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Saudi Arabian Mathematical Competitions

Geometry Difficulty 5.4 AIME, harder Prove it Saudi Arabia

In quadrilateral ABCDABCD, diagonals ACAC and BDBD intersect at OO. Denote by P,Q,R,SP, Q, R, S the orthogonal projections of OO onto AB,BC,CD,DAAB, BC, CD, DA, respectively. Prove that
PA⋅AB+RC⋅CD=12(AD2+BC2) PA \cdot AB + RC \cdot CD = \frac{1}{2}\left(AD^2 + BC^2\right)
if and only if
QB⋅BC+SD⋅DA=12(AB2+CD2). QB \cdot BC + SD \cdot DA = \frac{1}{2}\left(AB^2 + CD^2\right).

Solution

Figure 1

Denote by 1,2,…,81,2, \ldots, 8 the right angled triangles as in the above figure. Applying Pythagoras theorem successively in triangles 1,2,…,81,2, \ldots, 8, we get:
PA2+PO2−OA2=0PB2−PO2−OB2=0QB2+QO2−OB2=0QC2+QO2−OC2=0RC2+RO2−OC2=0RD2+RO2−OD2=0SD2+SO2−OD2=0SA2+SO2−OA2=0 \begin{aligned} & PA^2 + PO^2 - OA^2 = 0 \\ & PB^2 - PO^2 - OB^2 = 0 \\ & QB^2 + QO^2 - OB^2 = 0 \\ & QC^2 + QO^2 - OC^2 = 0 \\ & RC^2 + RO^2 - OC^2 = 0 \\ & RD^2 + RO^2 - OD^2 = 0 \\ & SD^2 + SO^2 - OD^2 = 0 \\ & SA^2 + SO^2 - OA^2 = 0 \end{aligned}
It follows
PA2−PB2+QB2−QC2+RC2−RD2+SD2−SA2=0. \begin{equation*} PA^2 - PB^2 + QB^2 - QC^2 + RC^2 - RD^2 + SD^2 - SA^2 = 0. \tag{1} \end{equation*}
The last relation is equivalent to:
PA2−(AB−PA)2+QB2−(BC−QB)2+RC2−(CD−RC)2+SD2−(DA−SD)2=0 \begin{gathered} PA^2 - (AB - PA)^2 + QB^2 - (BC - QB)^2 + RC^2 - (CD - RC)^2 \\ + SD^2 - (DA - SD)^2 = 0 \end{gathered}
hence
2(PA⋅AB+QB⋅BC+RC⋅CD+SD⋅DA)=AB2+BC2+CD2+DA2 \begin{gathered} 2(PA \cdot AB + QB \cdot BC + RC \cdot CD + SD \cdot DA) \\ = AB^2 + BC^2 + CD^2 + DA^2 \end{gathered}
and the conclusion follows.

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