https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Palisade&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T01:30:09ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_12&diff=1044222019 AIME I Problems/Problem 122019-03-15T02:39:21Z<p>Palisade: /* Solution */</p>
<hr />
<div>The 2019 AIME I takes place on March 13, 2019.<br />
<br />
==Problem 12==<br />
Given <math>f(z) = z^2-19z</math>, there are complex numbers <math>z</math> with the property that <math>z</math>, <math>f(z)</math>, and <math>f(f(z))</math> are the vertices of a right triangle in the complex plane with a right angle at <math>f(z)</math>. There are positive integers <math>m</math> and <math>n</math> such that one such value of <math>z</math> is <math>m+\sqrt{n}+11i</math>. Find <math>m+n</math>.<br />
<br />
==Solution==<br />
<br />
We will use the fact that segments <math>AB</math> and <math>BC</math> are perpendicular in the complex plane if and only if <math>\frac{a-b}{b-c}\in i\mathbb{R}</math>. To prove this, when dividing two complex numbers you subtract the angle of one from the other, and if the two are perpendicular, subtracting these angles will yield an imaginary number with no real part. <br />
<br />
Now to apply this: <br />
<cmath>\frac{f(z)-z}{f(f(z))-f(z)}\in i\mathbb{R}</cmath><br />
<cmath>\frac{z^2-19z-z}{(z^2-19z)^2-19(z^2-19z)-(z^2-19z)}</cmath><br />
<cmath>\frac{z^2-20z}{z^4-38z^3+341z^2+380z}</cmath><br />
<cmath>\frac{z(z-20)}{z(z+1)(z-19)(z-20)}</cmath><br />
<cmath>\frac{1}{(z+1)(z-19)}\in i\mathbb{R}</cmath><br />
<br />
The factorization of the nasty denominator above is made easier with the intuition that <math>(z-20)</math> must be a divisor for the problem to lead anywhere. Now we know <math>(z+1)(z-19)\in i\mathbb{R}</math> so using the fact that the imaginary part of <math>z</math> is <math>11i</math> and calling the real part r, <br />
<br />
<cmath>(r+1+11i)(r-19+11i)\in i\mathbb{R}</cmath><br />
<cmath>r^2-18r-140=0</cmath><br />
<br />
solving the above quadratic yields <math>r=9+\sqrt{221}</math> so our answer is <math>9+221=\boxed{230}</math><br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=I|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Palisadehttps://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_12&diff=1044072019 AIME I Problems/Problem 122019-03-15T02:18:02Z<p>Palisade: /* Problem 12 */</p>
<hr />
<div>The 2019 AIME I takes place on March 13, 2019.<br />
<br />
==Problem 12==<br />
Given <math>f(z) = z^2-19z</math>, there are complex numbers <math>z</math> with the property that <math>z</math>, <math>f(z)</math>, and <math>f(f(z))</math> are the vertices of a right triangle in the complex plane with a right angle at <math>f(z)</math>. There are positive integers <math>m</math> and <math>n</math> such that one such value of <math>z</math> is <math>m+\sqrt{n}+11i</math>. Find <math>m+n</math>.<br />
<br />
==Solution==<br />
==See Also==<br />
{{AIME box|year=2019|n=I|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Palisadehttps://artofproblemsolving.com/wiki/index.php?title=2003_AIME_I_Problems/Problem_10&diff=1011282003 AIME I Problems/Problem 102019-02-01T02:10:40Z<p>Palisade: /* Solutions */</p>
<hr />
<div>== Problem ==<br />
