Difference between revisions of "2009 AIME I Problems/Problem 12"
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In right <math>\triangle ABC</math> with hypotenuse <math>\overline{AB}</math>, <math>AC = 12</math>, <math>BC = 35</math>, and <math>\overline{CD}</math> is the altitude to <math>\overline{AB}</math>. Let <math>\omega</math> be the circle having <math>\overline{CD}</math> as a diameter. Let <math>I</math> be a point outside <math>\triangle ABC</math> such that <math>\overline{AI}</math> and <math>\overline{BI}</math> are both tangent to circle <math>\omega</math>. The ratio of the perimeter of <math>\triangle ABI</math> to the length <math>AB</math> can be expressed in the form <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | In right <math>\triangle ABC</math> with hypotenuse <math>\overline{AB}</math>, <math>AC = 12</math>, <math>BC = 35</math>, and <math>\overline{CD}</math> is the altitude to <math>\overline{AB}</math>. Let <math>\omega</math> be the circle having <math>\overline{CD}</math> as a diameter. Let <math>I</math> be a point outside <math>\triangle ABC</math> such that <math>\overline{AI}</math> and <math>\overline{BI}</math> are both tangent to circle <math>\omega</math>. The ratio of the perimeter of <math>\triangle ABI</math> to the length <math>AB</math> can be expressed in the form <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | ||
− | == Solution == | + | == Solution 1== |
+ | Let <math>O</math> be center of the circle and <math>P</math>,<math>Q</math> be the two points of tangent such that <math>P</math> is on <math>BI</math> and <math>Q</math> is on <math>AI</math>. We know that <math>AD:CD = CD:BD = 12:35</math>. | ||
+ | |||
+ | Since the ratios between corresponding lengths of two similar diagrams are equal, we can let <math>AD = 144, CD = 420</math> and <math>BD = 1225</math>. Hence <math>AQ = 144, BP = 1225, AB = 1369</math> and the radius <math>r = OD = 210</math>. | ||
+ | |||
+ | Since we have <math>\tan OAB = \frac {35}{24}</math> and <math>\tan OBA = \frac{6}{35}</math> , we have <math>\sin {(OAB + OBA)} = \frac {1369}{\sqrt {(1801*1261)}},</math><math>\cos {(OAB + OBA)} = \frac {630}{\sqrt {(1801*1261)}}</math>. | ||
+ | |||
+ | Hence <math>\sin I = \sin {(2OAB + 2OBA)} = \frac {2*1369*630}{1801*1261}</math>. let <math>IP = IQ = x</math> , then we have Area<math>(IBC)</math> = <math>(2x + 1225*2 + 144*2)*\frac {210}{2}</math> = <math>(x + 144)(x + 1225)* \sin {\frac {I}{2}}</math>. Then we get <math>x + 1369 = \frac {3*1369*(x + 144)(x + 1225)}{1801*1261}</math>. | ||
+ | |||
+ | Now the equation looks very complex but we can take a guess here. Assume that <math>x</math> is a rational number | ||
+ | (If it's not then the answer to the problem would be irrational which can't be in the form of <math>\frac {m}{n}</math>) | ||
+ | that can be expressed as <math>\frac {a}{b}</math> such that <math>(a,b) = 1</math>. Look at both sides; we can know that <math>a</math> has to be a multiple of <math>1369</math> and not of <math>3</math> and it's reasonable to think that <math>b</math> is divisible by <math>3</math> so that we can cancel out the <math>3</math> on the right side of the equation. | ||
+ | |||
+ | Let's see if <math>x = \frac {1369}{3}</math> fits. Since <math>\frac {1369}{3} + 1369 = \frac {4*1369}{3}</math>, and <math>\frac {3*1369*(x + 144)(x + 1225)}{1801*1261} = \frac {3*1369* \frac {1801}{3} * \frac {1261*4}{3}} {1801*1261} = \frac {4*1369}{3}</math>. Amazingly it fits! | ||
+ | |||
+ | Since we know that <math>3*1369*144*1225 - 1369*1801*1261 < 0</math>, the other solution of this equation is negative which can be ignored. Hence <math>x = 1369/3</math>. | ||
+ | |||
+ | Hence the perimeter is <math>1225*2 + 144*2 + \frac {1369}{3} *2 = 1369* \frac {8}{3}</math>, and <math>BC</math> is <math>1369</math>. Hence <math>\frac {m}{n} = \frac {8}{3}</math>, <math>m + n = 11</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | As in Solution <math>1</math>, let <math>P</math> and <math>Q</math> be the intersections of <math>\omega</math> with <math>BI</math> and <math>AI</math> respectively. | ||
