Difference between revisions of "2005 AIME II Problems/Problem 12"
Ramanan369 (talk | contribs) m (→Solution 1 (trigonometry)) |
m (→Solution 9) |
||
(41 intermediate revisions by 14 users not shown) | |||
Line 2: | Line 2: | ||
[[Square]] <math>ABCD </math> has [[center]] <math> O,\ AB=900,\ E </math> and <math> F </math> are on <math> AB </math> with <math> AE<BF </math> and <math> E </math> between <math> A </math> and <math> F, m\angle EOF =45^\circ, </math> and <math> EF=400. </math> Given that <math> BF=p+q\sqrt{r}, </math> where <math> p,q, </math> and <math> r </math> are [[positive]] [[integer]]s and <math> r </math> is not divisible by the [[square]] of any [[prime]], find <math> p+q+r. </math> | [[Square]] <math>ABCD </math> has [[center]] <math> O,\ AB=900,\ E </math> and <math> F </math> are on <math> AB </math> with <math> AE<BF </math> and <math> E </math> between <math> A </math> and <math> F, m\angle EOF =45^\circ, </math> and <math> EF=400. </math> Given that <math> BF=p+q\sqrt{r}, </math> where <math> p,q, </math> and <math> r </math> are [[positive]] [[integer]]s and <math> r </math> is not divisible by the [[square]] of any [[prime]], find <math> p+q+r. </math> | ||
__TOC__ | __TOC__ | ||
+ | |||
== Solutions == | == Solutions == | ||
=== Solution 1 (trigonometry) === | === Solution 1 (trigonometry) === | ||
Line 34: | Line 35: | ||
<math>x=250\pm 50\sqrt{7}</math>. Since <math>BF > AE,</math> we take the positive root, and our answer is <math>p+q+r = 250 + 50 + 7 = 307</math>. | <math>x=250\pm 50\sqrt{7}</math>. Since <math>BF > AE,</math> we take the positive root, and our answer is <math>p+q+r = 250 + 50 + 7 = 307</math>. | ||
− | === Solution 3 === | + | === Solution 3 (similar triangles)=== |
<asy> | <asy> | ||
size(3inch); | size(3inch); | ||
Line 50: | Line 51: | ||
<cmath>(OF)^2 = 450^2 + (450 - x)^2.</cmath> | <cmath>(OF)^2 = 450^2 + (450 - x)^2.</cmath> | ||
Setting these two expressions for <math>(OF)^2</math> equal and solving for <math>x</math> (it is helpful to scale the problem down by a factor of 50 first), we get <math>x = 250\pm 50\sqrt{7}</math>. Since <math>BF > AE</math>, we want the value <math>x = 250 + 50\sqrt{7}</math>, and the answer is <math>250 + 50 + 7 = \boxed{307}</math>. | Setting these two expressions for <math>(OF)^2</math> equal and solving for <math>x</math> (it is helpful to scale the problem down by a factor of 50 first), we get <math>x = 250\pm 50\sqrt{7}</math>. Since <math>BF > AE</math>, we want the value <math>x = 250 + 50\sqrt{7}</math>, and the answer is <math>250 + 50 + 7 = \boxed{307}</math>. | ||
+ | |||
+ | === Solution 4 (Abusing Stewart) === | ||
+ | Let <math>x = BF</math>, so <math>AE = 500-x</math>. Let <math>a = OE</math>, <math>b = OF</math>. Applying Stewart's Theorem on triangles <math>AOB</math> twice, first using <math>E</math> as the base point and then <math>F</math>, we arrive at the equations <cmath>(450 \sqrt{2})^2 (900) = 900(500-x)(400+x) + a^2 (900)</cmath> and <cmath>(450 \sqrt{2})^2 (900) = 900x(900-x) + b^2 (900)</cmath> Now applying law of sines and law of cosines on <math>\triangle EOF</math> yields <cmath>\frac{1}{2} ab \sin 45^{\circ} = \frac{4}{9} \times \frac{1}{4} \times 900^2 = 202500</cmath> and <cmath>a^2+b^2- 2 ab \cos 45^{\circ} = 160000</cmath> Solving for <math>ab</math> from the sines equation and plugging into the law of cosines equation yields <math>a^2+b^2 = 290000</math>. We now finish by adding the two original stewart equations and obtaining: <cmath>2(450\sqrt{2})^2 = (500-x)(400+x)+x(900-x)+520000</cmath> This is a quadratic which only takes some patience to solve for <math>x = 250 + 50\sqrt{7}</math> | ||
+ | |||
+ | === Solution 5 (Complex Numbers) === | ||
