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Explore the Back-Story**Problem**

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 $r$?

**Hint 1: **

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.

**Hint 2 - **

Find out $AC$ 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 $AX^2 = 13^2 - 12^2 = 5^2$ )

**Hint 3 - **

inradius and area is related 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 $**8$

Outstanding mathematics for brilliant school students.**Pre - RMO problems, discussions and other resources.** Go Back**Work with great problems from Mathematics Olympiads, Physics, Computer Science, Chemistry Olympiads and I.S.I. C.M.I. Entrance**. Click Here

**Problem**

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 $r$?

**Hint 1: **

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.

**Hint 2 - **

Find out $AC$ 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 $AX^2 = 13^2 - 12^2 = 5^2$ )

**Hint 3 - **

inradius and area is related 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 $**8$

Outstanding mathematics for brilliant school students.**Pre - RMO problems, discussions and other resources.** Go Back**Work with great problems from Mathematics Olympiads, Physics, Computer Science, Chemistry Olympiads and I.S.I. C.M.I. Entrance**. Click Here

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