Problem:
In triangle ABC, it is given that angles B and C are congruent. Points P and Q lie on AC and AB, respectively, so that AP=PQ=QB=BC. Angle ACB is r times as large as angle APQ, where r is a positive real number. Find the greatest integer that does not exceed 1000r.
Solution:
Let x denote the degree measure of angle QPB. Because angle AQP is exterior to isosceles triangle BQP. its measure is 2x, and angle PAQ has the same measure. Because angle BPC is exterior to triangle BPA, its measure is 3x. Let y denote the measure of angle PBC. It follows that the measure of angle ACB is x+y, and that 4x+2y=180. Two of the angles of triangle APQ have measure 2x, and thus the measure of angle APQ is 2y. It follows that AQ=2β
APβ
siny. Because AB=AC and AP=QB, it also follows that AQ=PC. Now apply the Law of Sines to triangle PBC to find that
BCsin3xβ=PCsinyβ=2β
APβ
sinysinyβ=2β
BC1β
because AP=BC. Hence sin3x=21β. This and 4x<180 imply that 3x=30 and x=10. Thus y=70 and r=2β
7010+70β=74β, so 1000r=571β+73β.
OR
Let u=β ACB=β ABC. Then β A=β AQP=180β2u,β APQ=4uβ180,β QBP= β QPB=90βu,β BQC=β BCQ=90β21βu,β CQP=25βuβ90, and β QCP=23βuβ90. Apply the Law of Sines to triangles β³PQ and CPQ to obtain
sin(180β2u)sin(4uβ180)β=APAQβ=PQPCβ=sin(23βuβ90)sin(25βuβ90)β
This is equivalent to
sin2uβsin4uβ=cos23βucos25βuβ
or β2cos2ucos23βu=cos25βu. Use the identity 2cosΞ±cosΞ²=cos(Ξ±βΞ²)+cos(Ξ±+Ξ²) to obtain cos27βu+cos25βu+cos21βu=0, and then again to obtain 2cos3ucos21βu+cos21βu=0. This implies that cos3u=β21β. so that u=40 or u=80. Because 4u must be greater than 180 , it follows that u=80. Thus β APQ=4uβ180=140 and r=4/7, as above.
The problems on this page are the property of the MAA's American Mathematics Competitions
