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 close to r?

Drawing a good picture is sometimes very helpful. Here is a computer drawn picture (using GeoGebra). But how do you construct this? First, notice that the quadrilateral is a kite. Diagonals of a kite bisect each other (

**Prove this!)**If X is the point of intersection then it is easy to show \( \Delta AXB \equiv \Delta AXD \) implying BX = XD = 12. Finally to find the incenter, draw the angle bisector of \( \angle ADC\). Wherever it meets AC, call that point I. It is the incenter (**Prove this too!)**To draw the incircle, drop a perpendicular from I to AD and draw a circle taking that perpendicular segment as radius, and I as the center. It is easy to find (using Pythagoras theorem) that AX = 5 and CX = 16. (Notice that \( \Delta AXB \) is right angled. Hence \( AX^2 + BX^2 = AB^2 \) or \( AX^2 = 13^2 – 12^2 = 5^2 \) )

Any polygon with an inscribable circle (a circle that touches each of its sides) has this beautiful property: \( Area = r \times s \) where r = radius of the inscribed circle and s = semiperimeter (half of the perimeter). Why is that true? Well, there is a simple argument that works for all polygons (in which circles can be inscribed). However we will describe the special case of this kite: Drop perpendiculars from incenter I to each of the sides of the kite. Since the incircle

**touches**each side, hence each side is a tangent to the incircle. Hence these perpendicular segments are clearly the radii of the incircle (after this was our*method of construction*) in the first place. Join ID and IB. The kite is thus divided into four triangles: red, green, yellow and blue. Notice that area of these triangles are \( \frac{1}{2} \times r \times AB, \frac{1}{2} \times r \times BC, \frac{1}{2} \times r \times CD, \frac{1}{2} \times r \times DA \) respectively. Adding the areas of the 4 triangles we will get the area of the kite which is: \( \frac{1}{2} \times r \times (AB + BC + CD + DA ) = r \times s \)**Thus the area of the kite is in radius times semiperimeter.**The exact same argument holds for any polygon with inscribable circles. The semiperimeter is \( \frac{1}{2} \times (13 + 13 + 20 + 20) = 33 \). Hence the area of the kite is 33r. But we also know that area of the kite is the product of diagonals by 2 (or just separately find out the area of the triangles ABC, ADC and add them). This gives the area to be \( \frac{1}{2} \times 24 \times 21 = 252 \) Equating we have 33r = 252 or r ~ 7.6. Its**closest integer is**8# Get Started with Math Olympiad Program

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