https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mapletree14&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T11:59:44ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_23&diff=598082008 AMC 12B Problems/Problem 232014-02-17T15:07:41Z<p>Mapletree14: </p>
<hr />
<div>==Problem 23==<br />
The sum of the base-<math>10</math> logarithms of the divisors of <math>10^n</math> is <math>792</math>. What is <math>n</math>?<br />
<br />
<math>\text{(A)}\ 11\qquad \text{(B)}\ 12\qquad \text{(C)}\ 13\qquad \text{(D)}\ 14\qquad \text{(E)}\ 15</math><br />
<br />
__TOC__<br />
==Solution==<br />
=== Solution 1 ===<br />
Every factor of <math>10^n</math> will be of the form <math>2^a \times 5^b , a\leq n , b\leq n</math>. Using the logarithmic property <math>\log(a \times b) = \log(a)+\log(b)</math>, it suffices to count the total number of 2's and 5's running through all possible <math>(a,b)</math>. For every factor <math>2^a \times 5^b</math>, there will be another <math>2^b \times 5^a</math>, so it suffices to count the total number of 2's occurring in all factors (because of this symmetry, the number of 5's will be equal). And since <math>\log(2)+\log(5) = \log(10) = 1</math>, the final sum will be the total number of 2's occurring in all factors of <math>10^n</math>.<br />
<br />
There are <math>n+1</math> choices for the exponent of 5 in each factor, and for each of those choices, there are <math>n+1</math> factors (each corresponding to a different exponent of 2), yielding <math>0+1+2+3...+n = \frac{n(n+1)}{2}</math> total 2's. The total number of 2's is therefore <math>\frac{n \cdot(n+1)^2}{2} = \frac{n^3+2n^2+n}{2}</math>. Plugging in our answer choices into this formula yields 11 (answer choice <math>\mathrm{(A)}</math>) as the correct answer.<br />
<br />
<br />
===Solution 2 (number-theoretic bash)===<br />
<br />
We are given <cmath> \log_{10}d_1 + \log_{10}d_2 + \ldots + \log_{10}d_k = 792 </cmath> The property <math>\log(ab) = \log(a)+\log(b)</math> now gives <cmath> \log_{10}(d_1 d_2\cdot\ldots d_k) = 792 </cmath> The product of the divisors is (from elementary number theory) <math>a^{d(n)/2}</math> where <math>d(n)</math> is the number of divisors. Note that <math>10^n = 2^n\cdot 5^n</math>, so* <math>d(n) = (n + 1)^2</math>. Substituting these values with <math>a = 10^n</math> in our equation above, we get <math>n(n + 1)^2 = 1584</math>, from whence we immediately obtain <math>\framebox{11 \,\mathrm{(A)}}</math> as the correct answer.<br />
<br />
<br />
*This is by use of a well-known formula for <math>d(n)</math>. <br />
<br />
=== Solution 3 ===<br />
For every divisor <math>d</math> of <math>10^n</math>, <math>d \le \sqrt{10^n}</math>, we have <math>\log d + \log \frac{10^n}{d} = \log 10^n = n</math>. There are <math>\left \lfloor \frac{(n+1)^2}{2} \right \rfloor</math> divisors of <math>10^n = 2^n \times 5^n</math> that are <math>\le \sqrt{10^n}</math>. After casework on the parity of <math>n</math>, we find that the answer is given by <math>n \times \frac{(n+1)^2}{2} = 792 \Longrightarrow n = 11\ \mathrm{(A)}</math>. <br />
<br />
==See Also==<br />
{{AMC12 box|year=2008|ab=B|num-b=22|num-a=24}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Mapletree14https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_14&diff=589652013 AIME I Problems/Problem 142014-01-30T03:04:15Z<p>Mapletree14: /* Solution 1 */</p>
<hr />
<div>== Problem 14 ==<br />
For <math>\pi \le \theta < 2\pi</math>, let<br />
<br />
<math>\begin{align*}</math><br />
<math>P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots</math><br />
<math>\end{align*}</math><br />
<br />
and<br />
<br />
<math>\begin{align*}</math><br />
<math>Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots</math><br />
<math>\end{align*}</math><br />
<br />
so that <math>\frac{P}{Q} = \frac{2\sqrt2}{7}</math>. Then <math>\sin\theta = -\frac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
== Solution ==<br />
===Solution 1===<br />
<math>\begin{align*}</math><br />
<math>P \sin\theta\ + Q \cos\theta\ = \cos\theta\ - \frac{1}{2}\ P</math><br />
<math>\end{align*}</math><br />
and<br />
<math>\begin{align*}</math><br />
<math>P \cos\theta\ + Q \sin\theta\ = -2(Q-1)</math><br />
<math>\end{align*}</math><br />
<br />
Solving for P, Q we have <br />
<br />
<br />
<math>\frac{P}{Q} = \frac{\cos\theta }{2 + \sin\theta } = \frac{2\sqrt2}{7}</math><br />
<br />
Square both side, and use polynomial rational root theorem to solve <math>\sin\theta</math><br />
<br />
<math>\sin\theta = -\frac{17}{19} </math><br />
<br />
The answer is <math>\boxed{036}</math><br />
<br />
===Solution 2===<br />
<math>\begin{align*}</math><br />
Use sum to product formulas to rewrite <math>P</math> and <math>Q</math> <br />
<math>\end{align*}</math><br />
<br />
<math>P \sin\theta\ + Q \cos\theta\ = \cos \theta\ - \frac{1}{4}\cos \theta + \frac{1}{8}\sin 2\theta + \frac{1}{16}\cos 3\theta - \frac{1}{32}\sin 4\theta + ... </math><br />
<br />
<math>\begin{align*}</math><br />
Therefore, <math>P \sin \theta - Q \cos \theta = -2P</math><br />
<math>\end{align*}</math><br />
<br />
<math>\begin{align*}</math><br />
Using <math>\frac{P}{Q} = \frac{2\sqrt2}{7}</math>, <math>Q = \frac{7}{2\sqrt2} P</math><br />
<math>\end{align*}</math><br />
<br />
<math>\begin{align*}</math><br />
Plug in to the previous equation and cancel out the "P" terms to get: <math>\sin\theta - \frac{7}{2\sqrt2} \cos\theta = -2</math><br />
<math>\end{align*}</math><br />
<br />
Then use the pythagorean identity to solve for <math>\sin\theta</math>, <math>\sin\theta = \boxed{036}</math><br />
<br />
===Solution 3===<br />
<br />
Note that <cmath>e^{i\theta}=\cos(\theta)+i\sin(\theta)</cmath><br />
<br />
Thus, the following identities follow immediately:<br />
<cmath>ie^{i\theta}=i(\cos(\theta)+i\sin(\theta))=-\sin(\theta)+i\cos(\theta)</cmath><br />
<cmath>i^2 e^{i\theta}=-e^{i\theta}=-\cos(\theta)-i\sin(\theta)</cmath><br />
<cmath>i^3 e^{i\theta}=\sin(\theta)-i\cos(\theta)</cmath><br />
<br />
