1990 AHSME Problems/Problem 28

Problem

A quadrilateral that has consecutive sides of lengths $70,90,130$ and $110$ is inscribed in a circle and also has a circle inscribed in it. The point of tangency of the inscribed circle to the side of length 130 divides that side into segments of length $x$ and $y$ . Find $|x-y|$ .

$\text{(A) } 12\quad \text{(B) } 13\quad \text{(C) } 14\quad \text{(D) } 15\quad \text{(E) } 16$

Solution

Let $A$ , $B$ , $C$ , and $D$ be the vertices of this quadrilateral such that $AB=70$ , $BC=110$ , $CD=130$ , and $DA=90$ . Let $O$ be the center of the incircle. Draw in the radii from the center of the incircle to the points of tangency. Let these points of tangency $X$ , $Y$ , $Z$ , and $W$ be on $AB$ , $BC$ , $CD$ , and $DA$ , respectively. Using the right angles and the fact that the $ABCD$ is cyclic, we see that quadrilaterals $AXOW$ and $OYCZ$ are similar.

Let $CZ$ have length $n$ . Chasing lengths, we find that $AX=AW=n-40$ . Using Brahmagupta's Formula we find that $ABCD$ has area $K=300\sqrt{1001}$ and from that we find, using that fact that $rs=K$ , where $r$ is the inradius and $s$ is the semiperimeter, $r=\frac{3}{2}\sqrt{1001}$ .

From the similarity we have $\frac{CY}{OX}=\frac{OY}{AX}$ Or, after cross multiplying and writing in terms of the variables, $n^2-40n-r^2=0$ Plugging in the value of $r$ and solving the quadratic gives $n=CZ=71.5$ , and from there we compute the desired difference to get $\fbox{(B) 13}$ .

1990 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1990 AHSME Problems/Problem 28

Problem

Solution

See also