Question

△ABC中，AB=AC=2，BC边上有100个不同的点p₁,p₂,…p₁₀₀;

                                        

                                           

Answer 1

解：作AD⊥BC于D，则BC=2BD=2CD．
根据勾股定理，得
AP_i²=AD²+DP_i²=AD²+（BD-BP_i）²=AD²+BD²-2BD•BP_i+BP_i²，
又P_iB•P_iC=P_iB•（BC-P_iB）=2BD•BP_i-BP_i²，
∴M_i=AD²+BD²=AB²=4，
∴M₁+M₂+…+M₁₀+M₁₀₀=4×100=400．