Consider, now, the sum <math>Q+iP</math>. It follows fairly immediately that:<br />
<br />
<cmath>Q+iP=1+\left(\frac{i}{2}\right)^1e^{i\theta}+\left(\frac{i}{2}\right)^2e^{2i\theta}+\ldots=\frac{1}{1-\frac{i}{2}e^{i\theta}}=\frac{2}{2-ie^{i\theta}}</cmath><br />
<cmath>Q+iP=\frac{2}{2-ie^{i\theta}}=\frac{2}{2-(-\sin(\theta)+i\cos(\theta))}=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}</cmath><br />
<br />
This follows straight from the geometric series formula and simple simplification. We can now multiply the denominator by it's complex conjugate to find:<br />
<br />
<cmath>Q+iP=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}\left(\frac{(2+\sin(\theta))+i\cos(\theta)}{(2+\sin(\theta))+i\cos(\theta)}\right)</cmath><br />
<cmath>Q+iP=\frac{2((2+\sin(\theta))+i\cos(\theta))}{(2+\sin(\theta))^2+\cos^2(\theta)}</cmath><br />
<br />
Comparing real and imaginary parts, we find:<br />
<cmath>\frac{P}{Q}=\frac{\cos(\theta)}{2+\sin(\theta)}=\frac{2\sqrt{2}}{7}</cmath><br />
<br />
Squaring this equation and letting <math>\sin^2(\theta)=x</math>:<br />
<br />
<math>\frac{P^2}{Q^2}=\frac{\cos^2(\theta)}{4+4\sin(\theta)+\sin^2(\theta)}=\frac{1-x^2}{4+4x+x^2}=\frac{8}{49}</math><br />
<br />
Clearing denominators and solving for <math>x</math> gives sine as <math>x=-\frac{17}{19}</math>.<br />
<br />
<math>017+019=\boxed{036}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2013|n=I|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Mapletree14https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_14&diff=589642013 AIME I Problems/Problem 142014-01-30T03:03:09Z<p>Mapletree14: </p>
<hr />
<div>== Problem 14 ==<br />
For <math>\pi \le \theta < 2\pi</math>, let<br />
<br />
<math>\begin{align*}</math><br />
<math>P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots</math><br />
<math>\end{align*}</math><br />
<br />
and<br />
<br />
<math>\begin{align*}</math><br />
<math>Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots</math><br />
<math>\end{align*}</math><br />
<br />
so that <math>\frac{P}{Q} = \frac{2\sqrt2}{7}</math>. Then <math>\sin\theta = -\frac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
== Solution ==<br />
===Solution 1===<br />
<math>\begin{align*}</math><br />
<math>P \sin\theta\ + Q \cos\theta\ = \cos\theta\ - \frac{1}{2}\ P</math><br />
<math>\end{align*}</math><br />
and<br />
<math>\begin{align*}</math><br />
<math>P \cos\theta\ + Q \sin\theta\ = -2(Q-1)</math><br />
<math>\end{align*}</math><br />
<br />
Solving for P, Q we have <br />
<br />
<br />
<math>\frac{P}{Q} = \frac{\cos\theta }{2 + \sin\theta } = \frac{2\sqrt2}{7}</math><br />
<br />
Square both side, and use polynomial rational root theorem to solve <math>sin\theta</math><br />
<br />
<math>\sin\theta = -\frac{17}{19} </math><br />
<br />
The answer is <math>\boxed{036}</math><br />
<br />
===Solution 2===<br />
<math>\begin{align*}</math><br />
Use sum to product formulas to rewrite <math>P</math> and <math>Q</math> <br />
<math>\end{align*}</math><br />
<br />
<math>P \sin\theta\ + Q \cos\theta\ = \cos \theta\ - \frac{1}{4}\cos \theta + \frac{1}{8}\sin 2\theta + \frac{1}{16}\cos 3\theta - \frac{1}{32}\sin 4\theta + ... </math><br />
<br />
<math>\begin{align*}</math><br />
Therefore, <math>P \sin \theta - Q \cos \theta = -2P</math><br />
<math>\end{align*}</math><br />
<br />
<math>\begin{align*}</math><br />
Using <math>\frac{P}{Q} = \frac{2\sqrt2}{7}</math>, <math>Q = \frac{7}{2\sqrt2} P</math><br />
<math>\end{align*}</math><br />
<br />
<math>\begin{align*}</math><br />
Plug in to the previous equation and cancel out the "P" terms to get: <math>\sin\theta - \frac{7}{2\sqrt2} \cos\theta = -2</math><br />
<math>\end{align*}</math><br />
<br />
Then use the pythagorean identity to solve for <math>\sin\theta</math>, <math>\sin\theta = \boxed{036}</math><br />
<br />
===Solution 3===<br />
<br />
Note that <cmath>e^{i\theta}=\cos(\theta)+i\sin(\theta)</cmath><br />
<br />
Thus, the following identities follow immediately:<br />
<cmath>ie^{i\theta}=i(\cos(\theta)+i\sin(\theta))=-\sin(\theta)+i\cos(\theta)</cmath><br />
<cmath>i^2 e^{i\theta}=-e^{i\theta}=-\cos(\theta)-i\sin(\theta)</cmath><br />
<cmath>i^3 e^{i\theta}=\sin(\theta)-i\cos(\theta)</cmath><br />
<br />
Consider, now, the sum <math>Q+iP</math>. It follows fairly immediately that:<br />
<br />
<cmath>Q+iP=1+\left(\frac{i}{2}\right)^1e^{i\theta}+\left(\frac{i}{2}\right)^2e^{2i\theta}+\ldots=\frac{1}{1-\frac{i}{2}e^{i\theta}}=\frac{2}{2-ie^{i\theta}}</cmath><br />
<cmath>Q+iP=\frac{2}{2-ie^{i\theta}}=\frac{2}{2-(-\sin(\theta)+i\cos(\theta))}=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}</cmath><br />
<br />
This follows straight from the geometric series formula and simple simplification. We can now multiply the denominator by it's complex conjugate to find:<br />
<br />
<cmath>Q+iP=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}\left(\frac{(2+\sin(\theta))+i\cos(\theta)}{(2+\sin(\theta))+i\cos(\theta)}\right)</cmath><br />
<cmath>Q+iP=\frac{2((2+\sin(\theta))+i\cos(\theta))}{(2+\sin(\theta))^2+\cos^2(\theta)}</cmath><br />
<br />
Comparing real and imaginary parts, we find:<br />
<cmath>\frac{P}{Q}=\frac{\cos(\theta)}{2+\sin(\theta)}=\frac{2\sqrt{2}}{7}</cmath><br />
<br />
Squaring this equation and letting <math>\sin^2(\theta)=x</math>:<br />
<br />
<math>\frac{P^2}{Q^2}=\frac{\cos^2(\theta)}{4+4\sin(\theta)+\sin^2(\theta)}=\frac{1-x^2}{4+4x+x^2}=\frac{8}{49}</math><br />
<br />
Clearing denominators and solving for <math>x</math> gives sine as <math>x=-\frac{17}{19}</math>.<br />
<br />
<math>017+019=\boxed{036}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2013|n=I|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Mapletree14https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_17&diff=588672013 AMC 12B Problems/Problem 172014-01-25T01:05:51Z<p>Mapletree14: </p>
<hr />
<div>==Problem==<br />
<br />
<br />
Let <math>a,b,</math> and <math>c</math> be real numbers such that <br />
<br />
<cmath>a+b+c=2, \text{ and} </cmath><br />
<cmath> a^2+b^2+c^2=12 </cmath><br />
<br />
