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Difference between revisions of "1990 AHSME Problems/Problem 28"

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== Solution ==
 
== Solution ==
  
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Let <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> be the vertices of this quadrilateral such that <math>AB=70</math>, <math>BC=110</math>, <math>CD=130</math>, and <math>DA=90</math>. Let <math>O</math> 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 <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> be on <math>AB</math>, <math>BC</math>, <math>CD</math>, and <math>DA</math>, respectively. Using the right angles and the fact that the <math>ABCD</math> is cyclic, we see that quadrilaterals <math>AXOW</math> and <math>OYCZ</math> are similar.   
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Let <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> be the vertices of this quadrilateral such that <math>AB=70</math>, <math>BC=110</math>, <math>CD=130</math>, and <math>DA=90</math>. Let <math>O</math> 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 <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> be on <math>AB</math>, <math>BC</math>, <math>CD</math>, and <math>DA</math>, respectively. Using the right angles and the fact that the <math>ABCD</math> is cyclic, we see that quadrilaterals <math>AXOW</math> and <math>OYCZ</math> are similar.   
  
 
Let <math>CZ</math> have length <math>n</math>. Chasing lengths, we find that <math>AX=AW=n-40</math>. Using Brahmagupta's Formula we find that <math>ABCD</math> has area <math>K=300\sqrt{1001}</math> and from that we find, using that fact that <math>rs=K</math>, where <math>r</math> is the inradius and <math>s</math> is the semiperimeter, <math>r=\frac{3}{2}\sqrt{1001}</math>.
 
Let <math>CZ</math> have length <math>n</math>. Chasing lengths, we find that <math>AX=AW=n-40</math>. Using Brahmagupta's Formula we find that <math>ABCD</math> has area <math>K=300\sqrt{1001}</math> and from that we find, using that fact that <math>rs=K</math>, where <math>r</math> is the inradius and <math>s</math> is the semiperimeter, <math>r=\frac{3}{2}\sqrt{1001}</math>.

Revision as of 20:05, 31 July 2016

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 $\fbox{B}$.

See also

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

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