[[Triangle]] <math> ABC </math> is [[isosceles triangle | isosceles]] with <math> AC = BC </math> and <math> \angle ACB = 106^\circ. </math> Point <math> M </math> is in the interior of the triangle so that <math> \angle MAC = 7^\circ </math> and <math> \angle MCA = 23^\circ. </math> Find the number of degrees in <math> \angle CMB. </math><br />
<center><asy><br />
pointpen = black; pathpen = black+linewidth(0.7); size(220);<br />
<br />
/* We will WLOG AB = 2 to draw following */<br />
<br />
pair A=(0,0), B=(2,0), C=(1,Tan(37)), M=IP(A--(2Cos(30),2Sin(30)),B--B+(-2,2Tan(23)));<br />
<br />
D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(A--D(MP("M",M))--B); D(C--M);<br />
</asy></center><br />
<br />
<br />
__TOC__<br />
== Solutions ==<br />
<center><asy><br />
pointpen = black; pathpen = black+linewidth(0.7); size(220);<br />
<br />
/* We will WLOG AB = 2 to draw following */<br />
<br />
pair A=(0,0), B=(2,0), C=(1,Tan(37)), M=IP(A--(2Cos(30),2Sin(30)),B--B+(-2,2Tan(23)));<br />
<br />
D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(A--D(MP("M",M))--B); D(C--M);<br />
</asy></center><br />
<br />
=== Solution 1 ===<br />
<center><asy><br />
pointpen = black; pathpen = black+linewidth(0.7); size(220);<br />
<br />
/* We will WLOG AB = 2 to draw following */<br />
<br />
pair A=(0,0), B=(2,0), C=(1,Tan(37)), M=IP(A--(2Cos(30),2Sin(30)),B--B+(-2,2Tan(23))), N=(2-M.x,M.y);<br />
<br />
D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(A--D(MP("M",M))--B); D(C--M); D(C--D(MP("N",N))--B--N--M,linetype("6 6")+linewidth(0.7)); <br />
</asy></center><br />
<br />
Take point <math>N</math> inside <math>\triangle ABC</math> such that <math>\angle CBN = 7^\circ</math> and <math>\angle BCN = 23^\circ</math>.<br />
<br />
<math>\angle MCN = 106^\circ - 2\cdot 23^\circ = 60^\circ</math>. Also, since <math>\triangle AMC</math> and <math>\triangle BNC</math> are congruent (by ASA), <math>CM = CN</math>. Hence <math>\triangle CMN</math> is an [[equilateral triangle]], so <math>\angle CNM = 60^\circ</math>.<br />
<br />
Then <math>\angle MNB = 360^\circ - \angle CNM - \angle CNB = 360^\circ - 60^\circ - 150^\circ = 150^\circ</math>. We now see that <math>\triangle MNB</math> and <math>\triangle CNB</math> are congruent. Therefore, <math>CB = MB</math>, so <math>\angle CMB = \angle MCB = \boxed{83^\circ}</math>.<br />
<br />
=== Solution 2 ===<br />
From the givens, we have the following [[angle]] [[measure]]s: <math>m\angle AMC = 150^\circ</math>, <math>m\angle MCB = 83^\circ</math>. If we define <math>m\angle CMB = \theta</math> then we also have <math>m\angle CBM = 97^\circ - \theta</math>. Then apply the [[Law of Sines]] to triangles <math>\triangle AMC</math> and <math>\triangle BMC</math> to get<br />
<br />
<cmath>\frac{\sin 150^\circ}{\sin 7^\circ} = \frac{AC}{CM} = \frac{BC}{CM} = \frac{\sin \theta}{\sin (97^\circ - \theta)}</cmath><br />
<br />
Clearing [[denominator]]s, evaluating <math>\sin 150^\circ = \frac 12</math> and applying one of our [[trigonometric identities]] to the result gives<br />
<br />
<cmath>\frac{1}{2} \cos (7^\circ - \theta )= \sin 7^\circ \sin \theta</cmath><br />
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and multiplying through by 2 and applying the [[double angle formula]] gives<br />
<br />
<cmath>\cos 7^\circ\cos\theta + \sin7^\circ\sin\theta = 2 \sin7^\circ \sin\theta</cmath><br />