What is the difference between the maximum and minimum possible values of <math>c</math>?<br />
<br />
<math> \text{(A) }2\qquad \text{ (B) }\frac{10}{3}\qquad \text{ (C) }4 \qquad \text{ (D) }\frac{16}{3}\qquad \text{ (E) }\frac{20}{3} </math><br />
<br />
==Solution 1==<br />
<math>a+b= 2-c</math>. Now, by Cauchy-Schwarz, we have that <math>(a^2+b^2) \ge \frac{(2-c)^2}{2}</math>. Therefore, we have that <math>\frac{(2-c)^2}{2}+c^2 \le 12</math>. We then find the roots of <math>c</math> that satisfy equality and find the difference of the roots. This gives the answer, <math>\boxed{\textbf{(D)} \ \frac{16}{3}}</math>.<br />
<br />
==Solution 2==<br />
This is similar to the first solution but is far more intuitive. From the given, we have <cmath> a + b = 2 - c \\a^2 + b^2 = 12 - c^2 </cmath> This immediately suggests use of the Cauchy-Schwarz inequality. By Cauchy, we have <cmath> 2\,(a^2 + b^2) \geq (a + b)^2</cmath> Substitution of the above results and some algebra yields <cmath> 3c^2 - 4c - 20 \leq 0 </cmath> This quadratic inequality is easily solved, and it is seen that equality holds for <math>c = -2</math> and <math>c = \frac{10}{3}</math>. <br />
<br />
The difference between these two values is <math>\boxed{\textbf{(D)} \ \frac{16}{3}}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2013|ab=B|num-b=16|num-a=18}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Mapletree14https://artofproblemsolving.com/wiki/index.php?title=1976_IMO_Problems/Problem_4&diff=567711976 IMO Problems/Problem 42013-07-30T15:45:10Z<p>Mapletree14: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Determine the greatest number, who is the product of some positive integers, and the sum of these numbers is <math>1976.</math><br />
<br />
== Solution ==<br />
=== Solution 1 ===<br />
Since <math>3*3=2*2*2+1</math>, 3's are more efficient than 2's. We try to prove that 3's are more efficient than anything:<br />
<br />
Let there be a positive integer <math>x</math>. If <math>3</math> is more efficient than <math>x</math>, then <math>x^3<3^x</math>. We try to prove that all integers greater than 3 are less efficient than 3:<br />
<br />
When <math>x</math> increases by 1, then the RHS is multiplied by 3. The other side is multiplied by <math>\dfrac{(x+1)^3}{x^3}</math>, and we must prove that this is less than 3 for all <math>x</math> greater than 3.<br />
<br />
<math>\dfrac{(x+1)^3}{x^3}<3\Rightarrow \dfrac{x+1}{x}<\sqrt[3]{3}\Rightarrow 1<(\sqrt[3]{3}-1)x</math><br />
<br />
<math>\dfrac{1}{\sqrt[3]{3}-1}<x</math><br />
<br />
Thus we need to prove that <math>\dfrac{1}{\sqrt[3]{3}-1}<4</math>. Simplifying, we get <math>5<4\sqrt[3]{3}\Rightarrow 125<64*3=192</math>, which is true. Working backwards, we see that all <math>x</math> greater than 3 are less efficient than 3, so we try to use the most 3's as possible:<br />
<br />
<math>\dfrac{1976}{3}=658.6666</math>, so the greatest product is <math>\boxed{3^{658}\cdot 2}</math>.<br />
<br />
===Solution 2===<br />
We demonstrate heuristically that 3's are the most efficient. <br />
As we know that the chosen numbers must be equal, by the AM-GM inequality, we wish to maximize <cmath> f(x) = \left(\frac{1976}{x}\right)^{x} </cmath><br />
Simple logarithmic differentiation shows that <math>\frac{S}{x} = e</math> maximizes the given function. As <math>e</math> is approximately 2.71828, we use 2's and 3's only. 3's are optimal, but we must use one 2. <br />
<br />
=== Solution 3 ===<br />
<br />
''Note: This solution uses the same strategy as Solution 1 (that having the largest possible number of three's is good), but approaches the proof in a different manner.''<br />
<br />
Let <math>f(S) \triangleq \prod_{x \in S} x</math>, and <math>g(S) = \sum_{x\in S} x</math>, where <math>S</math> is a multiset. We fix <math>g(S) = 1976</math>, and aim to maximize <math>f(S)</math>. Since <math>3(x-3) > x</math> for <math>x \geq 5</math>, we notice that <math>S</math> must only contain the integers <math>1,2,3</math> and <math>4</math>. We can replace any occurrences of <math>4</math> in <math>S</math> by replacing it with a couple of <math>2</math>'s, without changing <math>f(S)</math> or <math>g(S)</math>, so we may assume that <math>S</math> only contains the integers <math>1,2</math> and <math>3</math>. We may further assume that <math>S</math> contains at most one <math>1</math>, since any two <math>1</math>'s can be replaced by a <math>2</math> without changing <math>g(S)</math>, but with an increase in <math>f(S)</math>. If <math>S</math> contains exactly one <math>1</math>, then it must also contain at least one <math>2</math> (since <math>1976 = 2\;(\mathrm{mod }\;3)</math>). We can then replace this pair of a <math>1</math> and a <math>2</math> with a <math>3</math>, thus keeping <math>g(S)</math> constant, and increasing <math>f(S)</math>. Now we may assume that <math>S</math> contains only <math>2</math>'s and <math>3</math>'s. <br />
<br />
Now, as observed in the last solution, any triplet of <math>2</math>'s can be replaced by a couple of <math>3</math>'s, with <math>g(S)</math> constant, and an increase in <math>f(S)</math>. Thus, after repeating this operation, we will be left with at most two <math>2</math>'s. Since <math>g(S) = 1976</math>, and <math>1976 = 2\;(\mathrm{mod }\;3)</math>, we therefore get that <math>S</math> must have exactly one <math>2</math> (since we already showed it consists only of <math>2</math>'s and <math>3</math>'s). Thus, we get <math>\boxed{f(S) = 2\cdot 3^{658}}</math>.<br />
<br />
== See also ==<br />
{{IMO box|year=1976|num-b=3|num-a=5}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]</div>Mapletree14https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_8&diff=567701992 AIME Problems/Problem 82013-07-30T14:54:59Z<p>Mapletree14: /* Solution */</p>
<hr />
<div>== Problem ==<br />
