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Olympiad workbook
Exam
Olympiad workbook
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IOQM
Chapter
Circle

Olympiad workbook · IOQM · Circle

Let A B C D be a convex cyclic quadrilateral. Suppose P is a point...

Let ABCDA B C D be a convex cyclic quadrilateral. Suppose PP is a point in the plane of the quadrilateral such that the sum of its distances from the vertices of ABCD is the least. If {PA,PB,PC,PD}={3\{\mathrm{PA}, \mathrm{PB}, \mathrm{PC}, \mathrm{PD}\}=\{3, 4,6,8}4,6,8\} What is the maximum possible area of ABCDA B C D ?Let A B C D be a convex cyclic quadrilateral. Suppose P is a point in the plane of the quadrilateral such that the sum

Correct answer

55

Step-by-step solution

PP must be point of intersection of diagonals ACA C and BDB D Let A B C D be a convex cyclic quadrilateral. Suppose P is a point in the plane of the quadrilateral such that the sum
Let ∠APB=θ\quad \angle \mathrm{APB}=\theta, then area of △PAB=12×3×4×sin⁡θ\triangle \mathrm{PAB}=\frac{1}{2} \times 3 \times 4 \times \sin \theta area of △PAD=12×3×6×sin⁡(π−θ)\triangle \mathrm{PAD}=\frac{1}{2} \times 3 \times 6 \times \sin (\pi-\theta) area of △PDC=12×8×6×sin⁡θ\triangle \mathrm{PDC}=\frac{1}{2} \times 8 \times 6 \times \sin \theta area of △PCB=12×8×4×sin⁡(π−θ)\triangle \mathrm{PCB}=\frac{1}{2} \times 8 \times 4 \times \sin (\pi-\theta) ⇒\Rightarrow \quad area of quadrilateral ABCD is 12(12+18+48+32)sin⁡θ=(6+9+24+16)sin⁡θ\begin{aligned} & \frac{1}{2}(12+18+48+32) \sin \theta \\ & =(6+9+24+16) \sin \theta \end{aligned} Maximum area of quadrilateral ABCD is 6+9+24+16=556+9+24+16=55.
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