Problem:
In △ABC with AB=AC, point D lies strictly between A and C on side AC, and point E lies strictly between A and B on side AB such that AE=ED=DB=BC. The degree measure of ∠ABC is nm, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let x=∠BAC. Because △AED is isosceles, ∠ADE=x and ∠AED=180∘−2x. Then
∠DEB=180−∠AED=180−(180∘−2x)=2x.
Because △EDB is also isosceles, ∠EBD=2x and ∠EDB=180∘−4x.
Using the fact that AC is a straight line, we find ∠BDC=3x.
Since △BDC is also isosceles, ∠BCD=3x and ∠CBD=180∘−6x.
Since ∠ABC=∠ACB, we can equate 180−4x=3x, and it follows that ∠ABC= x=(7540)∘. The requested sum is 540+7=547.
The problems on this page are the property of the MAA's American Mathematics Competitions
