Problem:
Equilateral triangle ABC has been creased and folded so that vertex A now rests at A′ on BC as shown. If BA′=1 and A′C=2 then the length of crease PQ is
Answer Choices:
A. 58
B. 20721
C. 21+5
D. 813
E. 3
Solution:
Since ∠BA′P+∠A′PB+60∘=180∘=∠BA′P+60∘+∠QA′C, it follows that ∠A′PB=∠QA′C and thus △A′PB∼△QA′C. Let x=AP=A′P and y=QA=QA′. Then
QA′A′P=QCA′B=A′CPB, or yx=3−y1=23−x
Solve to obtain x=57 and y=47. Now apply the Law of Cosines to △PAQ,
PQ2=x2+y2−2xycos60∘=2549+1649−2049=40049⋅21,
which leads to PQ=20721.
OR
Let x=PA=PA′ and y=QA=QA′. Apply the Law of Cosines to △PBA′ to obtain x2=(3−x)2+1−2(3−x)cos60∘ which leads to x=7/5. Consider △QCA′ in a similar fashion to find y=7/4. Then complete the solution as above.