For any sequence of real numbers <math>A=(a_1,a_2,a_3,\ldots)</math>, define <math>\Delta A^{}_{}</math> to be the sequence <math>(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)</math>, whose <math>n^\mbox{th}_{}</math> term is <math>a_{n+1}-a_n^{}</math>. Suppose that all of the terms of the sequence <math>\Delta(\Delta A^{}_{})</math> are <math>1^{}_{}</math>, and that <math>a_{19}=a_{92}^{}=0</math>. Find <math>a_1^{}</math>.<br />
<br />
== Solution ==<br />
Note that the <math>\Delta</math>s are reminiscent of differentiation; from the condition <math>\Delta(\Delta{A}) = 1</math>, we are led to consider the differential equation<br />
<cmath> \frac{d^2 A}{dn^2} = 1 </cmath><br />
This inspires us to guess a quadratic with leading coefficient 1/2 as the solution; <br />
<cmath> a_{n} = \frac{1}{2}(n-19)(n-94) </cmath><br />
as we must have roots at <math>n = 19</math> and <math>n = 94</math>.<br />
<br />
Thus, <math>a_1=\frac{1}{2}(1-19)(1-92)=\boxed{819}</math>.<br />
<br />
== Solution 2 ==<br />
Let <math>\Delta^1 A=\Delta A</math>, and <math>\Delta^n A=\Delta(\Delta^{(n-1)}A)</math>.<br />
<br />
Note that in every sequence of <math>a_i</math>, <math>a_n=\binom{n-1}{1}\Delta a_n + \binom{n-1}{2}\Delta^2 a_n +\binom{n-1}{3}\Delta^3 a_n + ...</math><br />
<br />
Then <math>a_n=a_1 +\binom{n-1}{1}\Delta a_1 +\binom{n-1}{2}\Delta^2 a_1 +\binom{n-1}{3}\Delta^3 a_1 + ...</math><br />
<br />
Since <math>\Delta a_1 =a_2 -a_1</math>, <math>a_n=a_1 +\binom{n-1}{1}(a_2-a_1) +\binom{n-1}{2}\cdot 1=a_1 +n(a_2-a_1) +\binom{n-1}{2}</math><br />
<br />
<math>a_{19}=0=a_1+18(a_2-a_1)+\binom{18}{2}=18a_2-17a_1+153</math><br />
<br />
<math>a_{92}=0=a_1+91(a_2-a_1)+\binom{91}{2}=91a_2-90a_1+4095</math><br />
<br />
Solving, <math>a_1=\boxed{819}</math>.<br />
<br />
== Solution 3 ==<br />
Write out and add first <math>k-1</math> terms of the second finite difference sequence:<br />
<br />
<math>a_3+a_1-2*a_2=1</math><br />
<br />
<math>a_4+a_2-2*a_3=1</math><br />
<br />
…<br />
<br />
…<br />
<br />
…<br />
<br />
<math>a_k + a_{k-2} - 2*a_{k-1} = 1</math><br />
<br />
<math>a_{k+1} + a_{k-1} - 2*a_k = 1</math><br />
<br />
Adding the above <math>k-1</math> equations we get:<br />
<br />
<math>\boxed{(a_{k+1} - a_k) = k-1 + (a_2-a_1)} --- (1)</math><br />
<br />
Now taking sum <math>k = 1</math> to <math>18</math> in equation <math>(1)</math> we get:<br />
<math>18*(a_1-a_2) - a_1 = 153 --- (2)</math><br />
<br />
Now taking sum <math>k = 1</math> to <math>91</math> in equation <math>(1)</math> we get:<br />
<math>91*(a_1-a_2) - a_1 = 4095 --- (3)</math><br />
<br />
<math>18* (3) - 91*(2)</math> gives <math>a_1=\boxed{819}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1992|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Mapletree14https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_8&diff=567691992 AIME Problems/Problem 82013-07-30T14:54:07Z<p>Mapletree14: /* Solution */</p>
<hr />
<div>== Problem ==<br />
For any sequence of real numbers <math>A=(a_1,a_2,a_3,\ldots)</math>, define <math>\Delta A^{}_{}</math> to be the sequence <math>(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)</math>, whose <math>n^\mbox{th}_{}</math> term is <math>a_{n+1}-a_n^{}</math>. Suppose that all of the terms of the sequence <math>\Delta(\Delta A^{}_{})</math> are <math>1^{}_{}</math>, and that <math>a_{19}=a_{92}^{}=0</math>. Find <math>a_1^{}</math>.<br />
<br />
== Solution ==<br />
Note that the <math>\Delta</math>s are reminiscent of differentiation; from the condition <math>\Delta(\Delta{A}) = 1</math>, we are led to consider the differential equation<br />
<cmath> \frac{d^2 A}{dn^2} = 1 </cmath><br />
This inspires us to guess a quadratic with leading coefficient 1/2 as the solution; <br />
<cmath> a_{n} = \frac{1}{2}(n-19)(n-94) </cmath><br />
Thus, <math>a_1=\frac{1}{2}(1-19)(1-92)=\boxed{819}</math>.<br />
<br />
== Solution 2 ==<br />
Let <math>\Delta^1 A=\Delta A</math>, and <math>\Delta^n A=\Delta(\Delta^{(n-1)}A)</math>.<br />
<br />
Note that in every sequence of <math>a_i</math>, <math>a_n=\binom{n-1}{1}\Delta a_n + \binom{n-1}{2}\Delta^2 a_n +\binom{n-1}{3}\Delta^3 a_n + ...</math><br />
<br />
Then <math>a_n=a_1 +\binom{n-1}{1}\Delta a_1 +\binom{n-1}{2}\Delta^2 a_1 +\binom{n-1}{3}\Delta^3 a_1 + ...</math><br />
<br />
Since <math>\Delta a_1 =a_2 -a_1</math>, <math>a_n=a_1 +\binom{n-1}{1}(a_2-a_1) +\binom{n-1}{2}\cdot 1=a_1 +n(a_2-a_1) +\binom{n-1}{2}</math><br />
<br />
<math>a_{19}=0=a_1+18(a_2-a_1)+\binom{18}{2}=18a_2-17a_1+153</math><br />
<br />
<math>a_{92}=0=a_1+91(a_2-a_1)+\binom{91}{2}=91a_2-90a_1+4095</math><br />
<br />
Solving, <math>a_1=\boxed{819}</math>.<br />
<br />
== Solution 3 ==<br />
Write out and add first <math>k-1</math> terms of the second finite difference sequence:<br />
<br />
<math>a_3+a_1-2*a_2=1</math><br />
<br />
<math>a_4+a_2-2*a_3=1</math><br />
<br />
…<br />
<br />
…<br />
<br />
…<br />
<br />
<math>a_k + a_{k-2} - 2*a_{k-1} = 1</math><br />
<br />
<math>a_{k+1} + a_{k-1} - 2*a_k = 1</math><br />
<br />
Adding the above <math>k-1</math> equations we get:<br />
<br />
<math>\boxed{(a_{k+1} - a_k) = k-1 + (a_2-a_1)} --- (1)</math><br />
<br />
Now taking sum <math>k = 1</math> to <math>18</math> in equation <math>(1)</math> we get:<br />
<math>18*(a_1-a_2) - a_1 = 153 --- (2)</math><br />
<br />
Now taking sum <math>k = 1</math> to <math>91</math> in equation <math>(1)</math> we get:<br />
<math>91*(a_1-a_2) - a_1 = 4095 --- (3)</math><br />
<br />
<math>18* (3) - 91*(2)</math> gives <math>a_1=\boxed{819}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1992|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Mapletree14https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_8&diff=566562012 AIME I Problems/Problem 82013-07-16T00:16:24Z<p>Mapletree14: /* Alternative Solution (using calculus): think inside the box */</p>
<hr />
<div>==Problem 8==<br />
Cube <math>ABCDEFGH,</math> labeled as shown below, has edge length <math>1</math> and is cut by a plane passing through vertex <math>D</math> and the midpoints <math>M</math> and <math>N</math> of <math>\overline{AB}</math> and <math>\overline{CG}</math> respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form <math>\tfrac{p}{q},</math> where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q.</math><br />
