Difference between revisions of "2005 AIME II Problems/Problem 12"
(→Solution 4 (Abusing Stewart)) |
Alexlikemath (talk | contribs) (→Solution 5 (Complex Numbers)) |
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-MP8148 | -MP8148 | ||
+ | |||
+ | === Solution 6 === | ||
+ | <asy> | ||
+ | size(250); | ||
+ | pair A,B,C,D,O,E,F,G,H,K; | ||
+ | A = (0,0); | ||
+ | B = (900,0); | ||
+ | C = (900,900); | ||
+ | D = (0,900); | ||
+ | O = (450,450); | ||
+ | E = (600,0); | ||
+ | F = (150,0); | ||
+ | G = (-600,0); | ||
+ | H = (450,0); | ||
+ | K = (0,270); | ||
+ | draw(A--B--C--D--cycle); | ||
+ | draw(O--E); | ||
+ | draw(O--F); | ||
+ | draw(O--G); | ||
+ | draw(A--G); | ||
+ | draw(O--H); | ||
+ | label("O",O,N); | ||
+ | label("A",A,S); | ||
+ | label("B",B,SE); | ||
+ | label("C",C,NE); | ||
+ | label("D",D,NW); | ||
+ | label("E",E,SE); | ||
+ | label("F",F,S); | ||
+ | label("H",H,SW); | ||
+ | label("G",G,SW); | ||
+ | label("x",H--E,S); | ||
+ | label("K",K,NW); | ||
+ | </asy> | ||
+ | Let G be a point s.t. it lies on AB, and GOE is 90 degrees. Let H be foot of the altitude from O to AB. | ||
+ | Since GOE ~ OHE, <math>\frac{GO}{OE} = \frac{450}{x}</math>, and by [[Angle Bisector Theorem]], <math>\frac{GF}{FE} = \frac{450}{x}</math>. Thus, <math>GF = \frac{450 \cdot 400}{x}</math>. <math>AF = AH-FH = 50+x</math>, and <math>KA = EB</math> (90 degree rotation), and now we can bash on 2 similar triangles GAK ~ GHO. | ||
+ | |||
+ | <cmath>\frac{GA}{AK} = \frac{GH}{OH}</cmath> | ||
+ | <cmath>\frac{\frac{450 \cdot 400}{x}-50-x}{450-x} = \frac{\frac{450 \cdot 400}{x}+400-x}{450}</cmath> | ||
+ | I hope you like expanding | ||
+ | <cmath>x^2 - 850x + \frac{81000000}{x} = -450x - 22500 + \frac{81000000}{x}</cmath> | ||
+ | <cmath>x^2 - 400x + 22500 = 0</cmath> | ||
+ | Quadratic formula gives us | ||
+ | <cmath>x = 200 \pm 50 \sqrt{7}</cmath> | ||
+ | Since AE < BF | ||
+ | <cmath>x = 200 - 50 \sqrt{7}</cmath> | ||
+ | Thus, | ||
+ | <cmath>BF = 250 + 50 \sqrt{7}</cmath> | ||
+ | So, our answer is <math>\boxed{307}</math>. | ||
+ | |||
+ | -AlexLikeMath | ||
== See also == | == See also == |
Revision as of 01:08, 13 December 2019
Problem
Square has center
and
are on
with
and
between
and
and
Given that
where
and
are positive integers and
is not divisible by the square of any prime, find
Contents
Solutions
Solution 1 (trigonometry)
![[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), G=(4.5,9), O=(4.5,4.5); draw(A--B--C--D--A);draw(E--O--F);draw(G--O); dot(A^^B^^C^^D^^E^^F^^G^^O); label("\(A\)",A,(-1,1));label("\(B\)",B,(1,1));label("\(C\)",C,(1,-1));label("\(D\)",D,(-1,-1)); label("\(E\)",E,(0,1));label("\(F\)",F,(1,1));label("\(G\)",G,(-1,1));label("\(O\)",O,(1,-1)); label("\(x\)",E/2+G/2,(0,1));label("\(y\)",G/2+F/2,(0,1)); label("\(450\)",(O+G)/2,(-1,1)); [/asy]](http://latex.artofproblemsolving.com/8/0/8/808a781ca9b8e518e9c3eb70c404e134125a1765.png)
Let be the foot of the perpendicular from
to
. Denote
and
, and
(since
and
). Then
, and
.
By the tangent addition rule , we see that
Since
, this simplifies to
. We know that
, so we can substitute this to find that
.
Substituting again, we know have
. This is a quadratic with roots
. Since
, use the smaller root,
.
Now, . The answer is
.
Solution 2 (synthetic)
![[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), O=(4.5,4.5), G=O+(E-O)*dir(-90), J=O+(F-O)*dir(-90); draw(A--B--C--D--A);draw(E--O--F);draw(G--O--J);draw(F--G,linetype("4 4")); dot(A^^B^^C^^D^^E^^F^^G^^J^^O); label("\(A\)",A,(-1,1));label("\(B\)",B,(1,1));label("\(C\)",C,(1,-1));label("\(D\)",D,(-1,-1)); label("\(E\)",E,(0,1));label("\(F\)",F,(1,1));label("\(G\)",G,(1,0));label("\(J\)",J,(1,0));label("\(O\)",O,(1,-1)); label("\(x\)",(B+F)/2,(0,1)); label("\(400\)",(E+F)/2,(0,1)); label("\(900\)",(C+D)/2,(0,-1)); [/asy]](http://latex.artofproblemsolving.com/7/6/4/7643375bc11ea98eeb7c78a01a433681f95b800e.png)
Label , so
. Rotate
about
until
lies on
. Now we know that
therefore
also since
is the center of the square. Label the new triangle that we created
. Now we know that rotation preserves angles and side lengths, so
and
. Draw
and
. Notice that
since rotations preserve the same angles so
too. By SAS we know that
so
. Now we have a right
with legs
and
and hypotenuse
. By the Pythagorean Theorem,
and applying the quadratic formula we get that
. Since
we take the positive root, and our answer is
.
Solution 3 (similar triangles)
Let the midpoint of
be
and let
, so then
and
. Drawing
, we have
, so
By the Pythagorean Theorem on
,
Setting these two expressions for
equal and solving for
(it is helpful to scale the problem down by a factor of 50 first), we get
. Since
, we want the value
, and the answer is
.
Solution 4 (Abusing Stewart)
Let , so
. Let
,
. Applying Stewart's Theorem on triangles
twice, first using
as the base point and then
, we arrive at the equations
and
Now applying law of sines and law of cosines on
yields
and
Solving for
from the sines equation and plugging into the law of cosines equation yields
. We now finish by adding the two original stewart equations and obtaining:
This is a quadratic which only takes some patience to solve for
Solution 5 (Complex Numbers)
Let lower case letters be the complex numbers correspond to their respective upper case points in the complex plane, with , and
. Since
= 400,
. From
, we can deduce that the rotation of point
45 degrees counterclockwise,
, and the origin are collinear. In other words,
is a real number. Simplyfying using the fact that
, clearing the denominator, and setting the imaginary part equal to
, we eventually get the quadratic
which has solutions
. It is given that
, so
and
-MP8148
Solution 6
Let G be a point s.t. it lies on AB, and GOE is 90 degrees. Let H be foot of the altitude from O to AB.
Since GOE ~ OHE,
, and by Angle Bisector Theorem,
. Thus,
.
, and
(90 degree rotation), and now we can bash on 2 similar triangles GAK ~ GHO.
I hope you like expanding
Quadratic formula gives us
Since AE < BF
Thus,
So, our answer is
.
-AlexLikeMath
See also
2005 AIME II (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.