https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Epicfailiure&feedformat=atomAoPS Wiki - User contributions [en]2021-11-29T20:45:57ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_6&diff=455142012 AIME I Problems/Problem 62012-03-16T21:39:48Z<p>Epicfailiure: Created page with "Let <math>z</math> and <math>w</math> be complex numbers such that <math>z^{13} = w</math> and <math>w^{11} = z</math>. If the imaginary part of <math>z</math> can be written as ..."</p>
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<div>Let <math>z</math> and <math>w</math> be complex numbers such that <math>z^{13} = w</math> and <math>w^{11} = z</math>. If the imaginary part of <math>z</math> can be written as <math> \sin {\frac{m\pi}{n}}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>n</math>.</div>Epicfailiurehttps://artofproblemsolving.com/wiki/index.php?title=1994_USAMO_Problems/Problem_4&diff=450951994 USAMO Problems/Problem 42012-02-24T04:50:01Z<p>Epicfailiure: /* See Also */</p>
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<div>==Problem 4==<br />
Let <math>\, a_1, a_2, a_3, \ldots \,</math> be a sequence of positive real numbers satisfying <math>\, \sum_{j = 1}^n a_j \geq \sqrt {n} \,</math> for all <math>\, n \geq 1</math>. Prove that, for all <math>\, n \geq 1, \,</math><br />
<br />
<cmath>\sum_{j = 1}^n a_j^2 > \frac {1}{4} \left( 1 + \frac {1}{2} + \cdots + \frac {1}{n} \right).</cmath><br />
<br />
== Solution ==<br />
Note that if we try to minimize (a_j)^2, we would try to make the a_j as equal as possible. However, by the condition given in the problem, this isn't possible, the a_j's have to be an increasing sequence. Thinking of minimizing sequences, we realize that the optimal equation is a_n = \sqrt{n} - \sqrt{n-1} = 1/(\sqrt{n} + \sqrt{n-1}). Note that this is strictly greater than 1/(2\sqrt{n}). So it is greater than the sum of (1/\sqrt{4n})^2 over all n from 1 to infinity. So the sum we are looking to minimize is strictly greater than the sum of 1/4n over all n from 1 to infinity, which is what we wanted to do. <br />
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== See Also ==<br />
{{USAMO box|year=1994|num-b=3|num-a=5}}</div>Epicfailiurehttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_23&diff=371262011 AMC 10B Problems/Problem 232011-02-26T00:42:12Z<p>Epicfailiure: /* Solution */</p>
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<div>==Problem== <br />
<br />
What is the hundreds digit of 2011^2011?<br />
<br />
(A) 1 (B) 3 (C) 4 (D) 6 (E) 8<br />
<br />
==Solution==<br />
<br />
(2000 + 11) ^ 2011 mod 1000 \n<br />
<br />
11^2011 mod 1000 <br />
<br />
(10 + 1)^2011 mod 1000 <br />
<br />
2011C2 * 10^2 + 2011C1 * 10 + 1 mod 1000 <br />
<br />
500 + 110 + 1 mod 1000 <br />
<br />
611 mod 1000 <br />
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So we know the last three digits of 2011 ^ 2011 is 611, and so the hundreds digit is 6 (D).</div>Epicfailiurehttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_23&diff=371252011 AMC 10B Problems/Problem 232011-02-26T00:41:33Z<p>Epicfailiure: /* Solution */</p>
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<div>==Problem== <br />
<br />
What is the hundreds digit of 2011^2011?<br />
<br />
(A) 1 (B) 3 (C) 4 (D) 6 (E) 8<br />
<br />
==Solution==<br />
<br />
(2000 + 11) ^ 2011 mod 1000 \n<br />
<br />
11^2011 mod 1000 \n<br />
(10 + 1)^2011 mod 1000 \n<br />
2011C2 * 10^2 + 2011C1 * 10 + 1 mod 1000 \n<br />
500 + 110 + 1 mod 1000 \n<br />
611 mod 1000 \n<br />
<br />
So we know the last three digits of 2011 ^ 2011 is 611, and so the hundreds digit is 6 (D).</div>Epicfailiurehttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_23&diff=371242011 AMC 10B Problems/Problem 232011-02-26T00:40:48Z<p>Epicfailiure: Created page with '==Problem== What is the hundreds digit of 2011^2011? (A) 1 (B) 3 (C) 4 (D) 6 (E) 8 ==Solution== (2000 + 11) ^ 2011 mod 1000 11^2011 mod 1000 (10 + 1)^2011 mod 1000 2011C2 *…'</p>
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<div>==Problem== <br />
<br />
What is the hundreds digit of 2011^2011?<br />
<br />
(A) 1 (B) 3 (C) 4 (D) 6 (E) 8<br />
<br />
==Solution==<br />
<br />
(2000 + 11) ^ 2011 mod 1000<br />
11^2011 mod 1000<br />
(10 + 1)^2011 mod 1000<br />
2011C2 * 10^2 + 2011C1 * 10 + 1 mod 1000<br />
500 + 110 + 1 mod 1000<br />
611 mod 1000<br />
<br />
