Difference between revisions of "2002 AIME II Problems/Problem 14"
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The [[perimeter]] of triangle <math>APM</math> is <math>152</math>, and the angle <math>PAM</math> is a [[right angle]]. A [[circle]] of [[radius]] <math>19</math> with center <math>O</math> on <math>\overline{AP}</math> is drawn so that it is [[Tangent (geometry)|tangent]] to <math>\overline{AM}</math> and <math>\overline{PM}</math>. Given that <math>OP=m/n</math> where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers, find <math>m+n</math>. | The [[perimeter]] of triangle <math>APM</math> is <math>152</math>, and the angle <math>PAM</math> is a [[right angle]]. A [[circle]] of [[radius]] <math>19</math> with center <math>O</math> on <math>\overline{AP}</math> is drawn so that it is [[Tangent (geometry)|tangent]] to <math>\overline{AM}</math> and <math>\overline{PM}</math>. Given that <math>OP=m/n</math> where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers, find <math>m+n</math>. | ||
− | == Solution 1== | + | == Solution 1 == |
Let the circle intersect <math>\overline{PM}</math> at <math>B</math>. Then note <math>\triangle OPB</math> and <math>\triangle MPA</math> are similar. Also note that <math>AM = BM</math> by [[power of a point]]. Using the fact that the ratio of corresponding sides in similar triangles is equal to the ratio of their perimeters, we have | Let the circle intersect <math>\overline{PM}</math> at <math>B</math>. Then note <math>\triangle OPB</math> and <math>\triangle MPA</math> are similar. Also note that <math>AM = BM</math> by [[power of a point]]. Using the fact that the ratio of corresponding sides in similar triangles is equal to the ratio of their perimeters, we have | ||
<cmath>\frac{19}{AM} = \frac{152-2AM-19+19}{152} = \frac{152-2AM}{152}</cmath> | <cmath>\frac{19}{AM} = \frac{152-2AM-19+19}{152} = \frac{152-2AM}{152}</cmath> | ||
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The <math>x^3</math> cancels, yielding a quadratic. Solving yields <math>x = \frac{38}{3}</math>. | The <math>x^3</math> cancels, yielding a quadratic. Solving yields <math>x = \frac{38}{3}</math>. | ||
Add <math>19</math> to find <math>OP</math>, yielding <math>\frac{95}{3}</math> or <math>\boxed{098}</math>. | Add <math>19</math> to find <math>OP</math>, yielding <math>\frac{95}{3}</math> or <math>\boxed{098}</math>. | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | Let the foot of the perpendicular from <math>O</math> to <math>PM</math> be <math>D;</math> now <math>OD=19.</math> Also let <math>AM=x</math> and <math>PM=y.</math> This means that <math>OP=\frac{y}{x}\cdot 19</math>, since <math>O</math> is on the angle bisector of <math>\angle M.</math> | ||
+ | |||
+ | We have that <math>\tan(\angle AMO)=\frac{19}{x},</math> so | ||
+ | <cmath>\tan(\angle M)=\tan (2\cdot \angle AMO)=\frac{38x}{x^{2}-361}.</cmath> | ||
+ | |||
+ | However <math>\tan(\angle M)=\frac{PA}{AM}=\frac{PO+OA}{AM}=\frac{\frac{y}{x}\cdot 19 + 19}{x}</math>, so | ||
+ | <cmath>\frac{38x}{x^{2}-361}=19\cdot \frac{\frac{y}{x}+1}{x}</cmath> | ||
+ | <cmath>\frac{2x^{2}}{x^{2}-361}=\frac{y}{x}+1</cmath> | ||
+ | <cmath>\frac{x^{2}+361}{x^{2}-361}=\frac{y}{x}.</cmath> | ||
+ | <cmath>x\cdot \frac{x^{2}+361}{x^{2}-361}=y</cmath> | ||
+ | |||
+ | We now use the fact that the perimeter of <math>\triangle PAM</math> is <math>152</math>: | ||
+ | <cmath>PO+OA+AM+MP=152</cmath> | ||
+ | <cmath>\frac{y}{x}\cdot 19+19+x+y=152</cmath> | ||
+ | <cmath>19\left(\frac{x^{2}+361}{x^{2}-361}\right)+x\cdot \left(\frac{x^{2}+361}{x^{2}-361}\right)+x+19=152</cmath> | ||
+ | <cmath>(x+19)\left(\frac{x^{2}+361}{x^{2}-361}+\frac{x^{2}-361}{x^{2}-361}\right)=152</cmath> | ||
+ | <cmath>\frac{2x^{2}}{x-19}=152</cmath> | ||
+ | <cmath>x^{2}-76x+19\cdot 76=0.</cmath> | ||
+ | This quadratic factors as <math>(x-38)^{2}=0,</math> so <math>x=38</math>, and | ||
