Difference between revisions of "2011 AMC 12A Problems/Problem 24"
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By Pitot's Theorem, since <math>AB+CD=AD+BC</math>, there exists a circle tangent to all four sides of quadrilateral <math>ABCD</math>. Let <math>E</math>, <math>F</math>, <math>G</math>, and <math>H</math> be the points on <math>AB</math>, <math>BC</math>, <math>CD</math>, and <math>DA</math> respectively where the circle is tangent. Let the inscribed circle have center <math>O</math> and radius <math>r</math>. Note that <math>OE</math>, <math>OF</math>, <math>OG</math>, and <math>OH</math> are perpendicular to sides <math>AB</math>, <math>BC</math>, <math>CD</math>, and <math>DA</math>, respectively. Thus <math>\angle AOH=\angle AOE</math>, <math>\angle BOE=\angle BOF</math>, <math>\angle COF=\angle COG</math>, and <math>\angle DOG=\angle DOH</math>. | By Pitot's Theorem, since <math>AB+CD=AD+BC</math>, there exists a circle tangent to all four sides of quadrilateral <math>ABCD</math>. Let <math>E</math>, <math>F</math>, <math>G</math>, and <math>H</math> be the points on <math>AB</math>, <math>BC</math>, <math>CD</math>, and <math>DA</math> respectively where the circle is tangent. Let the inscribed circle have center <math>O</math> and radius <math>r</math>. Note that <math>OE</math>, <math>OF</math>, <math>OG</math>, and <math>OH</math> are perpendicular to sides <math>AB</math>, <math>BC</math>, <math>CD</math>, and <math>DA</math>, respectively. Thus <math>\angle AOH=\angle AOE</math>, <math>\angle BOE=\angle BOF</math>, <math>\angle COF=\angle COG</math>, and <math>\angle DOG=\angle DOH</math>. | ||
+ | |||
From <math>\angle AOH+\angle AOE+\angle BOE+\angle BOF+\angle COF+\angle COG+\angle DOG+\angle DOH=360^{\circ}</math>, we get the two equations <math>\angle AOE+\angle BOE+\angle COG+\angle DOG=180^{\circ}</math> and <math>\angle AOH+\angle DOH+\angle BOF+\angle COF=180^{\circ}</math>. These are equivalent to <math>\angle AOB+\angle COD=180^{\circ}</math> and <math>\angle AOD+\angle BOC=180^{\circ}</math>. | From <math>\angle AOH+\angle AOE+\angle BOE+\angle BOF+\angle COF+\angle COG+\angle DOG+\angle DOH=360^{\circ}</math>, we get the two equations <math>\angle AOE+\angle BOE+\angle COG+\angle DOG=180^{\circ}</math> and <math>\angle AOH+\angle DOH+\angle BOF+\angle COF=180^{\circ}</math>. These are equivalent to <math>\angle AOB+\angle COD=180^{\circ}</math> and <math>\angle AOD+\angle BOC=180^{\circ}</math>. | ||
+ | |||
The ratio of the area of <math>AOB</math> to <math>COD</math> is <math>2</math>, since the heights are both <math>r</math>. However, we can also express this ratio as <math>\frac{\frac{1}{2}\times AO\times BO\times \sin (AOB)}{\frac{1}{2}\times CO\times DO\times \sin (COD)}</math>. Since <math>\angle AOB=180^{\circ}-\angle COD</math>, <math>\sin (AOB)=\sin (COD)</math>. Thus <math>\frac{AO\times BO}{CO\times DO}=2</math>. Similarly, <math>\frac{AO\times DO}{BO\times CO}=\frac{12}{9}=\frac{4}{3}</math>. From these two equations we have <math>BO=\frac{\sqrt{6}}{2}\times DO</math>. | The ratio of the area of <math>AOB</math> to <math>COD</math> is <math>2</math>, since the heights are both <math>r</math>. However, we can also express this ratio as <math>\frac{\frac{1}{2}\times AO\times BO\times \sin (AOB)}{\frac{1}{2}\times CO\times DO\times \sin (COD)}</math>. Since <math>\angle AOB=180^{\circ}-\angle COD</math>, <math>\sin (AOB)=\sin (COD)</math>. Thus <math>\frac{AO\times BO}{CO\times DO}=2</math>. Similarly, <math>\frac{AO\times DO}{BO\times CO}=\frac{12}{9}=\frac{4}{3}</math>. From these two equations we have <math>BO=\frac{\sqrt{6}}{2}\times DO</math>. | ||
− | Now let <math>AH=AE=a</math>, <math>BE=BF=b</math>, <math>CF=CG=c</math>, and <math>DG=DH=d</math>. We thus have the equations <math>a+b=14</math>, <math>b+c=9</math>, <math>c+d=7</math>, and <math>d+a=12</math>. Solving gives us <math>a=8</math>, <math>b=6</math>, <math>c=3</math>, and <math>d=4</math>. By the Pythagorean theorem, we have <math>BO^2-b^2=DO^2-d^2</math>. This is <math>DO^2\times\frac{3}{2}-36=DO^2-16</math>, which gives us <math>DO^2=40</math>. Thus, <math>r=\sqrt{DO^2-d^2}=\sqrt{24}</math>, and <math>r=\boxed{\textbf{(C)}\ 2\sqrt{6}}</math>. | + | |