<br />
<center><asy>import cse5;<br />
unitsize(10mm);<br />
pathpen=black;<br />
dotfactor=3;<br />
<br />
pair A = (0,0), B = (3.8,0), C = (5.876,1.564), D = (2.076,1.564), E = (0,3.8), F = (3.8,3.8), G = (5.876,5.364), H = (2.076,5.364), M = (1.9,0), N = (5.876,3.465);<br />
pair[] dotted = {A,B,C,D,E,F,G,H,M,N};<br />
<br />
D(A--B--C--G--H--E--A);<br />
D(E--F--B);<br />
D(F--G);<br />
pathpen=dashed;<br />
D(A--D--H);<br />
D(D--C);<br />
<br />
dot(dotted);<br />
label("$A$",A,SW);<br />
label("$B$",B,S);<br />
label("$C$",C,SE);<br />
label("$D$",D,NW);<br />
label("$E$",E,W);<br />
label("$F$",F,SE);<br />
label("$G$",G,NE);<br />
label("$H$",H,NW);<br />
label("$M$",M,S);<br />
label("$N$",N,NE);<br />
<br />
</asy></center><br />
<br />
== Solution: think outside the box ==<br />
Define a coordinate system with <math>D</math> at the origin and <math>C,</math> <math>A,</math> and <math>H</math> on the <math>x</math>, <math>y</math>, and <math>z</math> axes respectively. The <math>D=(0,0,0),</math> <math>M=(.5,1,0),</math> and <math>N=(1,0,.5).</math> It follows that the plane going through <math>D,</math> <math>M,</math> and <math>N</math> has equation <math>2x-y-4z=0.</math> Let <math>Q = (1,1,.25)</math> be the intersection of this plane and edge <math>BF</math> and let <math>P = (1,2,0).</math> Now since <math>2(1) - 1(2) - 4(0) = 0,</math> <math>P</math> is on the plane. Also, <math>P</math> lies on the extensions of segments <math>DM,</math> <math>NQ,</math> and <math>CB</math> so the solid <math>DPCN = DMBCQN + MPBQ</math> is a right triangular pyramid. Note also that pyramid <math>MPBQ</math> is similar to <math>DPCN</math> with scale factor <math>.5</math> and thus the volume of solid <math>DMBCQN,</math> which is one of the solids bounded by the cube and the plane, is <math>[DPCN] - [MPBQ] = [DPCN] - \left(\frac{1}{2}\right)^3[DPCN] = \frac{7}{8}[DPCN].</math> But the volume of <math>DPCN</math> is simply the volume of a pyramid with base <math>1</math> and height <math>.5</math> which is <math>\frac{1}{3} \cdot 1 \cdot .5 = \frac{1}{6}.</math> So <math>[DMBCQN] = \frac{7}{8} \cdot \frac{1}{6} = \frac{7}{48}.</math> Note, however, that this volume is less than <math>.5</math> and thus this solid is the smaller of the two solids. The desired volume is then <math>[ABCDEFGH] - [DMBCQN] = 1 - \frac{7}{48} = \frac{41}{48} \rightarrow p+q = \boxed{089.}</math><br />
<br />
== Alternative Solution (using calculus): think inside the box ==<br />
Let <math>Q</math> be the intersection of the plane with edge <math>FB,</math> then <math>\triangle MQB</math> is similar to <math>\triangle DNC</math> and the volume <math>[DNCMQB]</math> is a sum of areas of cross sections of similar triangles running parallel to face <math>ABFE.</math> Let <math>x</math> be the distance from face <math>ABFE,</math> let <math>h</math> be the height parallel to <math>AB</math> of the cross-sectional triangle at that distance, and <math>a</math> be the area of the cross-sectional triangle at that distance. <math>A(x)=\frac{h^2}{4},</math> and <math>h=\frac{x+1}{2},</math> then <math>A=\frac{(x+1)^2}{16}</math>, and the volume <math>[DNCMQB]</math> is <math>\int^1_0{A(x)}\,\mathrm{d}x=\int^1_0{\frac{(x+1)^2}{16}}\,\mathrm{d}x=\frac{7}{48}.</math> Thus the volume of the larger solid is <math>1-\frac{7}{48}=\frac{41}{48} \rightarrow p+q = \boxed{089}</math><br />
<br />
== Alternative Solution think inside the box like a total nerd==<br />
<br />
If you memorized the formula for a frustum, then this problem is very trivial.<br />
<br />
The formula for a frustum is:<br />
<br />
<math>\frac{h_2b_2 -h_1b_1}3</math> where <math>b_i</math> is the area of the base and <math>h_i</math> is the height from the chopped of apex to the base.<br />
<br />
We can easily see that from symmetry, the area of the smaller front base is <math>\frac{1}{16}</math> and the area of the larger back base is <math>\frac{1}4</math><br />
<br />
Now to find the height of the apex.<br />
<br />
Extend the <math>DM</math> and (call the intersection of the plane with <math>FB</math> G) <math>NG</math> to meet at <math>x</math>. Now from similar triangles <math>XMG</math> and <math>XDN</math> we can easily find the total height of the triangle <math>XDN</math> to be <math>2</math><br />
<br />
Now straight from our formula, the area is <math>\frac{7}{48}</math> Thus the answer is:<br />
<br />
<math>1-Area \Longrightarrow \boxed{089}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2012|n=I|num-b=7|num-a=9}}<br />
<br />
{{MAA Notice}}</div>Mapletree14https://artofproblemsolving.com/wiki/index.php?title=1980_USAMO_Problems/Problem_5&diff=530981980 USAMO Problems/Problem 52013-06-26T22:39:20Z<p>Mapletree14: </p>
<hr />
<div>== Problem ==<br />
If <math>x, y, z</math> are reals such that <math>0\le x, y, z \le 1</math>, show that <math>\frac{x}{y + z + 1} + \frac{y}{z + x + 1} + \frac{z}{x + y + <br />
1} \le 1 - (1 - x)(1 - y)(1 - z)</math><br />
<br />
== Solution ==<br />
Rewrite the given inequality so that <math>1</math> is isolated on the right side. Set the left side to be <math>f(x, y, z)</math>. Now a routine computation shows<br />
<br />
<br />
<math>\frac{\partial^2 f}{\partial x^2} = \frac{2y}{(x + z + 1)^3} + \frac{2z}{(x + y + 1)^3}\geq 0 </math><br />
<br />
<br />
which shows that <math>f</math> is convex (concave up) in all three variables. Thus the maxima can only occur at the endpoints, i.e. if and only if <math>x, y, z \in \{0,1\}</math>. Checking all eight cases shows that the value of the expression cannot exceed 1.<br />
<br />
== See Also ==<br />
{{USAMO box|year=1980|num-b=4|after=Last Question}}<br />
<br />
[[Category:Olympiad Inequality Problems]]</div>Mapletree14https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_4&diff=521042013 AIME II Problems/Problem 42013-04-04T23:10:09Z<p>Mapletree14: </p>
<hr />