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and so <math>\cos 7^\circ \cos \theta = \sin 7^\circ \sin\theta \Longleftrightarrow \tan 7^{\circ} = \cot \theta</math>; since <math>0^\circ < \theta < 180^\circ</math>, we must have <math>\theta = 83^\circ</math>, so the answer is <math>\boxed{83}</math>.<br />
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=== Solution 3 ===<br />
[[Without loss of generality]], let <math>AC = BC = 1</math>. Then, using the [[Law of Sines]] in triangle <math>AMC</math>, we get <math>\frac {1}{\sin 150} = \frac {MC}{\sin 7}</math>, and using the sine addition formula to evaluate <math>\sin 150 = \sin (90 + 60)</math>, we get <math>MC = 2 \sin 7</math>. <br />
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Then, using the [[Law of Cosines]] in triangle <math>MCB</math>, we get <math>MB^2 = 4\sin^2 7 + 1 - 4\sin 7(\cos 83) = 1</math>, since <math>\cos 83 = \sin 7</math>. So triangle <math>MCB</math> is isosceles, and <math>\angle CMB = \boxed{83}</math>.<br />
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=== Solution 4 ===<br />
Note: A diagram would be much appreciated; I cannot make one since I'm bad at asymptote. Also, please make this less cluttered :) ~tauros<br />
<br />
First, take point <math>E</math> outside of <math>\triangle{ABC}</math> so that <math>\triangle{CEB}</math> is equilateral. Then, connect <math>A</math>, <math>C</math>, and <math>M</math> to <math>E</math>. Also, let <math>ME</math> intersect <math>AB</math> at <math>F</math>. <math>\angle{MCE} = 83^\circ - 60^\circ = 23^\circ</math>, <math>CE = AB</math>, and (trivially) <math>CM = CM</math>, so <math>\triangle{MCE} \cong <br />
\triangle{MCA}</math> by SAS congruence. Also, <math>\angle{CMA} = \angle{CME} = 150^\circ</math>, so <math>\angle{AME} = 60^\circ</math>, and <math>AM = ME</math>,<br />
making <math>\triangle{AME}</math> also equilateral. (it is isosceles with a <math>60^\circ</math> angle) <math>\triangle{MAF} \cong \triangle{EAF}</math> by SAS (<math>MA = AE</math>, <br />
<math>AF = AF</math>, and <math>m\angle{MAF} = m\angle{EAF} = 30^\circ</math>), and <math>\triangle{MAB} \cong \triangle{EAB}</math> by SAS (<math>MA = AE</math>, <math>AB = AB</math>, and <br />
<math>m\angle{MAB} = m\angle{EAB} = 30^\circ</math>). Thus, <math>\triangle{BME}</math> is isosceles, with <math>m\angle{BME} = m\angle{BEM} = 60^\circ + 7^\circ = 67^\circ</math>. Also, <math>\angle{EMB} + \angle {CMB} = \angle{CME} = 150^\circ</math>, so <math>\angle{CME} = 150^\circ - 67^\circ = \boxed{83^\circ}</math>.<br />
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=== Solution 5 (Ceva) ===<br />
<br />
Noticing that we have three concurrent cevians, we apply Ceva's theorem:<br />
<br />
<cmath>(\sin \angle ACM)(\sin \angle BAM)(\sin \angle CBM) = (\sin \angle CAM)(\sin \angle ABM)(\sin \angle BCM) </cmath><br />
<cmath>(\sin 23)(\sin 30)(\sin x) = (\sin 7)(\sin 37-x)(\sin 83)</cmath><br />
<br />
using the fact that <math>\sin 83 = \cos 7</math> and <math>(\sin 7)(\cos 7) = 1/2 (\sin 14)</math> we have:<br />
<br />
<cmath> (\sin 23)(\sin x) = (\sin 14)(\sin 37-x)</cmath><br />
<br />
By inspection, <math>x=14^\circ</math> works, so the answer is <math>180-83-14= \boxed{083}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2003|n=I|num-b=9|num-a=11}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
[[Category:Intermediate Trigonometry Problems]]<br />
