Difference between revisions of "1990 AHSME Problems/Problem 28"
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== Solution == | == Solution == | ||
− | + | 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>. | ||
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<cmath>\frac{CY}{OX}=\frac{OY}{AX}</cmath> | <cmath>\frac{CY}{OX}=\frac{OY}{AX}</cmath> | ||
Or, after cross multiplying and writing in terms of the variables, <cmath>n^2-40n-r^2=0</cmath> | Or, after cross multiplying and writing in terms of the variables, <cmath>n^2-40n-r^2=0</cmath> | ||
− | Plugging in the value of <math>r</math> and solving the quadratic gives <math>n=CZ=71.5</math>, and from there we compute the desired difference to get <math>\fbox{B}</math>. | + | Plugging in the value of <math>r</math> and solving the quadratic gives <math>n=CZ=71.5</math>, and from there we compute the desired difference to get <math>\fbox{(B) 13}</math>. |
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== See also == | == See also == |
Latest revision as of 11:51, 5 October 2019
Problem
A quadrilateral that has consecutive sides of lengths and
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
and
. Find
.
Solution
Let ,
,
, and
be the vertices of this quadrilateral such that
,
,
, and
. Let
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
,
,
, and
be on
,
,
, and
, respectively. Using the right angles and the fact that the
is cyclic, we see that quadrilaterals
and
are similar.
Let have length
. Chasing lengths, we find that
. Using Brahmagupta's Formula we find that
has area
and from that we find, using that fact that
, where
is the inradius and
is the semiperimeter,
.
From the similarity we have
Or, after cross multiplying and writing in terms of the variables,
Plugging in the value of
and solving the quadratic gives
, and from there we compute the desired difference to get
.
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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