Pre RMO 2018(IOQM), Question 2 incircle quadrilateral
Question 2, Pre RMO 2018-19(IOQM): In a quadrilateral ABCD, it is given that AB = AD = 13, BC= CD = 20, BD = 24. If r is the radius of the circle inscribable in the quadrilateral, then what is the integer closest to r?. Prerequisites: 1. Incenter, incircle and angle bisectors in triangle/polygons 2. Kite diagonals Solution : ABCD is a kite. So, A C ⊥ B D AC \perp BD . So, B O = O D = 12 cm BO = OD = 12 \text{ cm} . A O = 13 2 − 12 2 = 5 cm AO = \sqrt{13^2 - 12^2} = 5 \text{ cm} O C = 20 2 − 12 2 = 16 cm OC = \sqrt{20^2 - 12^2} = 16 \text{ cm} ar ( □ A B C D ) = 1 2 × A C × B D \text{ar}(\square ABCD) = \tfrac{1}{2} \times AC \times BD = 1 2 × 21 × 24 = \tfrac{1}{2} \times 21 \times 24 = 252 cm 2 = 252 \text{ cm}^2 r = Δ S = 252 33 = 7.63 cm r = \frac{\Delta}{S} = \frac{252}{33} = 7.63 \text{ cm} Closest integer = 8 = 8 .