+ | Let lower case letters be the complex numbers correspond to their respective upper case points in the complex plane, with <math>o = 0, a = -450 + 450i, b = 450 + 450i</math>, and <math>f = x + 450i</math>. Since <math>EF</math> = 400, <math>e = (x-400) + 450i</math>. From <math>\angle{EOF} = 45^{\circ}</math>, we can deduce that the rotation of point <math>F</math> 45 degrees counterclockwise, <math>E</math>, and the origin are collinear. In other words, <cmath>\dfrac{e^{i \frac{\pi}{4}} \cdot (x + 450i)}{(x - 400) + 450i}</cmath> is a real number. Simplyfying using the fact that <math>e^{i \frac{\pi}{4}} = \dfrac{\sqrt{2}}{2} + i \dfrac{\sqrt{2}}{2}</math>, clearing the denominator, and setting the imaginary part equal to <math>0</math>, we eventually get the quadratic <cmath>x^2 - 400x + 22500 = 0</cmath> which has solutions <math>x = 200 \pm 50\sqrt{7}</math>. It is given that <math>AE < BF</math>, so <math>x = 200 - 50\sqrt{7}</math> and <cmath>BF = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7} \Rightarrow \boxed{307}.</cmath> | ||
+ | |||
+ | -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 such that it lies on AB, and GOE is 90 degrees. Let H be foot of the altitude from O to AB. | ||
+ | Since <math>\triangle GOE \sim \triangle OHE</math>, <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 <math>\triangle GAK \sim \triangle GHO</math>. | ||
+ | |||
+ | <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 | ||
+ | |||
+ | === Solution 7 (Using a Circle) === | ||
+ | <center><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=(4.5-0.5*sqrt(7),7), O=(4.5,4.5), H=(4.5-0.5*sqrt(7),4.5), I=(0,4.5), J=(4.5-0.5*sqrt(7),9); | ||
+ | draw(A--B--C--D--A); | ||
+ | draw(E--O--F); | ||
+ | draw(J--G); | ||
+ | draw(E--G--F); | ||
+ | draw(G--H--O--G); | ||
+ | draw(I--H); | ||
+ | draw(circle(G,2*sqrt(2))); | ||
+ | markscalefactor=0.05; | ||
+ | draw(rightanglemark(E,G,F)); | ||
+ | dot(A^^B^^C^^D^^E^^F^^G^^H^^I^^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("\(H\)",H,(-1,1)); | ||
+ | label("\(I\)",I,(-1,0)); | ||
+ | label("\(J\)",J,(0,1)); | ||
+ | label("\(O\)",O,(1,-1)); | ||
+ | </asy></center> | ||
+ | |||
+ | We know that G is on the perpendicular bisector of <math>EF</math>, which means that <math>EJ=JF=200</math>, <math>EG=GF=200\sqrt{2}</math> and <math>GH=250</math>. Now, let <math>HO</math> be equal to <math>x</math>. We can set up an equation with the Pythagorean Theorem: | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \sqrt{x^2+250^2}&=(200\sqrt{2})^2 \\ | ||
+ | x^2+62500&=80000 \\ | ||
+ | x^2&=17500 \\ | ||
+ | x&=50\sqrt{7} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Now, since <math>IO=450</math>, | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | HI&=450-x \\ | ||
+ | &=450-50\sqrt{7} \\ | ||
+ | \end{align*} \\ | ||
+ | </cmath> | ||
+ | |||
+ | Since <math>HI=AJ</math>, we now have: | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | BF&=AB-AJ-JF \\ | ||
+ | &=900-(450-50\sqrt{7})-200 \\ | ||
+ | &=250+50\sqrt{7} \\ | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | This means that our answer would be <math>250+50+7=\boxed{307}</math> | ||
+ | |||
+ | ~Jerry_Guo | ||
+ | |||
+ | === Solution 8 (More Similar Triangles) === | ||
+ | |||
+ | Construct <math>BO, AO.</math> Let <math>\angle{FOB} = \alpha.</math> Also let <math>FB = x</math> then <math>AE = 500-x.</math> We then have from simple angle-chasing: | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \angle{BFO} = 135 - \alpha \\ | ||
+ | \angle{OFE} = 45 + \alpha \\ | ||
+ | \angle{EOA} = 45 - \alpha \\ | ||
+ | \angle{AEO} = 90 + \alpha \\ | ||
+ | \angle{OEF} = 90 - \alpha. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | From AA similarity we have <cmath>\triangle{EOB} \sim \triangle{EFO}.</cmath> This gives the ratios, | ||
+ | <cmath>\dfrac{400 + x}{EO} = \dfrac{450\sqrt{2}}{FO}.</cmath> | ||
+ | Similarly from AA similarity <cmath>\triangle{FOA} \sim \triangle{FEO}.</cmath> So we get the ratios <cmath>\dfrac{EO}{450\sqrt{2}} = \dfrac{FO}{900-x}.</cmath> We can multiply to get | ||
+ | <cmath>\dfrac{400 + x}{450\sqrt{2}} = \dfrac{450\sqrt{2}}{900 - x}.</cmath> Cross-multiplying reveals | ||
+ | <cmath>360000 + 500x - x^2 = 405000.</cmath> Bringing everything to one side we have <cmath>x^2 - 500x + 45000 = 0.</cmath> By the quadratic formula we get <cmath>x = \dfrac{500 + \sqrt{500^2 - 4\cdot45000}}{2} = \dfrac{500 + \sqrt{70000}}{2} = \dfrac{500 + 100\sqrt{7}}{2} = 250 + 50\sqrt{7}.</cmath> | ||
