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第一届数学奥林匹克国家集训班选拔考试暨第四届“祖冲之”杯初中数学邀请赛都有这样一道试题: 题目:正方形ABCD边长为1,AB、AD上各有一点P、Q.如果△APQ的周长为2,求∠PCQ的度数. 解:将△CDQ顺时针旋转90°使CD与CB重合,则Q点落在AB的延长线上,记此点为E.过C作PQ的垂线,垂足为F.显然Rt△CDQ≌Rt△CBE,于是EB=DQ.由AP+AQ+PQ=2知PQ=2-AQ-AP=(1-AP)+(1-AQ)=BP+DQ=BP+EB=EP.
The first mathematics Olympics national training class selection examination and the fourth “Zuchongzhi” Cup junior high school mathematics invitational tournament have such questions: Title: Square ABCD side length is 1, AB, AD each have a little P, Q. If △ APQ The circumference is 2, find the degree of PCQ. Solution: △ CDQ rotates clockwise by 90° so that CD and CB coincide, then the Q point falls on the extended line AB, remember this point is E. over C for PQ Vertical line, foot is F. Obviously RtΔCDQ≌Rt△CBE, then EB=DQ. From AP+AQ+PQ=2, PQ=2-AQ-AP=(1-AP)+(1-AQ) =BP+DQ=BP+EB=EP.