Geometry Problem

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

Click the figure below to see the complete problem 927.

## Sunday, September 22, 2013

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

Labels:
arc,
chord,
congruence,
diameter,
radius,
semicircle

Subscribe to:
Post Comments (Atom)

FG = 2R sin80°tan10°

ReplyDeletePQ = 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

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

Deletehttp://img849.imageshack.us/img849/7236/vn8u.png

ReplyDeleteDefine 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

So FG+PQ+MN+KL=OU+UT+TS+SA= radius R

The following facts could be easily found so i skip the proof:

ReplyDeleteEF//DG//CH//AI & FG//EH//DI//CJ//AB

So FEPG,EDMQ,DCKN,CAOL are parallelogram.

In addition, observe that

∆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.