## Sunday, September 22, 2013

### Geometry Problem 927: Semicircle, Diameter, Radius, Equal Arcs, Chord, Congruence

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 927.

1. FG = 2R sin80°tan10°
PQ = 2R sin60°tan10°
MN = 2R sin40°tan10°
KL = 2R sin20°tan10°

FG + PQ + MN + KL
= 2R tan10° [sin20°+sin40°+sin60°+sin80°]
= 2R tan10° [4 sin50°cos20°cos10°]
= 8R cos20°cos40°cos80°
= 8R [sin20°cos20°cos40°cos80°]/sin20°
= 4R [sin40°cos40°cos80°]/sin20°
= 2R [sin80°cos80°]/sin20°
= R sin160°/sin20°
= R

1. To Jacob: Thanks for your solution. Now, try an elementary geometry (Euclid's Elements.) solution.

2. http://img849.imageshack.us/img849/7236/vn8u.png

Define points C’, D’,E’,F’, C1, D1,E1, S, T and U as per sketch
Due to sysmetrical proprety, EFF’E’ is a isoceles trapezoid with its diagonals intersect at U on diameter AB.
∠ (EUF)=1/2(Arc(EF)+Arc(E’F’))= 20°
∠ (FUO)= ∠ (EUA)=80°
So UE//OF and UF//OG

Similarly we have SC//TD//UE//OF and SD//TE//UF//OG
From this proprety we have FG=OU, PQ=EE1=UT, MN=DD1=TS, KL=CC1=AS

3. The following facts could be easily found so i skip the proof:
EF//DG//CH//AI & FG//EH//DI//CJ//AB
So FEPG,EDMQ,DCKN,CAOL are parallelogram.
∆FEP~∆PDM~∆MCK~∆KAO~∆FOG~∆POQ~∆MON~∆KOL

Now the proof starts:

1.
OA = OF = R
So, ∆KAO = ∆FOG -> OK=FG

2.
EF=FG=OK (from 1)
So, ∆FEP = ∆KOL -> FP=KL

3.
CK=CL-KL=OA-KL=OF-FP=OP
So, ∆MCK=∆POQ -> MK = PQ

4.
∠MOD=∠FOD=40=∠DOA=∠MDO, so DM=MO
So, ∆PDM=∆MON -> PM = MN

Summing up the result of 1,2,3,4,
FG + PQ + MN + KL
= OK + FP + MK + PM
= OK + KM + MP + PF
=R

q.e.d.

4. From F, E, D, C, draw // to OG and OF => FG=F'O (paralelogram)
draw from P an // QG => EP'=E'F' and so on AO we get all segm