Problem:
In △RED,RD=1,∠DRE=75∘ and ∠RED=45∘. Let M be the midpoint of segment RD. Point C lies on side ED such that RC⊥EM. Extend segment DE through E to point A such that CA=AR. Then AE=ca−b, where a and c are relatively prime positive integers, and b is a positive integer. Find a+b+c.
Solution:
Let N be the midpoint of CR and P be the intersection of EM and CR. In isosceles triangle ARC, median AN is perpendicular to the base CR, implying that AN∥EM. Note that MN is a midline of △RCD, from which it follows that NM∥CD and CD=2MN. Therefore MNAE is a parallelogram, and CD=2MN=2AE.
Angle EDR=180∘−(75∘+45∘)=60∘. Let ∠DEM=x. Then ∠REM=45∘−x,∠EMR=60∘+x,∠CRD=90∘−∠EMR=30∘−x, and, because △ECP is a right triangle, ∠ACR=90∘−x. Applying the Law of Sines in △REM and △MED gives
EMRM=sin75∘sin(45∘−x) and MDEM=sinxsin60∘
Multiplying the two equations together yields
1=sin75∘sinxsin(45∘−x)sin60∘ or sin60∘sin75∘=sinxsin(45∘−x).