So we know the last three digits of 2011 ^ 2011 is 611, and so the hundreds digit is 6 (D).</div>Epicfailiurehttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_18&diff=371232011 AMC 10B Problems/Problem 182011-02-26T00:36:29Z<p>Epicfailiure: Created page with '==Problem== There is a rectangle ABCD with AB = 6 and BC = 3. A point M lies on AB such that angles CMD and AMD are congruent. What is the measure of angle CMD? (A) 15 (B) 30 (…'</p>
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<div>==Problem==<br />
<br />
There is a rectangle ABCD with AB = 6 and BC = 3. A point M lies on AB such that angles CMD and AMD are congruent. What is the measure of angle CMD?<br />
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(A) 15 (B) 30 (C) 45 (D) 60 (E) 75<br />
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==Solution==<br />
<br />
Using alternate interior angles and the fact that AB and CD are parallel because it is a rectangle, the angles AMD and MDC are congruent. Using the transitive property and the given angular congruence, MDC and CMD are congruent, and so CD = CM = 6. So sin(BMC) = 3/6 = 0.5, and angle BMC is 30 degrees. Then AMD = 75 degrees (E).</div>Epicfailiurehttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_11&diff=371222011 AMC 10B Problems/Problem 112011-02-26T00:30:38Z<p>Epicfailiure: Created page with '==Problem== There are 52 people in a room. The statement " At least n people in the room have birthdays in the same month " is true for n. What is the largest possible value of …'</p>
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<div>==Problem==<br />
<br />
There are 52 people in a room. The statement " At least n people in the room have birthdays in the same month " is true for n. What is the largest possible value of n?<br />
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(A) 2 (B) 3 (C) 4 (D) 5 (E) 6<br />
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==Solution==<br />
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By the Pigeon Hole Principle, this the 52/12 rounded up. That is 5 (D).</div>Epicfailiurehttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_2&diff=371212011 AMC 10B Problems/Problem 22011-02-26T00:28:04Z<p>Epicfailiure: </p>
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<div>==Problem==<br />
<br />
Josanna's test scores to date are 90, 80, 70, 60, and 85. Her goal is to raise her test average at least 3 points with her next test. What is the minimum test score she would need to accomplish this goal?<br />
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(A) 80 (B) 82 (C) 85 (D) 90 (E) 95<br />
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<br />
==Solution==<br />
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The average of her current scores is 77. To raise it 3 points, she needs an average of 80, and so after her 6 tests, a sum of 480. Her current sum is 385, so she needs a 480 - 385 = 95 (E).</div>Epicfailiurehttps://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_1&diff=370851993 AHSME Problems/Problem 12011-02-23T01:08:43Z<p>Epicfailiure: Created page with '==Problem== ==Solution== Plugging in to the equation, we have 1^-1 - -1^2 + 2^1 = 2.'</p>
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<div>==Problem==<br />
<br />
==Solution==<br />
<br />
Plugging in to the equation, we have 1^-1 - -1^2 + 2^1 = 2.</div>Epicfailiurehttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_14&diff=370812006 AMC 12A Problems/Problem 142011-02-22T18:19:40Z<p>Epicfailiure: /* Solution */</p>
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<div>{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #14]] and [[2006 AMC 10A Problems/Problem 22|2006 AMC 10A #22]]}}<br />
<br />
== Problem ==<br />
<br />
Two farmers agree that pigs are worth <math>300</math> dollars and that goats are worth <math>210</math> dollars. When one farmer owes the other money, he pays the debt in pigs or goats, with "change" received in the form of goats or pigs as necessary. (For example, a <math>390</math> dollar debt could be paid with two pigs, with one goat received in change.) What is the amount of the smallest positive debt that can be resolved in this way?<br />
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<math> \mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 10\qquad \mathrm{(C) \ } 30\qquad \mathrm{(D) \ } 90\qquad \mathrm{(E) \ } 210</math><br />