<div>In the Cartesian plane let <math>A = (1,0)</math> and <math>B = \left( 2, 2\sqrt{3} \right)</math>. Equilateral triangle <math>ABC</math> is constructed so that <math>C</math> lies in the first quadrant. Let <math>P=(x,y)</math> be the center of <math>\triangle ABC</math>. Then <math>x \cdot y</math> can be written as <math>\tfrac{p\sqrt{q}}{r}</math>, where <math>p</math> and <math>r</math> are relatively prime positive integers and <math>q</math> is an integer that is not divisible by the square of any prime. Find <math>p+q+r</math>.<br />
<br />
==Solution 1 ==<br />
The distance from point <math>A</math> to point <math>B</math> is <math> \sqrt{13}</math>. The vector that starts at point A and ends at point B is given by <math>B - A = (1, 2\sqrt{3})</math>. Since the center of an equilateral triangle, <math>P</math>, is also the intersection of the perpendicular bisectors of the sides of the triangle, we need first find the equation for the perpendicular bisector to <math>\overline{AB}</math>. The line perpendicular to <math>\overline{AB}</math> through the midpoint, <math>M = (\dfrac{3}{2},\sqrt{3})</math>, <math>\overline{AB}</math> can be parameterized by <math> (\dfrac{2\sqrt{3}}{\sqrt{13}}, \dfrac{-1}{\sqrt{13}}) t + (\dfrac{3}{2},\sqrt{3})</math>. At this point, it is useful to note that <math>\Delta BMP</math> is a 30-60-90 triangle with <math>\overline{MB}</math> measuring <math>\dfrac{\sqrt{13}}{2}</math>. This yields the length of <math>\overline{MP}</math> to be <math>\dfrac{\sqrt{13}}{2\sqrt{3}}</math>. Therefore, <math>P =( \dfrac{2\sqrt{3}}{\sqrt{13}},\dfrac{-1}{\sqrt{13}})(\dfrac{\sqrt{13}}{2\sqrt{3}}) + (\dfrac{3}{2},\sqrt{3}) = (\dfrac{5}{2}, \dfrac{5}{2\sqrt{3}})</math>. Therefore <math>xy = \dfrac{25\sqrt{3}}{12}</math> yielding an answer of <math> p + q + r = 25 + 3 + 12 = \boxed{040}</math>.<br />
<br />
<br />
==Solution 2==<br />
<br />
Rather than considering the Cartesian plane, we use complex numbers. Thus A is 1 and B is <math>2 + 2\sqrt{3}i</math>. <br />
<br />
Recall that a rotation of <math>\theta</math> radians counterclockwise is equivalent to multiplying a complex number by <math>e^{i\theta}</math>, but here we require a clockwise rotation, so we multiply by <math>e^{-\frac{i\pi}{3}}</math> to obtain C. Upon averaging the coordinates of A, B, and C, we obtain the coordinates of P, viz. <math>\left(\frac{5}{2}, \frac{5\sqrt{3}}{6}\right)</math>.<br />
<br />
Therefore <math>xy</math> is <math>\frac{25\sqrt{3}}{12}</math> and the answer is <math>25 + 12 + 3 = \boxed{040}</math>.</div>Mapletree14https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_25&diff=519822011 AMC 12A Problems/Problem 252013-03-30T00:24:15Z<p>Mapletree14: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Triangle <math>ABC</math> has <math>\angle BAC = 60^{\circ}</math>, <math>\angle CBA \leq 90^{\circ}</math>, <math>BC=1</math>, and <math>AC \geq AB</math>. Let <math>H</math>, <math>I</math>, and <math>O</math> be the orthocenter, incenter, and circumcenter of <math>\triangle ABC</math>, respectively. Assume that the area of pentagon <math>BCOIH</math> is the maximum possible. What is <math>\angle CBA</math>?<br />
<br />
<math><br />
\textbf{(A)}\ 60^{\circ} \qquad<br />
\textbf{(B)}\ 72^{\circ} \qquad<br />
\textbf{(C)}\ 75^{\circ} \qquad<br />
\textbf{(D)}\ 80^{\circ} \qquad<br />
\textbf{(E)}\ 90^{\circ} </math><br />
<br />
== Solution ==<br />
<br />
<br /><br />
'''Solution:'''<br />
<br />
1) Let's draw a circle with center <math>O</math> (which will be the circumcircle of <math>\triangle ABC</math>. Since <math>\angle BAC = 60^{\circ}</math>, <math>\overline{BC}</math> is a chord that intercept an arc of <math>120 ^{\circ}</math><br />
<br />
2) Draw any chord that can be <math>BC</math>, and let's define that as unit length.<br />
<br />
3) Draw the diameter <math>\perp</math> to <math>BC</math>. Let's call the interception of the diameter with <math>BC</math> <math>M</math> (because it is the midpoint) and interception with the circle <math>X</math>.<br />
<br />
4) Note that OMB and XMC is fixed, hence the area is a constant. Thus, <math>XOIHC</math> also achieved maximum area.<br />
<br />
<br /><br />
'''Lemma:''' <br />
<br />
<math>m\angle BOC = m \angle BIC = m \angle BHC = 120^{\circ}</math><br />
<br />
For <math>m\angle BOC</math>, we fixed it to <math>120^{\circ}</math> when we drew the diagram.<br />
<br />
Let <math>m\angle ABC = \beta</math>, <math>m\angle ACB = \gamma</math><br />
<br />
<br /><br />
Now, let's isolate the points <math>A</math>,<math>B</math>,<math>C</math>, and <math>I</math>.<br />
<br />
<math>m\angle IBC = \frac{\beta}{2}</math>, <math>m\angle ICB = \frac{\gamma}{2}</math><br />
<br />
<math>m\angle BIC = 180^{\circ} - \frac{\beta}{2} - \frac{\gamma}{2} = 180^{\circ} - ({180^{\circ} - 120^{\circ})= 120 ^{\circ}</math><br />
<br />
<br /><br />
Now, lets isolate the points <math>A</math>,<math>B</math>,<math>C</math>, and <math>H</math>.<br />
<br />
<math>m\angle HBC = \beta - 30^{\circ}</math>, <math>m\angle HCB = \gamma - 30^{\circ}</math><br />
<br />
<math>m\angle BHC = 180^{\circ} - \beta - \gamma + 60^{\circ} = 240^{\circ} - 120^{\circ} = 120^{\circ}</math><br />
<br />
<br /><br />
Lemma proven. The lemma yields that BOIHC is a cyclic pentagon.<br />
<br />
Since we got that XOIHC also achieved maximum area,<br />
<br />
Let <math>m\angle XOI = x_1</math>, <math>m\angle OIH = x_2</math>, <math>m\angle IHC = x_3</math>, and the radius is <math>R</math> (which will drop out.)<br />
<br />
then area = <math>\frac{r^2}{2}(\sin x_1 + \sin x_2 + \sin x_3)</math>, where <math>x_1 + x_2 + x_3 = 60^\circ</math><br />
<br />
So we want to maximize <math>f(x_1, x_2) = \sin x_1 + \sin x_2 + \sin x_3</math>, Note that <math>x_3 = 60 ^\circ - x_1 - x_2</math>.<br />
<br />
Let's do some multivariable calculus.<br />
<br />
<math>f_{x_1} = \cos x_1 - \cos (x_3)</math>, <math>f_{x_2} = \cos x_2 - \cos (x_3)</math><br />
<br />