+ | Now let <math>AH=AE=a</math>, <math>BE=BF=b</math>, <math>CF=CG=c</math>, and <math>DG=DH=d</math>. We thus have the equations <math>a+b=14</math>, <math>b+c=9</math>, <math>c+d=7</math>, and <math>d+a=12</math>. Solving gives us <math>a=8</math>, <math>b=6</math>, <math>c=3</math>, and <math>d=4</math>. By the Pythagorean theorem, we have <math>BO^2-b^2=DO^2-d^2</math>. This is <math>DO^2\times\frac{3}{2}-36=DO^2-16</math>, which gives us <math>DO^2=40</math>. Thus, <math>r=\sqrt{DO^2-d^2}=\sqrt{24}</math>, and <math>r=\boxed{\textbf{(C)}\ 2\sqrt{6}}</math>. ~Awesome_guy | ||
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=23|num-a=25|ab=A}} | {{AMC12 box|year=2011|num-b=23|num-a=25|ab=A}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:29, 28 December 2019
Contents
Problem
Consider all quadrilaterals such that
,
,
, and
. What is the radius of the largest possible circle that fits inside or on the boundary of such a quadrilateral?
Solution
Solution 1
Note as above that ABCD must be tangential to obtain the circle with maximal radius. Let ,
,
, and
be the points on
,
,
, and
respectively where the circle is tangent. Let
and
. Since the quadrilateral is cyclic,
and
. Let the circle have center
and radius
. Note that
,
,
, and
are right angles.
Hence ,
,
, and
.
Therefore, and
.
Let . Then
,
,
, and
. Using
and
we have
, and
. By equating the value of
from each,
. Solving we obtain
so that
.
Solution 2
To maximize the radius of the circle, we also need to maximize its area. To maximize the area of the circle, the quadrilateral must be tangential (have an incircle). In a tangential quadrilateral, the sum of opposite sides is equal to the semiperimeter of the quadrilateral. , so this particular quadrilateral has an incircle. By definition, given
side lengths, a cyclic quadrilateral has the maximum area of any quadrilateral with those side lengths. Therefore, to maximize the area of the quadrilateral and thus the incircle, we assume that this quadrilateral is cyclic.
For cyclic quadrilaterals, Brahmagupta's formula gives the area as where
is the semiperimeter and
and
are the side lengths. Breaking it up into
triangles, we see the area of a tangential quadrilateral is also equal to
. Equate these two equations.
Substituting
, the semiperimeter, and
, the area and solving for
,we get
.
Solution 3 (Trigonometry)
By Pitot's Theorem, since , there exists a circle tangent to all four sides of quadrilateral
. Thus, we need to find the radius of this circle.
Let the circle be tangent to ,
,
, and
at points
,
,
, and
, respectively. Also, let
. Then
also equals
. Let the center of the circle be
. Observe that
bisects angle
, so
. Moreover,
. But
, so
, and we find that
. Hence,
.
Similarly, , and
. Therefore,
. Putting this together with the above equation yields
Thus,
, and
.
Solution 4 (Areas)
By Pitot's Theorem, since , there exists a circle tangent to all four sides of quadrilateral
. Let
,
,
, and
be the points on
,
,
, and
respectively where the circle is tangent. Let the inscribed circle have center
and radius
. Note that
,
,
, and
are perpendicular to sides
,
,
, and
, respectively. Thus
,
,
, and
.
From , we get the two equations
and
. These are equivalent to
and
.
The ratio of the area of to
is
, since the heights are both
. However, we can also express this ratio as
. Since
,
. Thus
. Similarly,
. From these two equations we have
.
Now let ,
,
, and
. We thus have the equations
,
,
, and
. Solving gives us
,
,
, and
. By the Pythagorean theorem, we have
. This is
, which gives us
. Thus,
, and
. ~Awesome_guy
See also
2011 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 | |
All AMC 12 Problems and Solutions |
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