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== Solution ==<br />
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Solution 1: <br />
<br />
The problem can be restated as an equation of the form <math>300p + 210g = x</math>, where <math>p</math> is the number of pigs, <math>g</math> is the number of goats, and <math>x</math> is the positive debt. The problem asks us to find the lowest ''x'' possible. ''p'' and ''g'' must be [[integer]]s, which makes the equation a [[Diophantine equation]]. The [[Euclidean algorithm]] tells us that there are integer solutions to the Diophantine equation <math>am + bn = c</math>, where <math>c</math> is the [[greatest common divisor]] of <math>a</math> and <math>b</math>, and no solutions for any smaller <math>c</math>. Therefore, the answer is the greatest common divisor of 300 and 210, which is 30, <math>\mathrm{(C) \ }</math><br />
<br />
Alternatively, note that <math>300p + 210g = 30(10p + 7g)</math> is divisible by 30 no matter what <math>p</math> and <math>g</math> are, so our answer must be divisible by 30. In addition, three goats minus two pigs give us <math>630 - 600 = 30</math> exactly. Since our theoretical best can be achieved, it must really be the best, and the answer is <math>\mathrm{(C) \ }</math>.<br />
debt that can be resolved.<br />
<br />
Solution 2:<br />
<br />
Let us think of it as if the farmers would like to pay off a certain debt, then the farmers take turns giving each other goats and pigs until it is resolved. Let us suppose they start at 0 debt, and they want to get to the smallest debt possible. Each time a pig or goat is exchanged, the remainder of the debt when divided by 30 stays the same. This is because a goat and pig are both worth multiples of 30. Since the debt is positive, the smallest possible achievable debt is 30 (C).<br />
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== See also ==<br />
{{AMC12 box|year=2006|ab=A|num-b=13|num-a=15}}<br />
{{AMC10 box|year=2006|ab=A|num-b=21|num-a=23}}<br />
<br />
[[Category:Introductory Number Theory Problems]]</div>Epicfailiurehttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_18&diff=370612003 AMC 12A Problems/Problem 182011-02-21T22:56:47Z<p>Epicfailiure: /* Solution */</p>
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<div>== Problem ==<br />
Let <math>n</math> be a <math>5</math>-digit number, and let <math>q</math> and <math>r</math> be the quotient and the remainder, respectively, when <math>n</math> is divided by <math>100</math>. For how many values of <math>n</math> is <math>q+r</math> divisible by <math>11</math>? <br />
<br />
<math> \mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090 </math><br />
<br />
== Solution ==<br />
<br />
Solution 1:<br />
<br />
When a <math>5</math>-digit number is divided by <math>100</math>, the first <math>3</math> digits become the quotient, <math>q</math>, and the last <math>2</math> digits become the remainder, <math>r</math>. <br />
<br />
Therefore, <math>q</math> can be any integer from <math>100</math> to <math>999</math> inclusive, and <math>r</math> can be any integer from <math>0</math> to <math>99</math> inclusive. <br />
<br />
For each of the <math>9\cdot10\cdot10=900</math> possible values of <math>q</math>, there are at least <math>\lfloor \frac{100}{11} \rfloor = 9</math> possible values of <math>r</math> such that <math>q+r \equiv 0\pmod{11}</math>. <br />
<br />
Since there is <math>1</math> "extra" possible value of <math>r</math> that is congruent to <math>0\pmod{11}</math>, each of the <math>\lfloor \frac{900}{11} \rfloor = 81</math> values of <math>q</math> that are congruent to <math>0\pmod{11}</math> have <math>1</math> more possible value of <math>r</math> such that <math>q+r \equiv 0\pmod{11}</math>. <br />
<br />
Therefore, the number of possible values of <math>n</math> such that <math>q+r \equiv 0\pmod{11}</math> is <math>900\cdot9+81\cdot1=8181 \Rightarrow B</math>.<br />
<br />
Solution 2:<br />
<br />
Let the 5 digit number be abcde, where a, b, c, d, and e are digits. Then q = abc, and r = de. Then the digits of q +r theoretically are a, then b+d, then c + e. If this is to be divisible by 11, we can use the divisibility check for 11 to say that a + c + e - b -d is divisible by 11. If we revert this again, but we want all the digits to be single digit, then we get abcde again. So if q + r is to be divisible by 11, then the number we started with, abcde, must also be divisible by 11. There are 90000 total possible values of abcde, of which 8181 of them are divisible by 11. So for those numbers q + r is divisible by 11, so we have 8181 (B).<br />