If the partial derivatives with respect to <math>x_1</math> and <math>x_2</math> are zero, then <math>x_1 = x_2 = x_3 = 20^\circ</math>, and it is very easy to show that <math>f(x_1, x_2)</math> is the maximum with the second derivative test (left for the reader).<br />
<br />
<br /><br />
Now, we need to verify that such a situation exists and find the angle for this situation.<br />
<br />
Let's extend <math>AI</math> to the direction of <math>X</math>, since <math>AI</math> is the angle bisector, <math>AI</math> should intersection the midpoint of the arc, which is <math>X</math>. Hence, if such a case exists, <math>m\angle AXB = m \angle ACB = 40 ^\circ</math>, which yields <math>m\angle CBA = 80 ^\circ</math>.<br />
<br />
If the angle is <math>80 ^\circ</math>, it is clear that since <math>I</math> and <math>H</math> are on the second circle (follows from the lemma). <math>I</math> will be at the right place. <math>H</math> can be easily verified too.<br />
<br />
<br /><br />
Hence, the answer is <math>(D) 80</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|num-b=24|after=Last Problem|ab=A}}</div>Mapletree14https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_25&diff=519812011 AMC 12A Problems/Problem 252013-03-30T00:23:08Z<p>Mapletree14: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Triangle <math>ABC</math> has <math>\angle BAC = 60^{\circ}</math>, <math>\angle CBA \leq 90^{\circ}</math>, <math>BC=1</math>, and <math>AC \geq AB</math>. Let <math>H</math>, <math>I</math>, and <math>O</math> be the orthocenter, incenter, and circumcenter of <math>\triangle ABC</math>, respectively. Assume that the area of pentagon <math>BCOIH</math> is the maximum possible. What is <math>\angle CBA</math>?<br />
<br />
<math><br />
\textbf{(A)}\ 60^{\circ} \qquad<br />
\textbf{(B)}\ 72^{\circ} \qquad<br />
\textbf{(C)}\ 75^{\circ} \qquad<br />
\textbf{(D)}\ 80^{\circ} \qquad<br />
\textbf{(E)}\ 90^{\circ} </math><br />
<br />
== Solution ==<br />
<br />
<br /><br />
'''Solution:'''<br />
<br />
1) Let's draw a circle with center <math>O</math> (which will be the circumcircle of <math>\triangle ABC</math>. Since <math>\angle BAC = 60^{\circ}</math>, <math>\overline{BC}</math> is a chord that intercept an arc of <math>120 ^{\circ}</math><br />
<br />
2) Draw any chord that can be <math>BC</math>, and lets define that as unit length.<br />
<br />
3) Draw the diameter <math>\perp</math> to <math>BC</math>. Let's call the interception of the diameter with <math>BC</math> <math>M</math> (because it is the midpoint) and interception with the circle <math>X</math>.<br />
<br />
4) Note that OMB and XMC is fixed, hence the area is a constant. Thus, <math>XOIHC</math> also achieved maximum area.<br />
<br />
<br /><br />
'''Lemma:''' <br />
<br />
<math>m\angle BOC = m \angle BIC = m \angle BHC = 120^{\circ}</math><br />
<br />
For <math>m\angle BOC</math>, we fixed it to <math>120^{\circ}</math> when we drew the diagram.<br />
<br />
Let <math>m\angle ABC = \beta</math>, <math>m\angle ACB = \gamma</math><br />
<br />
<br /><br />
Now, let's isolate the points <math>A</math>,<math>B</math>,<math>C</math>, and <math>I</math>.<br />
<br />
<math>m\angle IBC = \frac{\beta}{2}</math>, <math>m\angle ICB = \frac{\gamma}{2}</math><br />
<br />
<math>m\angle BIC = 180^{\circ} - \frac{\beta}{2} - \frac{\gamma}{2} = 180^{\circ} - ({180^{\circ} - 120^{\circ})= 120 ^{\circ}</math><br />
<br />
<br /><br />
Now, lets isolate the points <math>A</math>,<math>B</math>,<math>C</math>, and <math>H</math>.<br />
<br />
<math>m\angle HBC = \beta - 30^{\circ}</math>, <math>m\angle HCB = \gamma - 30^{\circ}</math><br />
<br />
<math>m\angle BHC = 180^{\circ} - \beta - \gamma + 60^{\circ} = 240^{\circ} - 120^{\circ} = 120^{\circ}</math><br />
<br />
<br /><br />
Lemma proven. The lemma yields that BOIHC is a cyclic pentagon.<br />
<br />
Since we got that XOIHC also achieved maximum area,<br />
<br />
Let <math>m\angle XOI = x_1</math>, <math>m\angle OIH = x_2</math>, <math>m\angle IHC = x_3</math>, and the radius is <math>R</math> (which will drop out.)<br />
<br />
then area = <math>\frac{r^2}{2}(\sin x_1 + \sin x_2 + \sin x_3)</math>, where <math>x_1 + x_2 + x_3 = 60^\circ</math><br />
<br />
So we want to maximize <math>f(x_1, x_2) = \sin x_1 + \sin x_2 + \sin x_3</math>, Note that <math>x_3 = 60 ^\circ - x_1 - x_2</math>.<br />
<br />
Let's do some multivariable calculus.<br />
<br />
<math>f_{x_1} = \cos x_1 - \cos (x_3)</math>, <math>f_{x_2} = \cos x_2 - \cos (x_3)</math><br />
<br />
If the partial derivatives with respect to <math>x_1</math> and <math>x_2</math> are zero, then <math>x_1 = x_2 = x_3 = 20^\circ</math>, and it is very easy to show that <math>f(x_1, x_2)</math> is the maximum with the second derivative test (left for the reader).<br />
<br />
<br /><br />
Now, we need to verify that such situation exist and find the angle for this situation.<br />
<br />
Let's extend <math>AI</math> to the direction of <math>X</math>, since <math>AI</math> is the angle bisector, <math>AI</math> should intersection the midpoint of the arc, which is <math>X</math>. Hence, if such a case exists, <math>m\angle AXB = m \angle ACB = 40 ^\circ</math>, which yields <math>m\angle CBA = 80 ^\circ</math>.<br />
<br />
If the angle is <math>80 ^\circ</math>, it is clear that since <math>I</math> and <math>H</math> are on the second circle (follows from the lemma). <math>I</math> will be at the right place. <math>H</math> can be easily verified too.<br />
<br />
<br /><br />
Hence, the answer is <math>(D) 80</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|num-b=24|after=Last Problem|ab=A}}</div>Mapletree14https://artofproblemsolving.com/wiki/index.php?title=Calculus&diff=51156Calculus2013-02-21T00:40:08Z<p>Mapletree14: </p>
<hr />
<div>The discovery of the branch of [[mathematics]] known as '''calculus''' was motivated by two classical problems: how to find the [[slope]] of the [[tangent line]] to a curve at a [[point]] and how to find the [[area]] bounded by a curve. What is surprising is that these two problems are fundamentally connected and, together with the notion of limits, can be used to analyze instantaneous [[rate]]s of change, accumulations of change, [[volume]]s of irregular [[solid]]s, and much more.<br />