+ | <cmath>\frac{y}{x}=\frac{38^{2}+361}{38^{2}-361}=\frac{5}{3}</cmath> | ||
+ | <cmath>OP=\frac{y}{x}\cdot 19=\frac{95}{3}\to \boxed{98.}</cmath> | ||
+ | |||
+ | ==Solution 4== | ||
+ | Let <math>K</math> be the foot of the altitude from <math>O</math> to <math>PM,</math> and notice how <math>OK</math> is a radius of the circle. Also, we have that <math>\angle OKM=\angle OAM=90^\circ,</math> <math>OK=OA=90,</math> and <math>OM=OM,</math> so <math>\triangle OKM</math> is congruent to <math>\triangle OAM.</math> This means that because <math>\angle PMO=\angle KMO=\angle AMO,</math> <math>O</math> is the foot of the angle bisector from <math>M.</math> | ||
+ | |||
+ | Then, let <math>k=\angle AMO=\angle PMO.</math> We have that | ||
+ | <cmath>\angle POK=180^\circ-\angle KOM-\angle AOM=2k=\angle PMA,</cmath> | ||
+ | so <math>\triangle PKO</math> is similar to <math>\triangle PAM.</math> Let <math>a=AM=KM,</math> and let <math>p=152</math> be the perimeter of <math>\triangle PAM.</math> By similarity, we have that the perimeter of <math>\triangle PKO</math> is <math>\frac{19}{a}\cdot p,</math> so | ||
+ | <cmath>PK+PO=\dfrac{19}{a}\cdot p-19,</cmath> | ||
+ | so | ||
+ | <cmath>p=PK+PO+OA+AM+MK=\dfrac{19}{a}\cdot p-19+19+2a.</cmath> | ||
+ | Solving for <math>p,</math> we have that | ||
+ | <cmath>p=\dfrac{2a}{1-\frac{19}{a}}</cmath> | ||
+ | and using <math>p=152</math> we find that | ||
+ | <cmath>a^2-76a-38^2=(a-38)^2=0</cmath> | ||
+ | so <math>a=38.</math> | ||
+ | |||
+ | Finally, let <math>b=PO.</math> We have by the angle bisector theorem that <math>PM=2b,</math> so using the perimeter information one more time we get <math>3b+57=152</math> and solving gives us | ||
+ | <cmath>b=\dfrac{95}{3}\implies \boxed{098}.</cmath> | ||
+ | |||
+ | ~BS2012 | ||
== See also == | == See also == |
Latest revision as of 20:41, 3 February 2025
Problem
The perimeter of triangle is
, and the angle
is a right angle. A circle of radius
with center
on
is drawn so that it is tangent to
and
. Given that
where
and
are relatively prime positive integers, find
.
Solution 1
Let the circle intersect at
. Then note
and
are similar. Also note that
by power of a point. Using the fact that the ratio of corresponding sides in similar triangles is equal to the ratio of their perimeters, we have
Solving,
. So the ratio of the side lengths of the triangles is 2. Therefore,
so
and
Substituting for
, we see that
, so
and the answer is
.
Solution 2
Reflect triangle across line
, creating an isoceles triangle. Let
be the distance from the top of the circle to point
, with
as
. Given the perimeter is 152, subtracting the altitude yields the semiperimeter
of the isoceles triangle, as
. The area of the isoceles triangle is:
Now use similarity, draw perpendicular from to
, name the new point
. Triangle
is similar to triangle
, by AA Similarity. Equating the legs, we get:
Solving for , it yields
.
The cancels, yielding a quadratic. Solving yields
.
Add
to find
, yielding
or
.
Solution 3
Let the foot of the perpendicular from to
be
now
Also let
and
This means that
, since
is on the angle bisector of
We have that so
However , so
We now use the fact that the perimeter of is
:
This quadratic factors as
so
, and
Solution 4
Let be the foot of the altitude from
to
and notice how
is a radius of the circle. Also, we have that
and
so
is congruent to
This means that because
is the foot of the angle bisector from
Then, let We have that
so
is similar to
Let
and let
be the perimeter of
By similarity, we have that the perimeter of
is
so
so
Solving for
we have that
and using
we find that
so
Finally, let We have by the angle bisector theorem that
so using the perimeter information one more time we get
and solving gives us
~BS2012
See also
2002 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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