<br />
== See Also ==<br />
*[[2003 AMC 12A Problems]]<br />
*[[2003 AMC 12A/Problem 17|Previous Problem]]<br />
*[[2003 AMC 12A/Problem 19|Next Problem]]<br />
<br />
[[Category:Introductory Number Theory Problems]]</div>Epicfailiurehttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_23&diff=370602003 AMC 12A Problems/Problem 232011-02-21T22:50:54Z<p>Epicfailiure: /* Solution */</p>
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<div><br />
==Problem==<br />
<br />
==Solution==<br />
<br />
We want to find the number of perfect square factors in the product of all the factorials of numbers from 1 - 9. We can write this out and take out the factorials, and then find a prime factorization of the entire product. We can also find this prime factorization by finding the number of times each factor is repeated in each factorial. Anyway, this comes out to be equal to 2^30 * 3^13 * 5^5 * 7^3. To find the amount of perfect square factors, we realize that each exponent in the prime factorization must be even. This gives us 16*7*3*2 = 672 (B).<br />
<br />
== See Also ==<br />
{{AMC12 box|year=2003|ab=A|num-b=22|num-a=24}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Epicfailiurehttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_23&diff=370592003 AMC 12A Problems/Problem 232011-02-21T22:49:26Z<p>Epicfailiure: /* Solution */</p>
<hr />
<div><br />
==Problem==<br />
<br />
==Solution==<br />
<br />
We want to find the number of perfect square factors in the product of all the factorials of numbers from 1 - 9. We can write this out and take out the factorials, and then find a prime factorization of the entire product. We can also find this prime factorization by finding the number of times each factor is repeated in each factorial. Anyway, this comes out to be equal to 2^30 * 3^13 * 5^5 * 7^3. To find the amount of perfect square factors, we realize that each exponent in the prime factorization must be even. This gives us 16*8*3*2 = 768.<br />
<br />
== See Also ==<br />
{{AMC12 box|year=2003|ab=A|num-b=22|num-a=24}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Epicfailiurehttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_23&diff=370582003 AMC 12A Problems/Problem 232011-02-21T22:47:03Z<p>Epicfailiure: /* See Also */</p>
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<div><br />
==Problem==<br />
<br />
==Solution==<br />
<br />
We want to find the number of perfect square factors in the product of all the factorials of numbers from 1 - 9. We can write this out and take out the factorials, and then find a prime factorization of the entire product. We can also find this prime factorization by finding the number of times each factor is repeated in each factorial. Anyway, this comes out to be equal to <br />
<br />
== See Also ==<br />
{{AMC12 box|year=2003|ab=A|num-b=22|num-a=24}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Epicfailiurehttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_19&diff=370572003 AMC 12A Problems/Problem 192011-02-21T22:12:56Z<p>Epicfailiure: Created page with '==Problem== ==Solution== If we take the parabola ax^2 + bx + c and reflect it over the x - axis, we have the parabola -ax^2 - bx - c. Without loss of generality, let us say tha…'</p>
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<div>==Problem==<br />
<br />
==Solution==<br />
<br />
If we take the parabola ax^2 + bx + c and reflect it over the x - axis, we have the parabola -ax^2 - bx - c. Without loss of generality, let us say that the parabola is translated 5 units to the left, and the reflection to the right. Then:<br />
<br />
f(x) = a(x+5)^2 + b(x+5) + c = ax^2 + 10ax + 25a + bx +5b +c = ax^2 + (10a+b)x + 25a + 5b + c.<br />
g(x) = -a(x-5)^2 - b(x-5) - c = -ax^2 + 10ax -bx - 25a + 5b - c<br />
<br />
Adding them up:<br />
<br />
(f+g)(x) = 20ax + 10b which is a line with slope 20a. We know this is not horizontal, as a is not equal to 0. This is true because if a = 0, then the graph of ax^2 + bx +c would not be a parabola as stated in the problem.</div>Epicfailiurehttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_16&diff=370542003 AMC 12A Problems/Problem 162011-02-21T22:03:28Z<p>Epicfailiure: /* Problem */</p>
