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August 28, 2018

Area and inradius - Pre RMO 2018 Problem 2 Discussion

[et_pb_section fb_built="1" admin_label="Blog Hero" _builder_version="3.22" use_background_color_gradient="on" background_color_gradient_start="rgba(114,114,255,0.24)" background_color_gradient_end="#ffffff" background_blend="multiply" custom_padding="0|0px|0|0px|false|false" animation_style="slide" animation_direction="top" animation_intensity_slide="2%" locked="off"][et_pb_row _builder_version="3.25" background_size="initial" background_position="top_left" background_repeat="repeat" custom_margin="|||" custom_padding="27px|0px|27px|0px" custom_width_px="1280px"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="3.27.4" text_text_color="#474ab6" text_line_height="1.9em" background_size="initial" background_position="top_left" background_repeat="repeat" text_orientation="center" max_width="540px" module_alignment="center" locked="off"]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?
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PRMO 2018 Problem 2

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.
[/et_pb_tab][et_pb_tab title="Hint 2 - Find out AC" _builder_version="3.13"] 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 \( AX^2 = 13^2 - 12^2 = 5^2 \) )
[/et_pb_tab][et_pb_tab title="Hint 3 - inradius and area is related" _builder_version="3.13"]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:

PRMO problem 2 2

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
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