+ | |||
+ | First, by pythagorean theorem, <math>AB = \sqrt{12^2+35^2} = 37</math>. Now the area of <math>ABC</math> is <math>1/2*12*35 = 1/2*37*CD</math>, so <math>CD=\frac{420}{37}</math> and the inradius of <math>\triangle ABI</math> is <math>r=\frac{210}{37}</math>. | ||
+ | |||
+ | Now from <math>\triangle CDB \sim \triangle ACB</math> we find that <math>\frac{BC}{BD} = \frac{AB}{BC}</math> so <math>BD = BC^2/AB = 35^2/37</math> and similarly, <math>AD = 12^2/37</math>. | ||
+ | |||
+ | Note <math>IP=IQ=x</math>, <math>BP=BD</math>, and <math>AQ=AD</math>. So we have <math>AI = 144/37+x</math>, <math>BI = 1225/37+x</math>. Now we can compute the area of <math>\triangle ABI</math> in two ways: by heron's formula and by inradius times semiperimeter, which yields | ||
+ | |||
+ | <math>rs=210/37(37+x) = \sqrt{(37+x)(37-144/37)(37-1225/37)x}</math> | ||
+ | |||
+ | <math>210/37(37+x) = 12*35/37 \sqrt{x(37+x)}</math> | ||
+ | |||
+ | <math>37+x = 2 \sqrt{x(x+37)}</math> | ||
+ | |||
+ | <math>x^2+74x+1369 = 4x^2 + 148x</math> | ||
+ | |||
+ | <math>3x^2 + 74x - 1369 = 0</math> | ||
+ | |||
+ | The quadratic formula now yields <math>x=37/3</math>. Plugging this back in, the perimeter of <math>ABI</math> is <math>2s=2(37+x)=2(37+37/3) = 37(8/3)</math> so the ratio of the perimeter to <math>AB</math> is <math>8/3</math> and our answer is <math>8+3=\boxed{011}</math> | ||
+ | |||
+ | ==Solution 3== | ||
+ | As in Solution <math>2</math>, let <math>P</math> and <math>Q</math> be the intersections of <math>\omega</math> with <math>BI</math> and <math>AI</math> respectively. | ||
+ | |||
+ | Recall that the distance from a point outside a circle to that circle is the same along both tangent lines to the circle drawn from the point. | ||
+ | |||
+ | Recall also that the length of the altitude to the hypotenuse of a right-angle triangle is the geometric mean of the two segments into which it cuts the hypotenuse. | ||
+ | |||
+ | Let <math>x = \overline{AD} = \overline{AQ}</math>. Let <math>y = \overline{BD} = \overline{BP}</math>. Let <math>z = \overline{PI} = \overline{QI}</math>. The semi-perimeter of <math>ABI</math> is <math>x + y + z</math>. | ||
+ | Since the lengths of the sides of <math>ABI</math> are <math>x + y</math>, <math>y + z</math> and <math>x + z</math>, the square of its area by Heron's formula is <math>(x+y+z)xyz</math>. | ||
+ | |||
+ | The radius <math>r</math> of <math>\omega</math> is <math>\overline{CD}/2</math>. Therefore <math>r^2 = xy/4</math>. As <math>\omega</math> is the in-circle of <math>ABI</math>, the area of <math>ABI</math> is also <math>r(x+y+z)</math>, and so the square area is <math>r^2(x+y+z)^2</math>. | ||
+ | |||
+ | Therefore <cmath>(x+y+z)xyz = r^2(x+y+z)^2 = \frac{xy(x+y+z)^2}{4}</cmath> Dividing both sides by <math>xy(x+y+z)/4</math> we get: <cmath>4z = (x+y+z),</cmath> and so <math>z = (x+y)/3</math>. The semi-perimeter of <math>ABI</math> is therefore <math>\frac{4}{3}(x+y)</math> and the whole perimeter is <math>\frac{8}{3}(x+y)</math>. Now <math>x + y = \overline{AB}</math>, so the ratio of the perimeter of <math>ABI</math> to the hypotenuse <math>\overline{AB}</math> is <math>8/3</math> and our answer is <math>8+3=\boxed{011}</math> | ||
+ | |||
+ | == Solution 4 == | ||
<asy> | <asy> | ||
− | size( | + | size(300); |
− | + | defaultpen(linewidth(0.4)+fontsize(10)); | |
− | + | pen s = linewidth(0.8)+fontsize(8); | |
− | |||
− | |||
− | real | + | pair A,B,C,D,O,X; |
− | + | C=origin; | |
− | + | A=(0,12); | |
− | + | B=(18,0); | |
− | + | D=foot(C,A,B); | |
− | + | O = (C+D)/2; | |
+ | real r = length(D-C)/2; | ||
+ | path c = CR(O, r); | ||
+ | pair OA = (O+A)/2; | ||
+ | real rA = length(A-O)/2; | ||
+ | pair Ap = OP(CR(OA,rA), c); | ||
+ | pair OB = (O+B)/2; | ||
+ | real rB = length(B-O)/2; | ||
+ | pair Bp = OP(CR(OB,rB), c); | ||
+ | X=extension(A,Ap,B,Bp); | ||
+ | draw(A--B--C--A, s); | ||