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<div><br />
== Problem ==<br />
<br />
A point P is chosen at random from the equilateral triangle ABC. What is the probability that triangle ACP has a greater area than triangles BCP and ABP?<br />
<br />
== Solution==<br />
<br />
Solution 1:<br />
<br />
After we pick point P, we realize that ABC is symmetric for this purpose, and so the probability that ACP is the greatest area, or ABP or BCP, are all the same. Since they add to 1, the probability that ACP has the greatest area is 1/3 (C).<br />
<br />
Solution 2:<br />
<br />
We will use an approach of geometric probability to solve this problem. Let us take point P, and draw the perpendiculars to AB, BC, and AC, and call the feet of these perpendiculars D, E, and F respectively. The area of triangle ACP is simply 1/2 * AC * PF. Similarly we can find the area of triangles BCP and ABP. If we add these up and realize that it equals the area of the entire triangle, we see that no matter where we choose P, PD + PE + PF = the height of the triangle. Setting the area of triangle ACP greater than ABP and BCP, we want PF to be the largest of PF, PD, and PE. We then realize that PF = PD = PE when P is the orthocenter of ABC. Let us call the orthocenter of the triangle Q. If we want PF to be the largest of the three, by testing points we realize that P must be in the interior of quadrilateral QFCE. So our probability (using geometric probability) is the area of QFCE divided by the area of ABC. We will now show that the three quadrilaterals, QFCE, QEBD, and QDAF are congruent. As the definition of point Q yields, QF = QD = QE. Since ABC is equilateral, Q is also the circumcenter of ABC, so QA = QB = QC. Using the Pythagorean theorem, BD = DA = AF = FC = CE = EB. Also, angles BDQ, BEQ, CEQ, CFQ, AFQ, and ADQ are all equal to 90 degrees by the definition of an altitude. Also, angles DBE, FCE, DAF are all equal to 60 degrees as equilateral triangles are also equiangular. It is now clear that QFCE, QFAD, QEBD are all congruent. Summing up these areas gives us the area of ABC. QFCE contributes to a third of that area, as they are all congruent, so the ratio of the areas of QFCE to ABC is 1/3 (C).</div>Epicfailiurehttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_16&diff=370532003 AMC 12A Problems/Problem 162011-02-21T22:00:56Z<p>Epicfailiure: /* = Problem */</p>
<hr />
<div><br />
== Problem ==<br />
<br />
<br />
<br />
== Solution==<br />
<br />
Solution 1:<br />
<br />
After we pick point P, we realize that ABC is symmetric for this purpose, and so the probability that ACP is the greatest area, or ABP or BCP, are all the same. Since they add to 1, the probability that ACP has the greatest area is 1/3 (C).<br />
<br />
Solution 2:<br />
<br />
We will use an approach of geometric probability to solve this problem. Let us take point P, and draw the perpendiculars to AB, BC, and AC, and call the feet of these perpendiculars D, E, and F respectively. The area of triangle ACP is simply 1/2 * AC * PF. Similarly we can find the area of triangles BCP and ABP. If we add these up and realize that it equals the area of the entire triangle, we see that no matter where we choose P, PD + PE + PF = the height of the triangle. Setting the area of triangle ACP greater than ABP and BCP, we want PF to be the largest of PF, PD, and PE. We then realize that PF = PD = PE when P is the orthocenter of ABC. Let us call the orthocenter of the triangle Q. If we want PF to be the largest of the three, by testing points we realize that P must be in the interior of quadrilateral QFCE. So our probability (using geometric probability) is the area of QFCE divided by the area of ABC. We will now show that the three quadrilaterals, QFCE, QEBD, and QDAF are congruent. As the definition of point Q yields, QF = QD = QE. Since ABC is equilateral, Q is also the circumcenter of ABC, so QA = QB = QC. Using the Pythagorean theorem, BD = DA = AF = FC = CE = EB. Also, angles BDQ, BEQ, CEQ, CFQ, AFQ, and ADQ are all equal to 90 degrees by the definition of an altitude. Also, angles DBE, FCE, DAF are all equal to 60 degrees as equilateral triangles are also equiangular. It is now clear that QFCE, QFAD, QEBD are all congruent. Summing up these areas gives us the area of ABC. QFCE contributes to a third of that area, as they are all congruent, so the ratio of the areas of QFCE to ABC is 1/3 (C).</div>Epicfailiurehttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_16&diff=370522003 AMC 12A Problems/Problem 162011-02-21T22:00:39Z<p>Epicfailiure: Created page with ' == Problem = == Solution== Solution 1: After we pick point P, we realize that ABC is symmetric for this purpose, and so the probability that ACP is the greatest area, or AB…'</p>