+ | Therefore | ||
+ | <cmath>p+q+r = 250 + 50 + 7 = \boxed{307}.</cmath> | ||
+ | ~aa1024 | ||
+ | |||
+ | |||
+ | === Solution 9 === | ||
+ | We use ratio lemma and Stewart's theorem: | ||
+ | Connect <math>OA, OE, OF, OB</math> and let <math>AE = x</math> and <math>BF = 500 - x.</math> | ||
+ | Let angle <math>AOE = y,</math> hence <math>BOF = 45 - y.</math> | ||
+ | Now, we apply Stewart's theorem in triangles <math>AOF</math> and <math>BOE</math> to get <math>OE</math> and <math>OF</math> in terms of <math>x</math> | ||
+ | finally, calculate <math>x/400</math> and <math>500-x/400</math> using ratio lemma to find <math>x</math> and <math>y</math> | ||
+ | |||
+ | === Solution 10(Similar Triangles) === | ||
+ | Draw AO, OB, and extend OB to D. Let <math>\angle{FOB} = \alpha.</math> Then, after angle chasing, we find that <cmath>\angle{AEB} = 90 + \alpha</cmath>. | ||
+ | Using this, we draw a line perpendicular to <math>AB</math> at <math>E</math> to meet <math>BD</math> at <math>M</math>. Since <math>\angle{MEO} = \alpha</math> and <math>\angle{EMO} = 45</math>, we have that <cmath>\triangle{EMO} \sim \triangle{OBF}</cmath> | ||
+ | Let <math>FB = x</math>. Then <math>EM = 400+x</math>. Since <math>FB/BO = \frac{x}{450\sqrt{2}}</math>, and <math>MO/EM = FB/OB</math>, we have <cmath>MO = \frac{(400+x)x}{450\sqrt{2}}</cmath> | ||
+ | Since <math>\triangle{EBM}</math> is a <math>45-45-90</math> triangle, <cmath>(400+x)\sqrt{2} = 450 | ||
+ | \sqrt{2} + \frac{(400+x)x}{450\sqrt{2}}</cmath> | ||
+ | Solving for <math>x</math>, we get that <math>x=250 +- 50s\sqrt{7}</math>, but since <math>FB>AE</math>, <math>FB = 250+50\sqrt{7}</math>, thus <cmath>p+q+r=\boxed{307}</cmath> | ||
+ | -dchang0524 | ||
== See also == | == See also == |
Latest revision as of 13:14, 23 July 2024
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 such that it lies on AB, and GOE is 90 degrees. Let H be foot of the altitude from O to AB.
Since
,
, and by Angle Bisector Theorem,
. Thus,
.
, and
(90 degree rotation), and now we can bash on 2 similar triangles
.
I hope you like expanding
Quadratic formula gives us
Since AE < BF
Thus,
So, our answer is
.
-AlexLikeMath
Solution 7 (Using a Circle)
![[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=(4.5-0.5*sqrt(7),7), O=(4.5,4.5), H=(4.5-0.5*sqrt(7),4.5), I=(0,4.5), J=(4.5-0.5*sqrt(7),9); draw(A--B--C--D--A); draw(E--O--F); draw(J--G); draw(E--G--F); draw(G--H--O--G); draw(I--H); draw(circle(G,2*sqrt(2))); markscalefactor=0.05; draw(rightanglemark(E,G,F)); dot(A^^B^^C^^D^^E^^F^^G^^H^^I^^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("\(H\)",H,(-1,1)); label("\(I\)",I,(-1,0)); label("\(J\)",J,(0,1)); label("\(O\)",O,(1,-1)); [/asy]](http://latex.artofproblemsolving.com/7/1/3/713c04919c6f36e471616770a350f6b25006331c.png)
We know that G is on the perpendicular bisector of , which means that
,
and
. Now, let
be equal to
. We can set up an equation with the Pythagorean Theorem:
Now, since ,
Since , we now have:
This means that our answer would be
~Jerry_Guo
Solution 8 (More Similar Triangles)
Construct Let
Also let
then
We then have from simple angle-chasing:
From AA similarity we have
This gives the ratios,
Similarly from AA similarity
So we get the ratios
We can multiply to get
Cross-multiplying reveals
Bringing everything to one side we have
By the quadratic formula we get
Therefore
~aa1024
Solution 9
We use ratio lemma and Stewart's theorem:
Connect and let
and
Let angle
hence
Now, we apply Stewart's theorem in triangles
and
to get
and
in terms of
finally, calculate
and
using ratio lemma to find
and
Solution 10(Similar Triangles)
Draw AO, OB, and extend OB to D. Let Then, after angle chasing, we find that
.
Using this, we draw a line perpendicular to
at
to meet
at
. Since
and
, we have that
Let
. Then
. Since
, and
, we have
Since
is a
triangle,
Solving for
, we get that
, but since
,
, thus
-dchang0524
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.