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Geometry Difficulty 4.5 AIME Prove it Brazil

ABCDABCD is a convex quadrilateral. E,F,G,HE, F, G, H are the midpoints of sides AB,BC,CD,DAAB, BC, CD, DA respectively. Find the point PP such that area PHAE=area PEBF=area PFCG=area PGDHPHAE = \text{area } PEBF = \text{area } PFCG = \text{area } PGDH.

Solution

Let MM be the midpoint of ACAC, NN the midpoint of BDBD. Take PP so that PMPM is parallel to BDBD and PNPN is parallel to ACAC. Now EMEM is a midline of ABC\triangle ABC, so area AEM=area ABC4\text{area } AEM = \frac{\text{area } ABC}{4}. Similarly, area AHM=area ADC4\text{area } AHM = \frac{\text{area } ADC}{4}. So area MHAE=area ABCD4\text{area } MHAE = \frac{\text{area } ABCD}{4}. But MPMP is parallel to EHEH, so area MEH=area PEH\text{area } MEH = \text{area } PEH. Hence

area PHAE=area PEH+area AEH=area MEH+area AEH=area MHAE=area ABCD4 \begin{align*} \text{area } PHAE &= \text{area } PEH + \text{area } AEH \\ &= \text{area } MEH + \text{area } AEH \\ &= \text{area } MHAE = \frac{\text{area } ABCD}{4} \end{align*}

Similarly for area PEBF\text{area } PEBF and the other quadrilaterals.

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