+ | draw(C--D^^B--O--A^^Ap--O--X, gray+0.25); | ||
+ | draw(c^^A--X--B); | ||
− | + | dot("$A$", A, N); | |
− | + | dot("$B$", B, SE); | |
− | + | dot("$C$", C, SW); | |
− | + | dot("$D$", D, 0.2*(D-C)); | |
− | + | dot("$I$", X, 0.5*(X-C)); | |
− | + | dot("$P$", Ap, 0.3*(Ap-O)); | |
− | + | dot("$Q$", Bp, 0.3*(Bp-O)); | |
− | + | dot("$O$", O, W); | |
− | + | label("$\beta$",B,10*dir(157)); | |
− | + | label("$\alpha$",A,5*dir(-55)); | |
− | + | label("$\theta$",X,5*dir(55)); | |
− | |||
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− | label( | ||
− | |||
− | |||
− | |||
− | label( | ||
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− | label( | ||
</asy> | </asy> | ||
− | + | Let <math>AP=AD=x</math>, let <math>BQ=BD=y</math>, and let <math>IP=IQ=z</math>. Let <math>OD=r</math>. We find <math>AB=37</math>. Let <math>\alpha</math>, <math>\beta</math>, and <math>\theta</math> be the angles <math>OAD</math>, <math>OBD</math>, and <math>OPI</math> respectively. Then <math>\alpha + \beta + \theta = 90^\circ</math>, so <cmath>\theta = 90^\circ - (\alpha+\beta).</cmath> | |
− | <cmath>\tan \alpha=\frac{ | + | The perimeter of <math>\triangle ABI</math> is <math>2(x+y+z)=2(37+z)</math>. The desired ratio is then |
− | + | <cmath>\rho = 2\left(1+\frac z{37}\right)</cmath> | |
− | + | We need to find <math>z</math>. In <math>\triangle OPI</math>, <math>z=r\cot\theta = r\tan (\alpha+\beta)</math>. We get <cmath>\tan\alpha = \frac{OD}{AD} = \frac 12 \frac{CD}{AD} = \frac 12 \tan A = \frac 12 \frac{BC}{AC} = \frac{35}{24}.</cmath> Similarly, <math>\tan\beta = \tfrac 6{35}</math>. Then <cmath>z = r\cdot \tan (\alpha+\beta) = r\cdot \frac{\tan\alpha + \tan\beta}{1-\tan\alpha\tan\beta}= \frac{37^2\cdot r}{18\cdot 35}</cmath> | |
− | + | Computing <math>[ABC]</math> in two ways we get <math>CD = \tfrac{12\cdot 35}{37}</math>, so <math>r=\tfrac{6\cdot 35}{37}</math>. Using this value of <math>r</math> we get <math>z=\tfrac {37}3</math>. Thus <cmath>\rho = 2\left(1+\frac 1{3}\right) = \frac 8{3},</cmath> | |
− | < | + | and <math>8+3=\boxed{011}</math>. |
− | \ | + | |
− | + | == Solution 5 == | |
− | + | This solution is not a real solution and is solving the problem with a ruler and compass. | |
− | + | ||
− | + | Draw <math>AC = 4.8, BC = 14, AB = 14.8</math>. Then, drawing the tangents and intersecting them, we get that <math>IA</math> is around <math>6.55</math> and <math>IB</math> is around <math>18.1</math>. We then find the ratio to be around <math>\frac{39.45}{14.8}</math>. Using long division, we find that this ratio is approximately 2.666, which you should recognize as <math>\frac{8}{3}</math>. Since this seems reasonable, we find that the answer is <math>\boxed{11}</math> ~ilp | |
− | + | ||
− | + | == Solution 6== | |
− | + | Denoting three tangents has length <math>h_1,h_2,h_3</math> while <math>h_1,h_3</math> lies on <math>AB</math> with <math>h_1>h_3</math>.The area of <math>ABC</math> is <math>1/2*12*35 = 1/2*37*CD</math>, so <math>CD=\frac{420}{37}</math> and the inradius of <math>\triangle ABI</math> is <math>r=\frac{210}{37}</math>.As we know that the diameter of the circle is the height of <math>\triangle ACB</math> from <math>C</math> to <math>AB</math>. Assume that <math>\tan\alpha=\frac{h_1}{r}</math> and <math>\tan\beta=\frac{h_3}{r}</math> and <math>\tan\omega=\frac{h_2}{r}</math>. But we know that <math>\tan(\alpha+\beta)=-\tan(180-\alpha-\beta)=-\tan\omega</math> According to the basic computation, we can get that <math>\tan(\alpha)=\frac{35}{6}</math>; <math>\tan(\beta)=\frac{24}{35}</math> | |
− | < | + | So we know that <math>\tan(\omega)=\frac{1369}{630}</math> according to the tangent addition formula. Hence, it is not hard to find that the length of <math>h_2</math> is <math>\frac{37}{3}</math>. According to basic addition and division, we get the answer is <math>\frac{8}{3}</math> which leads to <math>8+3=\boxed{11}</math> ~bluesoul |
− | |||
− | < | ||
− | |||
− | < | ||
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− | < | ||