<hr />
<div><br />
== Problem =<br />
<br />
<br />
<br />
== Solution==<br />
<br />
Solution 1:<br />
<br />
After we pick point P, we realize that ABC is symmetric for this purpose, and so the probability that ACP is the greatest area, or ABP or BCP, are all the same. Since they add to 1, the probability that ACP has the greatest area is 1/3 (C).<br />
<br />
Solution 2:<br />
<br />
We will use an approach of geometric probability to solve this problem. Let us take point P, and draw the perpendiculars to AB, BC, and AC, and call the feet of these perpendiculars D, E, and F respectively. The area of triangle ACP is simply 1/2 * AC * PF. Similarly we can find the area of triangles BCP and ABP. If we add these up and realize that it equals the area of the entire triangle, we see that no matter where we choose P, PD + PE + PF = the height of the triangle. Setting the area of triangle ACP greater than ABP and BCP, we want PF to be the largest of PF, PD, and PE. We then realize that PF = PD = PE when P is the orthocenter of ABC. Let us call the orthocenter of the triangle Q. If we want PF to be the largest of the three, by testing points we realize that P must be in the interior of quadrilateral QFCE. So our probability (using geometric probability) is the area of QFCE divided by the area of ABC. We will now show that the three quadrilaterals, QFCE, QEBD, and QDAF are congruent. As the definition of point Q yields, QF = QD = QE. Since ABC is equilateral, Q is also the circumcenter of ABC, so QA = QB = QC. Using the Pythagorean theorem, BD = DA = AF = FC = CE = EB. Also, angles BDQ, BEQ, CEQ, CFQ, AFQ, and ADQ are all equal to 90 degrees by the definition of an altitude. Also, angles DBE, FCE, DAF are all equal to 60 degrees as equilateral triangles are also equiangular. It is now clear that QFCE, QFAD, QEBD are all congruent. Summing up these areas gives us the area of ABC. QFCE contributes to a third of that area, as they are all congruent, so the ratio of the areas of QFCE to ABC is 1/3 (C).</div>Epicfailiurehttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12B_Problems/Problem_24&diff=370452005 AMC 12B Problems/Problem 242011-02-21T18:26:26Z<p>Epicfailiure: </p>
<hr />
<div>{{empty}}<br />
== Problem ==<br />
<br />
All three vertices of an equilateral triangle are on the parabola y = x^2, and one of its sides has a slope of 2. The x-coordinates of the three vertices have a sum of m/n, where m and n are relatively prime positive integers. What is the value of m + n? <br />
== Solution ==<br />
<br />
Let the points be (a,a^2), (b,b^2) and (c,c^2). Using elementary calculus concepts and the fact that they lie on y = x^2,<br />
<br />
a+b = the slope of AB,<br />
b+c = the slope of BC,<br />
a+c = the slope of AC.<br />
<br />
So the value that we need to find is simply the sum of the slopes of the three sides of the triangle divided by 2. WLOG, let AB be the side that has the smallest angle with the positive x-axis. Let J be an arbitrary point with the coordinates (1, 0). Let us translate the triangle so A is at the origin. Then tan(BOJ) = 2. Using the fact that the slope of a line is equal to the tangent of the angle formed by the line and the positive x- axis, the value we now need to find is simply,<br />
<br />
(tan(BOJ) + tan(BOJ+60) + tan(BOJ-60)) / 2. <br />
<br />
Using tan(BOJ) = 2, and basic trigonometric identities, this simplifies to 3/11, so the answer is 3 + 11 = 14 (A). <br />
<br />
<br />
<br />
<br />
== See also ==<br />
* [[2005 AMC 12B Problems]]</div>Epicfailiurehttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12B_Problems/Problem_24&diff=370442005 AMC 12B Problems/Problem 242011-02-21T18:25:03Z<p>Epicfailiure: </p>
<hr />
<div>{{empty}}<br />
== Problem ==<br />
<br />
