2011 AMC 12A Problem 24

Question

Consider all quadrilaterals

ABCD

such that

AB = 14,

BC = 9,

CD = 7,

and

DA = 12.

What is the radius of the largest possible circle that fits inside or on the boundary of such a quadrilateral?

\sqrt{15}

\sqrt{21}

2\sqrt{6}

5

2\sqrt{7}

If a circle of radius

r

fits, splitting from its center shows the quadrilateral’s area is at least

21r

Bretschneider’s inequality bounds the area by the cyclic case; the equality case is also tangential because opposite side sums are equal

Suppose a circle of radius

r

centered at

X

fits in one of the quadrilaterals. If

h_1,h_2,h_3,h_4

are the distances from

X

to the four side lines, then each

h_i\ge r.

Splitting the quadrilateral into four triangles gives

\begin{aligned} K&=\dfrac12(14h_1+9h_2 \\ &\qquad+7h_3+12h_4) \\ &\ge21r. \end{aligned}

Bretschneider’s inequality bounds the area of any quadrilateral with these sides by the cyclic case:

\begin{aligned} K^2&\le(21-14)(21-9) \\ &\qquad\cdot(21-7)(21-12) \\ &=7\cdot12\cdot14\cdot9, \\ K&\le42\sqrt6. \end{aligned}

Therefore

r\le \frac{K}{21}\le2\sqrt6.

Equality is attainable: the cyclic quadrilateral with these sides is also tangential because

14+7=9+12,

and its incircle has radius

\frac{K}{21}=2\sqrt6.

Thus, the correct answer is C.

Problem 24 in other years