== See also == | == See also == | ||
− | |||
{{AIME box|year=2009|n=I|num-b=11|num-a=13}} | {{AIME box|year=2009|n=I|num-b=11|num-a=13}} | ||
+ | [[Category: Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 18:51, 25 November 2022
Contents
Problem
In right with hypotenuse
,
,
, and
is the altitude to
. Let
be the circle having
as a diameter. Let
be a point outside
such that
and
are both tangent to circle
. The ratio of the perimeter of
to the length
can be expressed in the form
, where
and
are relatively prime positive integers. Find
.
Solution 1
Let be center of the circle and
,
be the two points of tangent such that
is on
and
is on
. We know that
.
Since the ratios between corresponding lengths of two similar diagrams are equal, we can let and
. Hence
and the radius
.
Since we have and
, we have
.
Hence . let
, then we have Area
=
=
. Then we get
.
Now the equation looks very complex but we can take a guess here. Assume that is a rational number
(If it's not then the answer to the problem would be irrational which can't be in the form of
)
that can be expressed as
such that
. Look at both sides; we can know that
has to be a multiple of
and not of
and it's reasonable to think that
is divisible by
so that we can cancel out the
on the right side of the equation.
Let's see if fits. Since
, and
. Amazingly it fits!
Since we know that , the other solution of this equation is negative which can be ignored. Hence
.
Hence the perimeter is , and
is
. Hence
,
.
Solution 2
As in Solution , let
and
be the intersections of
with
and
respectively.
First, by pythagorean theorem, . Now the area of
is
, so
and the inradius of
is
.
Now from we find that
so
and similarly,
.
Note ,
, and
. So we have
,
. Now we can compute the area of
in two ways: by heron's formula and by inradius times semiperimeter, which yields
The quadratic formula now yields . Plugging this back in, the perimeter of
is
so the ratio of the perimeter to
is
and our answer is
Solution 3
As in Solution , let
and
be the intersections of
with
and
respectively.
Recall that the distance from a point outside a circle to that circle is the same along both tangent lines to the circle drawn from the point.
Recall also that the length of the altitude to the hypotenuse of a right-angle triangle is the geometric mean of the two segments into which it cuts the hypotenuse.
Let . Let
. Let
. The semi-perimeter of
is
.
Since the lengths of the sides of
are
,
and
, the square of its area by Heron's formula is
.
The radius of
is
. Therefore
. As
is the in-circle of
, the area of
is also
, and so the square area is
.
Therefore Dividing both sides by
we get:
and so
. The semi-perimeter of
is therefore
and the whole perimeter is
. Now
, so the ratio of the perimeter of
to the hypotenuse
is
and our answer is
Solution 4
Let
, let
, and let
. Let
. We find
. Let
,
, and
be the angles
,
, and
respectively. Then
, so
The perimeter of
is
. The desired ratio is then
We need to find
. In
,
. We get
Similarly,
. Then
Computing
in two ways we get
, so
. Using this value of
we get
. Thus
and
.
Solution 5
This solution is not a real solution and is solving the problem with a ruler and compass.
Draw . Then, drawing the tangents and intersecting them, we get that
is around
and
is around
. We then find the ratio to be around
. Using long division, we find that this ratio is approximately 2.666, which you should recognize as
. Since this seems reasonable, we find that the answer is
~ilp
Solution 6
Denoting three tangents has length while
lies on
with
.The area of
is
, so
and the inradius of
is
.As we know that the diameter of the circle is the height of
from
to
. Assume that
and
and
. But we know that
According to the basic computation, we can get that
;
So we know that
according to the tangent addition formula. Hence, it is not hard to find that the length of
is
. According to basic addition and division, we get the answer is
which leads to
~bluesoul
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.