{{MAA Notice}}</div>Palisadehttps://artofproblemsolving.com/wiki/index.php?title=1984_AHSME_Problems/Problem_29&diff=1009371984 AHSME Problems/Problem 292019-01-27T05:58:26Z<p>Palisade: /* Solution */</p>
<hr />
<div>==Problem==<br />
Find the largest value for <math> \frac{y}{x} </math> for pairs of [[real numbers]] <math> (x, y) </math> which satisfy <math> (x-3)^2+(y-3)^2=6 </math>.<br />
<br />
<math> \mathrm{(A) \ }3+2\sqrt{2} \qquad \mathrm{(B) \ }2+\sqrt{3} \qquad \mathrm{(C) \ } 3\sqrt{3} \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 6+2\sqrt{3} </math><br />
<br />
==Solution==<br />
Let <math> \frac{y}{x}=k </math>, so that <math> y=kx </math>. Substituting this into the given [[equation]] <math> (x-3)^2+(y-3)^2=6 </math> yields <math> (x-3)^2+(kx-3)^2=6 </math>. Multiplying this out and forming it into a [[quadratic]] yields <math> (1+k^2)x^2+(-6-6k)x+12=0 </math>. <br />
<br />
We want <math> x </math> to be a [[real number]], so we must have the [[discriminant]] <math> \geq0 </math>. <br />
The discriminant is <math> (-6-6k)^2-4(1+k^2)(12)=36k^2+72k+36-48-48k^2=-12k^2+72k-12 </math>. Therefore, we must have <math> -12k^2+72k-12\geq0 </math>, or <math> k^2-6k+1\leq0 </math>. The [[Root (polynomial)|roots]] of this quadratic, using the [[quadratic formula]], are <math> 3\pm2\sqrt{2} </math>, so the quadratic can be [[Factoring|factored]] as <math> (k-(3-2\sqrt{2}))(k-(3+2\sqrt{2}))\leq0 </math>. We can now separate this into <math> 3 </math> cases:<br />
<br />
'''Case 1: <math> k<3-2\sqrt{2} </math>'''<br />
Then, both [[Term|terms]] in the factored quadratic are negative, so the inequality doesn't hold.<br />
<br />
'''Case 2: <math> 3-2\sqrt{2}<k<3+2\sqrt{2} </math>'''<br />
Then, the first term is positive and the second is negative and the second is positive, so the inequality holds.<br />
<br />
'''Case 3: <math> k>3+2\sqrt{2} </math>'''<br />
Then, both terms are positive, so the inequality doesn't hold.<br />
<br />
Also, when <math> k=3-2\sqrt{2} </math> or <math> k=3+2\sqrt{2} </math>, the equality holds.<br />
<br />
Therefore, we must have <math> 3-2\sqrt{2}\leq k\leq3+2\sqrt{2} </math>, and the maximum value of <math> k=\frac{y}{x} </math> is <math> {3+2\sqrt{2}}, \boxed{\text{A}} </math>.<br />
<br />
==Solution 2==<br />
The equation represents a circle of radius <math>\sqrt{6}</math> centered at <math>A=(3,3)</math>. To find the maximal <math>k</math> with <math>y=kx</math> is equivalent to finding the maximum slope of a line passing through the origin <math>O</math> and intersecting the circle. The steepest such line is tangent to the circle at some point <math>X</math>. We have <math>XA=\sqrt{6}</math>, <math>OA=\sqrt{18}</math>, <math>\angle OXA = 90</math> because the line is tangent to the circle. Using the pythagorean theorem, we have <math>OX=\sqrt{12}</math>. <br />
<br />
The slope we are looking for is equivalent to <math>\tan (\theta + 45)</math> where <math>\angle AOX = \theta</math>. Using tangent addition,<br />
<br />
<cmath> \tan (\theta + 45)= \frac{\tan \theta + \tan 45}{1-\tan\theta\tan 45} = \frac{\frac{1}{\sqrt{2}}+1}{1-\frac{1}{\sqrt{2}}}=3+2\sqrt{3}</cmath><br />
<br />
So <math>\boxed{A}</math> is the answer<br />
<br />
==See Also==<br />
{{AHSME box|year=1984|num-b=28|num-a=30}}<br />
{{MAA Notice}}</div>Palisade