All three vertices of an equilateral triangle are on the parabola , and one of its sides has a slope of . The -coordinates of the three vertices have a sum of , where and are relatively prime positive integers. What is the value of ?<br />
{{All three vertices of an equilateral triangle are on the parabola y = x^2, and one of its sides has a slope of 2. The x-coordinates of the three vertices have a sum of m/n, where m and n are relatively prime positive integers. What is the value of m + n? }}<br />
== Solution ==<br />
<br />
Let the points be (a,a^2), (b,b^2) and (c,c^2). Using elementary calculus concepts and the fact that they lie on y = x^2,<br />
<br />
a+b = the slope of AB,<br />
b+c = the slope of BC,<br />
a+c = the slope of AC.<br />
<br />
So the value that we need to find is simply the sum of the slopes of the three sides of the triangle divided by 2. WLOG, let AB be the side that has the smallest angle with the positive x-axis. Let J be an arbitrary point with the coordinates (1, 0). Let us translate the triangle so A is at the origin. Then tan(BOJ) = 2. Using the fact that the slope of a line is equal to the tangent of the angle formed by the line and the positive x- axis, the value we now need to find is simply,<br />
<br />
(tan(BOJ) + tan(BOJ+60) + tan(BOJ-60)) / 2. <br />
<br />
Using tan(BOJ) = 2, and basic trigonometric identities, this simplifies to 3/11, so the answer is 3 + 11 = 14 (A). <br />
<br />
<br />
<br />
<br />
== See also ==<br />
* [[2005 AMC 12B Problems]]</div>Epicfailiurehttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12B_Problems/Problem_24&diff=370432005 AMC 12B Problems/Problem 242011-02-21T18:21:49Z<p>Epicfailiure: </p>
<hr />
<div>{{empty}}<br />
== Problem ==<br />
{{problem}}<br />
== Solution ==<br />
<br />
Let the points be (a,a^2), (b,b^2) and (c,c^2). Using elementary calculus concepts and the fact that they lie on y = x^2,<br />
<br />
a+b = the slope of AB,<br />
b+c = the slope of BC,<br />
a+c = the slope of AC.<br />
<br />
So the value that we need to find is simply the sum of the slopes of the three sides of the triangle divided by 2. WLOG, let AB be the side that has the smallest angle with the positive x-axis. Let J be an arbitrary point with the coordinates (1, 0). Let us translate the triangle so A is at the origin. Then tan(BOJ) = 2. Using the fact that the slope of a line is equal to the tangent of the angle formed by the line and the positive x- axis, the value we now need to find is simply,<br />
<br />
(tan(BOJ) + tan(BOJ+60) + tan(BOJ-60)) / 2. <br />
<br />
Using tan(BOJ) = 2, and basic trigonometric identities, this simplifies to 3/11, so the answer is 3 + 11 = 14 (A). <br />
<br />
<br />
<br />
<br />
== See also ==<br />
* [[2005 AMC 12B Problems]]</div>Epicfailiurehttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems&diff=329432000 AMC 8 Problems2009-11-17T00:21:50Z<p>Epicfailiure: </p>
<hr />
<div>==Problem 1==<br />
Aunt Anna is <math>42</math> years old. Caitlin is <math>5</math> years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin?<br />
<br />
<math><br />
\mathrm{(A)}\ 15<br />
\qquad<br />
\mathrm{(B)}\ 16<br />
\qquad<br />
\mathrm{(C)}\ 17<br />
\qquad<br />
\mathrm{(D)}\ 21<br />
\qquad<br />
\mathrm{(E)}\ 37<br />
</math><br />
<br />
[[2000 AMC 8 Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
Which of these numbers is less than its reciprocal?<br />
<br />
<math><br />
\mathrm{(A)}\ -2<br />
\qquad<br />
\mathrm{(B)}\ -1<br />
\qquad<br />
\mathrm{(C)}\ 0<br />
\qquad<br />
\mathrm{(D)}\ 1<br />
\qquad<br />
\mathrm{(E)}\ 2<br />
</math><br />
<br />
[[2000 AMC 8 Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
How many whole numbers lie in the interval between <math>\frac{5}{3}</math> and <math>2\pi?</math><br />
<br />
<br />
<math><br />
\mathrm{(A)}\ 2<br />
\qquad<br />
\mathrm{(B)}\ 3<br />
\qquad<br />
\mathrm{(C)}\ 4<br />
\qquad<br />
\mathrm{(D)}\ 5<br />
\qquad<br />
\mathrm{(E)}\ infinitely\ many<br />
</math><br />
<br />
[[2000 AMC 8 Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
In <math>1960</math> only <math>5\%</math> of the working adults in Carlin City worked at home. By <math>1970</math> the "at-home" work force had increased to <math>8\%</math>. In <math>1980</math> there were approximately <math>15\%</math> working at home, and in <math>1990</math> there were <math>30\%</math>. The graph that best illustrates this is:<br />
<br />
[[2000 AMC 8 Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
