Maths Olympiad Prep

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, 2011

Geometry Difficulty 5.6 AIME, harder Prove it South Africa

Let PP be a point inside triangle ABCABC. Construct the points P1P_1, P2P_2, P3P_3 such that PP1BCPP_1 \perp BC, PP2CAPP_2 \perp CA, PP3ABPP_3 \perp AB and BP3=BP1BP_3 = BP_1, CP2=CP1CP_2 = CP_1. Prove that AP3=AP2AP_3 = AP_2.

Solution

Place the points AA, BB and CC in the complex plane such that PP is at the origin. Then we have the following equations:
PP1BC    BCP1=CˉBˉPˉ1(3) PP_1 \perp BC \implies \frac{B-C}{P_1} = \frac{\bar{C} - \bar{B}}{\bar{P}_1} \qquad (3)
PP2AC    ACP2=CˉAˉPˉ2(4) PP_2 \perp AC \implies \frac{A-C}{P_2} = \frac{\bar{C} - \bar{A}}{\bar{P}_2} \qquad (4)
PP3AB    BAP3=AˉBˉPˉ3(5) PP_3 \perp AB \implies \frac{B-A}{P_3} = \frac{\bar{A} - \bar{B}}{\bar{P}_3} \qquad (5)
BP1=BP3    (BP1)(BˉPˉ1)=(BP3)(BˉPˉ3)    BPˉ3+BˉP3P3Pˉ3=BPˉ1+BˉP1P1Pˉ1 \begin{align} BP_1 = BP_3 &\implies (B - P_1)(\bar{B} - \bar{P}_1) = (B - P_3)(\bar{B} - \bar{P}_3) \nonumber \\ &\implies B\bar{P}_3 + \bar{B}P_3 - P_3\bar{P}_3 = B\bar{P}_1 + \bar{B}P_1 - P_1\bar{P}_1 \tag{6} \end{align}
CP1=CP2    (CP1)(CˉPˉ1)=(CP2)(CˉPˉ2)    CPˉ2+CˉP2P2Pˉ2=CPˉ1+CˉP1P1Pˉ1 \begin{align} CP_1 = CP_2 &\implies (C - P_1)(\bar{C} - \bar{P}_1) = (C - P_2)(\bar{C} - \bar{P}_2) \nonumber \\ &\implies C\bar{P}_2 + \bar{C}P_2 - P_2\bar{P}_2 = C\bar{P}_1 + \bar{C}P_1 - P_1\bar{P}_1 \tag{7} \end{align}
Then,
AP22AP32=(AP2)(AˉPˉ2)(AP3)(AˉPˉ3)=APˉ3+AˉP3APˉ2AˉP2+P2Pˉ2P3Pˉ3=BPˉ3+BˉP3CPˉ2CˉP2+P2P22P3Pˉ3(using (4) and (5))=(BPˉ3+BˉP3P3Pˉ3)(CPˉ2+CˉP2P2Pˉ2) \begin{align*} & AP_2^2 - AP_3^2 \\ &= (A - P_2)(\bar{A} - \bar{P}_2) - (A - P_3)(\bar{A} - \bar{P}_3) \\ &= A\bar{P}_3 + \bar{A}P_3 - A\bar{P}_2 - \bar{A}P_2 + P_2\bar{P}_2 - P_3\bar{P}_3 \\ &= B\bar{P}_3 + \bar{B}P_3 - C\bar{P}_2 - \bar{C}P_2 + P_2P_2 - 2P_3\bar{P}_3 \quad \text{(using (4) and (5))} \\ &= (B\bar{P}_3 + \bar{B}P_3 - P_3\bar{P}_3) - (C\bar{P}_2 + \bar{C}P_2 - P_2\bar{P}_2) \end{align*}
=(BPˉ1+BˉP1P1Pˉ1)(CPˉ1+CˉP1P2Pˉ1(using (6) and (7))=(BC)Pˉ1(CˉBˉ)P1=0,(using (3)) \begin{aligned} &= (B\bar{P}_1 + \bar{B}P_1 - P_1\bar{P}_1) - (C\bar{P}_1 + \bar{C}P_1 - P_2\bar{P}_1 \quad \text{(using (6) and (7))} \\ &= (B - C)\bar{P}_1 - (\bar{C} - \bar{B})P_1 \\ &= 0, \quad \text{(using (3))} \end{aligned}
as required.

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