In an equiangular pentagon, the sum of the squares of the side lengths equals 308, and the sum of the squares of the diagonal lengths equals 800. The square of the perimeter of the pentagon can be expressed as mn, where m and n are positive integers and n is not divisible by the square of any prime. Find m+n.
Let AC=d1,BD=d2,CE=d3,DA=d4,EB=d5 S=s12+s22+s32+s42+s52=308D=d12+d22+d32+d42+d52=800(s1+s2+s3+s4+s5)2=?Each Angle=5180(5−2)=108
Let A=s1s2+s2s3+s3s4+s4s5+s5s1B=s1s3+s2s4+s3s5+s4s1+s5s2 Then (s1+s2+s3+s4+s5)2=(s12+s22+s32+s42+s52)+2A+2B=308+2A+2B
By law of cosines in △ABC,△BCD,△CDE,△DEA,△EAB d12=s12+s22−2s1s2cos108d22=s22+s32−2s2s3cos108d32=s32+s42−2s3s4cos108d42=s42+s52−2s4s5cos108d52=s52+s12−2s5s1cos108⇒D=2S−2Acos108800=2(308)−2Acos108A=2cos108−184=cos7292=45−192=5−192⋅4=(5−1)(5+1)92⋅4(5+1)=92(5+1)
v1+v2+v3+v4+v5=0⟹∣v1+v2+v3+v4+v5∣2=0⟹(v1+v2+v3+v4+v5)⋅(v1+v2+v3+v4+v5)=0⟹∣v1∣2+∣v2∣2+⋯+∣v5∣2+2v1⋅v2+⋯+2v4⋅v5=0∣vi∣2=si2v1⋅v2=∣v1∣∣v2∣cos72=s1s2cos72v1⋅v3=∣v1∣∣v3∣cos144=s1s3cos144=−s1s3cos36 By Hint 2 S+2Acos72−2Bcos36=0 By Hint 3 308+2(92(5+1))(45−1)−2B(41+5)=0308+184−2B(1+5)=0⇒2B(1+5)=492⇒B=1+5984B=4984(5−1)=246(5−1)
By Hints 2,3,4 (s1+s2+s3+s4+s5)2=308+2A+2B=308+2(92(5+1))+2(246(5−1))=308+1845+184+4925−492=6765=mnAns=m+n=676+5=681
