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Problem 1737

National Olympiad, first round
Geometry Difficulty 6.1 Prove it Iranian Mathematical Olympiad · Iran

Point PP lies inside of parallelogram ABCDABCD. Perpendicular lines to PA,PB,PCPA, PB, PC and PDPD through A,B,CA, B, C and DD construct convex quadrilateral XYZTXYZT. Prove that the area of XYZTXYZT is not less than twice the area of ABCDABCD.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

At first we shall prove following lemma:
Lemma. If point A′A' is antipode of vertex AA in circumcircle of triangle ABCABC, then
SABC−SA′BC=12∣BC∣2cot⁡∠A, S_{ABC} - S_{A'BC} = \frac{1}{2} |BC|^2 \cot \angle A,
where SXYZS_{XYZ} denotes the area of triangle XYZXYZ.
Proof. Let HH be the orthocenter of triangle ABCABC. Notice that
SABC−SA′BC=SABC−SHBC=12BC⋅AH=12∣BC∣2cot⁡∠A. S_{ABC} - S_{A'BC} = S_{ABC} - S_{HBC} = \frac{1}{2} BC \cdot AH = \frac{1}{2} |BC|^2 \cot \angle A.
Figure 1
Figure 2

Now let AB=aAB = a, AD=bAD = b, ∠APB=∠P1\angle APB = \angle P_1, ∠BPC=∠P2\angle BPC = \angle P_2, ∠CPD=∠P3\angle CPD = \angle P_3 and ∠DPA=∠P4\angle DPA = \angle P_4. From the lemma, it's easy to see that
2SABCD−SXYZT=a22(cot⁡∠P1+cot⁡∠P3)+b22(cot⁡∠P2+cot⁡∠P4)=a22(sin⁡(∠P1+∠P3)sin⁡∠P1⋅sin⁡∠P3)+b22(sin⁡(∠P2+∠P4)sin⁡∠P2⋅sin⁡∠P4) 2S_{ABCD} - S_{XYZT} = \frac{a^2}{2} (\cot \angle P_1 + \cot \angle P_3) + \frac{b^2}{2} (\cot \angle P_2 + \cot \angle P_4) \\ = \frac{a^2}{2} \left( \frac{\sin (\angle P_1 + \angle P_3)}{\sin \angle P_1 \cdot \sin \angle P_3} \right) + \frac{b^2}{2} \left( \frac{\sin (\angle P_2 + \angle P_4)}{\sin \angle P_2 \cdot \sin \angle P_4} \right)
Without loss of generality, we can assume ∠P1+∠P3>180∘\angle P_1 + \angle P_3 > 180^\circ (If ∠P1+∠P3=180∘\angle P_1 + \angle P_3 = 180^\circ we are done) so sin⁡(∠P1+∠P3)<0\sin(\angle P_1 + \angle P_3) < 0. Let points K,L,MK, L, M and NN be the foot of the perpendicular lines from PP to AB,BC,CDAB, BC, CD and DADA, respectively. We need to prove that
⇔a22(sin⁡(∠P1+∠P3)sin⁡∠P1⋅sin⁡∠P3)+b22(sin⁡(∠P2+∠P4)sin⁡∠P2⋅sin⁡∠P4)<0 \Leftrightarrow \frac{a^2}{2} \left( \frac{\sin (\angle P_1 + \angle P_3)}{\sin \angle P_1 \cdot \sin \angle P_3} \right) + \frac{b^2}{2} \left( \frac{\sin (\angle P_2 + \angle P_4)}{\sin \angle P_2 \cdot \sin \angle P_4} \right) < 0
⇔a2sin⁡∠P1⋅sin⁡∠P3>b2sin⁡∠P2⋅sin⁡∠P4 \Leftrightarrow \frac{a^2}{\sin \angle P_1 \cdot \sin \angle P_3} > \frac{b^2}{\sin \angle P_2 \cdot \sin \angle P_4}
⇔APsin⁡∠ABP⋅CPsin⁡∠CDP>APsin⁡∠ADP⋅CPsin⁡∠CBP \Leftrightarrow \frac{AP}{\sin \angle ABP} \cdot \frac{CP}{\sin \angle CDP} > \frac{AP}{\sin \angle ADP} \cdot \frac{CP}{\sin \angle CBP}
⇔sin⁡∠\Leftrightarrow \sin \angle CBP ⋅sin⁡∠\cdot \sin \angle ADP > sin⁡∠\sin \angle ABP ⋅sin⁡∠\cdot \sin \angle CDP
⇔BP⋅sin⁡∠CBP⋅DP⋅sin⁡∠ADP>BP⋅sin⁡∠ABP⋅DP⋅sin⁡∠CDP \Leftrightarrow BP \cdot \sin \angle CBP \cdot DP \cdot \sin \angle ADP > BP \cdot \sin \angle ABP \cdot DP \cdot \sin \angle CDP
⇔PL⋅PN>PK⋅PM \Leftrightarrow PL \cdot PN > PK \cdot PM
⇔∠LKN+∠LMN>∠KLM+∠KNM \Leftrightarrow \angle LKN + \angle LMN > \angle KLM + \angle KNM
⇔∠APB+∠CPD>∠APD+∠BPC \Leftrightarrow \angle APB + \angle CPD > \angle APD + \angle BPC
The latest assertion is true, so the proof is completed. ■

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.