Each principal of Lincoln High School serves exactly one <math>3</math>-year term. What is the maximum number of principals this school could have during an <math>8</math>-year period?<br />
<br />
<math><br />
\mathrm{(A)}\ 2<br />
\qquad<br />
\mathrm{(B)}\ 3<br />
\qquad<br />
\mathrm{(C)}\ 4<br />
\qquad<br />
\mathrm{(D)}\ 5<br />
\qquad<br />
\mathrm{(E)}\ 8<br />
</math><br />
<br />
[[2000 AMC 8 Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
Figure <math>ABCD</math> is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded <math>L</math>-shaped region is<br />
<br />
<math><br />
\mathrm{(A)}\ 7<br />
\qquad<br />
\mathrm{(B)}\ 10<br />
\qquad<br />
\mathrm{(C)}\ 12.5<br />
\qquad<br />
\mathrm{(D)}\ 14<br />
\qquad<br />
\mathrm{(E)}\ 15<br />
</math><br />
<br />
<br />
[[2000 AMC 8 Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
What is the minimum possible product of three different numbers of the set <math>\{-8.-6,-4,0,3,5,7\}?</math><br />
<br />
<math><br />
\mathrm{(A)}\ -336<br />
\qquad<br />
\mathrm{(B)}\ -280<br />
\qquad<br />
\mathrm{(C)}\ -210<br />
\qquad<br />
\mathrm{(D)}\ -192<br />
\qquad<br />
\mathrm{(E)}\ 0<br />
</math><br />
<br />
[[2000 AMC 8 Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
Three dice with faces numbered <math>1</math> through <math>6</math> are stacked as shown. Seven of the eighteen faces are visible, leaving eleven faces hidden (back, bottom, between). The total number of dots <math>NOT</math> visible in this view is<br />
<br />
<math><br />
\mathrm{(A)}\ 21<br />
\qquad<br />
\mathrm{(B)}\ 22<br />
\qquad<br />
\mathrm{(C)}\ 31<br />
\qquad<br />
\mathrm{(D)}\ 41<br />
\qquad<br />
\mathrm{(E)}\ 53<br />
</math><br />
<br />
[[2000 AMC 8 Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
Three-digit powers of <math>2</math> and <math>5</math> are used in this <math>cross-number</math> puzzle. What is the only possible digit for the outlined square?<br />
<br />
<math>ACROSS\ DOWN</math><br />
<br />
<math>2)\ 2^m \qquad\ 1)\ 5^n</math><br />
<br />
<br />
<math><br />
\mathrm{(A)}\ 0<br />
\qquad<br />
\mathrm{(B)}\ 2<br />
\qquad<br />
\mathrm{(C)}\ 4<br />
\qquad<br />
\mathrm{(D)}\ 6<br />
\qquad<br />
\mathrm{(E)}\ 8<br />
</math><br />
<br />
[[2000 AMC 8 Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
Ara and Shea were once the same height. Since then Shea has grown <math>20\%</math> while Ara has grow half as many inches as Shea. Shea is now <math>60</math> inches tall. How tall, in inches, is Ara now?<br />
<br />
<math><br />
\mathrm{(A)}\ 48<br />
\qquad<br />
\mathrm{(B)}\ 51<br />
\qquad<br />
\mathrm{(C)}\ 52<br />
\qquad<br />
\mathrm{(D)}\ 54<br />
\qquad<br />
\mathrm{(E)}\ 55<br />
</math><br />
<br />
[[2000 AMC 8 Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
The number <math>64</math> has the property that it is divisible by its units digit. How many whole numbers between <math>10</math> and <math>50</math> have this property?<br />
<br />
<br />
<math><br />
\mathrm{(A)}\ 15<br />
\qquad<br />
\mathrm{(B)}\ 16<br />
\qquad<br />
\mathrm{(C)}\ 17<br />
\qquad<br />
\mathrm{(D)}\ 18<br />
\qquad<br />
\mathrm{(E)}\ 20<br />
</math><br />
<br />
[[2000 AMC 8 Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
==Problem 13==<br />
==Problem 14==<br />
What is the units digit of <math>19^{19} + 99^{99}?</math><br />
<br />
<math><br />
\mathrm{(A)}\ 0<br />
\qquad<br />
\mathrm{(B)}\ 1<br />
\qquad<br />
\mathrm{(C)}\ 2<br />
\qquad<br />
\mathrm{(D)}\ 8<br />
\qquad<br />
\mathrm{(E)}\ 9<br />
</math><br />
<br />
[[2000 AMC 8 Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
==Problem 16==<br />
In order for Mateen to walk a kilometer <math>(1000m)</math> in his rectangular backyard, he must walk the length <math>25</math> times or walk its perimeter <math>10</math> times. What is the area of Mateen's backyard in square meters?<br />
<br />
<math><br />
\mathrm{(A)}\ 40<br />
\qquad<br />
\mathrm{(B)}\ 200<br />
\qquad<br />
\mathrm{(C)}\ 400<br />
\qquad<br />
\mathrm{(D)}\ 500<br />
\qquad<br />
\mathrm{(E)}\ 1000<br />
</math><br />
<br />
[[2000 AMC 8 Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
==Problem 18==<br />
==Problem 19==<br />
==Problem 20==<br />
==Problem 21==<br />
==Problem 22==<br />
==Problem 23==<br />
==Problem 24==<br />
==Problem 25==</div>Epicfailiure