m) which can be solved to get m=15 deg as before.
Now, having proven my bonafides, can someone please show me how such problems can be solved more elegantly by plane geometry alone?
Ajit: ajitathle@gmail.com

2. http://geometri-problemleri.blogspot.com/2009/05/problem-21-ve-cozumu.html

3. http://ahmetelmas.files.wordpress.com/2010/05/cozumlu-ornekler.pdf

4. the solution is uploaded to the following link:

5. CE perpendicular to AB and BE=EH. So , angle AHC=3x. Because, angle BDA=3x ,
BDCH , is a cyclic quadrilateral , therefore , angle BHD=x. The triangle AEC, is rectangular and AD=DC. So, ED=DA=DC, and angle BED=2x, therefore, angle EDH=x. So ,DE=EH=EB .Then angle BDH=90 degrees and 3x+2x+x=90 .Therefore x=15 degrees.
image : http://img248.imageshack.us/img248/8017/geogebra4.png

1. Excellent solution Michael

6. 7. Assume length of AC=2
Note triangle ABC and BDC are similar.
AC/BC=BC/DC
2/BC=BC/1
Hence BC=√2

Construct point E which lies on BC and <EDC=90. By joining AE, triangle EBA is again similar to ABC and BDC, and EA=EC.Let length of AE and EC=a
AC/BC=AE/AB
2/√2=AE/AB
AB=a/√2

BC/AB=AB/BE
√2/AB=AB/BE
BE=AB^2/√2
BE=(a/√2)^2/√2
BE=(√2)a^2/4

BE+EC=√2
(√2)a^2/4 + a - √2=0 (quadratic equation)
a=1.03527618
In triangle EDC,
cos x=1/a
x=15

8. 9. Let BE be perpendicular to AC, E on AC. Extend CA to F so that BF = BC. Let BE = h

Then 2a^2 = b^2 ......(1) since BC is a tangent to circle ABD

Also AF = AB = c and so AE = (b-c)/2 and hence CE = (b+c)/2 ......(2)

Therefore
h^2 = a^2 - {(b+c)/2}^2 = c^2 - {(b-c)/2)^2} and so 2a^2 = 2c^2 + 2bc = b^2 ...(2) upon substituting for 2a^2 from (1)

Therefore h^2 = c^2 - {(b-c)/2)^2} = c^2 - (1/4)(b^2 - 2bc + c^2) = c^2 - 3c^2/4 = c^2/4

So h = c/2 and hence ABE is a 30-60-90 Triangle and 2x = 30 and x = 15

Sumith Peiris
Moratuwa
Sri Lanka

10. Consider triangle BCD
sin2x/CD=sinx/BD
CD=BDsin2x/sinx

Consider triangle ABD

sin2x/sinx=sin5x/sin2x
2cosxsin2x=sin5x
sin3x+sinx=sin5x
sin5x-sin3x=sinx
2cos4xsinx=sinx
2cos4x=1
cos4x=1/2
x=15

11. Geometry Solution 2

Let the bisector of <CBD meet AC at E. Let the foot of the perpendicular from B to AC be F.

Then AB = BE = CE = c, DE = b/2-c and DF = c/2

Tr.s BCD & ABC are similar so a/(b/2) = b/a = c/d so b = a.srqt2 and d = c.sqrt2

Hence Tr. BFD is right isosceles and < ADB = 3x = 45

Therefore x = 15

Sumith Peiris
Moratuwa
Sri Lanka