<br />
[[Limit]]s are heavily used in calculus. The formal notion of a limit is what differentiates calculus from precalculus mathematics. <br />
<br />
The subject dealing with the rigorous foundations of calculus is called [[analysis]], specifically [[real analysis]].<br />
<br />
==History==<br />
Calculus was compiled into one mathematical science by [[Isaac Newton]] in 1665 and 1666. (Before this, some individual calculus ideas had been discovered by earlier mathematicians). However, [[Gottfried Leibniz]], whom did the same work independently a few years later, published his work earlier than Newton. This sparked an argument over who first discovered calculus. It is now known that Newton did discover calculus first, but Leibniz invented the majority of the notation we use today. <br />
<br />
== Calculus in Math Competitions ==<br />
The use of calculus in pre-collegiate [[mathematics competitions]] is generally frowned upon. However, many [[Physics competitions | physics competitions]] require it, as does the [[William Lowell Putnam Mathematical Competition|William Lowell Putnam competition]].<br />
<br />
There are a number of high school math contests that have a calculus round, or require calculus. These include:<br />
<br />
- The Harvard-MIT Invitational Tournament (HMIT) <br />
<br />
- The Johns-Hopkins Mathematics Tournament<br />
<br />
- The Rice University Mathematics Tournament<br />
<br />
<br />
None of the [[AMC]] competitions leading up to the [[IMO]] require it, nor does the [[ARML]], even though calculus solutions are still permitted. Online high school competitions, such as the [[iTest]], will occasionally require it, but generally not.<br />
<br />
== See also ==<br />
* [[Derivative]]<br />
* [[Limit]]<br />
* [[Integral]] (It is suggested that you look at derivative before this)<br />
<br />
=== Topics ===<br />
* [[Chain Rule]]<br />
* [[Implicit differentiation]]<br />
* [[Fundamental Theorem of Calculus]]<br />
<br />
[[Category:Calculus]]</div>Mapletree14https://artofproblemsolving.com/wiki/index.php?title=Calculus&diff=51155Calculus2013-02-21T00:39:31Z<p>Mapletree14: </p>
<hr />
<div>The discovery of the branch of [[mathematics]] known as '''calculus''' was motivated by two classical problems: how to find the [[slope]] of the [[tangent line]] to a curve at a [[point]] and how to find the [[area]] bounded by a curve. What is surprising is that these two problems are fundamentally connected and, together with the notion of limits, can be used to analyze instantaneous [[rate]]s of change, accumulations of change, [[volume]]s of irregular [[solid]]s, and much more.<br />
<br />
[[Limit]]s are heavily used in calculus. The formal notion of a limit is what differentiates calculus from precalculus mathematics. <br />
<br />
The subject dealing with the rigorous foundations of calculus is called [[analysis]], specifically [[real analysis]].<br />
<br />
==History==<br />
Calculus was compiled into one mathematical science by [[Isaac Newton]] in 1665 and 1666. (Before this, some individual calculus ideas had been discovered by earlier mathematicians). However, [[Gottfried Leibniz]], whom did the same work independently a few years later, published his work earlier than Newton. This sparked an argument over who first discovered calculus. It is now known that Newton did discover calculus first, but Leibniz invented the majority of the notation we use today. <br />
<br />
== Calculus in Math Competitions ==<br />
The use of calculus in pre-collegiate [[mathematics competitions]] is generally frowned upon. However, many [[Physics competitions | physics competitions]] require it, as does the [[William Lowell Putnam Mathematical Competition|William Lowell Putnam competition]].<br />
<br />
There are a number of high school math contests that have a calculus round, or require calculus. These include:<br />
<br />
- The Harvard-MIT Invitational Tournament (HMIT)<br />
- The Johns-Hopkins Mathematics Tournament<br />
- The Rice University Mathematics Tournament<br />
<br />
<br />
None of the [[AMC]] competitions leading up to the [[IMO]] require it, nor does the [[ARML]], even though calculus solutions are still permitted. Online high school competitions, such as the [[iTest]], will occasionally require it, but generally not.<br />
<br />
== See also ==<br />
* [[Derivative]]<br />
* [[Limit]]<br />
* [[Integral]] (It is suggested that you look at derivative before this)<br />
<br />
=== Topics ===<br />
* [[Chain Rule]]<br />
* [[Implicit differentiation]]<br />
* [[Fundamental Theorem of Calculus]]<br />
<br />
[[Category:Calculus]]</div>Mapletree14https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_8&diff=506272005 AIME I Problems/Problem 82013-02-02T10:32:48Z<p>Mapletree14: /* Solution */</p>
<hr />
<div>== Problem ==<br />
The [[equation]] <math> 2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1 </math> has three [[real]] [[root]]s. Given that their sum is <math> \frac mn </math> where <math> m </math> and <math> n </math> are [[relatively prime]] [[positive integer]]s, find <math> m+n. </math><br />
<br />
== Solution ==<br />
Let <math>y = 2^{111x}</math>. Then our equation reads <math>\frac{1}{4}y^3 + 4y = 2y^2 + 1</math> or <math>y^3 - 8y^2 + 16y - 4 = 0</math>. Thus, if this equation has roots <math>r_1, r_2</math> and <math>r_3</math>, by [[Vieta's formulas]] we have <math>r_1\cdot r_2\cdot r_3 = 4</math>. Let the corresponding values of <math>x</math> be <math>x_1, x_2</math> and <math>x_3</math>. Then the previous statement says that <math>2^{111\cdot(x_1 + x_2 + x_3)} = 4</math> so that taking a [[logarithm]] gives <math>111(x_1 + x_2 + x_3) = 2</math> and <math>x_1 + x_2 + x_3 = \frac{2}{111}</math>. Thus the answer is <math>111 + 2 = \boxed{113}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2005|n=I|num-b=7|num-a=9}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Mapletree14