id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
|---|---|---|---|---|---|
6,101 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_4 | 7 | In triangle $ABC$ $AB=20$ and $AC=11$ . The angle bisector of $\angle A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of the intersection of $AC$ and $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime posit... | Assume $ABC$ is a right triangle at $A$ . Line $AD = x$ and $BC = \tfrac{-11}{20}x + 11$ . These two lines intersect at $D$ which have coordinates $(\frac{220}{31},\frac{220}{31})$ and thus $M$ has coordinates $(\frac{110}{31},\frac{110}{31})$ . Thus, the line $BM = \tfrac{11}{51} \cdot (20-x)$ . When $x = 0$ $P$ has $... | 51 |
6,102 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_5 | 1 | The sum of the first $2011$ terms of a geometric sequence is $200$ . The sum of the first $4022$ terms is $380$ . Find the sum of the first $6033$ terms. | Since the sum of the first $2011$ terms is $200$ , and the sum of the first $4022$ terms is $380$ , the sum of the second $2011$ terms is $180$ .
This is decreasing from the first 2011, so the common ratio is less than one.
Because it is a geometric sequence and the sum of the first 2011 terms is $200$ , second $2011$ ... | 542 |
6,103 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_5 | 2 | The sum of the first $2011$ terms of a geometric sequence is $200$ . The sum of the first $4022$ terms is $380$ . Find the sum of the first $6033$ terms. | Solution by e_power_pi_times_i
The sum of the first $2011$ terms can be written as $\dfrac{a_1(1-k^{2011})}{1-k}$ , and the first $4022$ terms can be written as $\dfrac{a_1(1-k^{4022})}{1-k}$ . Dividing these equations, we get $\dfrac{1-k^{2011}}{1-k^{4022}} = \dfrac{10}{19}$ . Noticing that $k^{4022}$ is just the squa... | 542 |
6,104 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_5 | 3 | The sum of the first $2011$ terms of a geometric sequence is $200$ . The sum of the first $4022$ terms is $380$ . Find the sum of the first $6033$ terms. | The sum of the first 2011 terms of the sequence is expressible as $a_1 + a_1r + a_1r^2 + a_1r^3$ .... until $a_1r^{2010}$ . The sum of the 2011 terms following the first 2011 is expressible as $a_1r^{2011} + a_1r^{2012} + a_1r^{2013}$ .... until $a_1r^{4021}$ . Notice that the latter sum of terms can be expressed as $(... | 542 |
6,105 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_8 | 1 | Let $z_1,z_2,z_3,\dots,z_{12}$ be the 12 zeroes of the polynomial $z^{12}-2^{36}$ . For each $j$ , let $w_j$ be one of $z_j$ or $i z_j$ . Then the maximum possible value of the real part of $\sum_{j=1}^{12} w_j$ can be written as $m+\sqrt{n}$ where $m$ and $n$ are positive integers. Find $m+n$ | The twelve dots above represent the $12$ roots of the equation $z^{12}-2^{36}=0$ . If we write $z=a+bi$ , then the real part of $z$ is $a$ and the real part of $iz$ is $-b$ . The blue dots represent those roots $z$ for which the real part of $z$ is greater than the real part of $iz$ , and the red dots represent those r... | 784 |
6,106 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_8 | 2 | Let $z_1,z_2,z_3,\dots,z_{12}$ be the 12 zeroes of the polynomial $z^{12}-2^{36}$ . For each $j$ , let $w_j$ be one of $z_j$ or $i z_j$ . Then the maximum possible value of the real part of $\sum_{j=1}^{12} w_j$ can be written as $m+\sqrt{n}$ where $m$ and $n$ are positive integers. Find $m+n$ | The equation $z^{12}-2^{36}=0$ can be factored as follows: \[(z^6-2^{18})(z^6+2^{18})=0\] \[(z^2-2^6)(z^2+2^6)({(z^2+2^6)}^2-z^2\cdot2^6)({(z^2-2^6)}^2+z^2\cdot2^6)=0\] \[(z^2-2^6)(z^2+2^6)(z^2+2^6-z\cdot2^3)(z^2+2^6+z\cdot2^3) (z^2-2^6-iz\cdot2^3)(z^2-2^6+i z\cdot2^3)=0\]
Since this is a 12th degree equation, there ar... | 784 |
6,107 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_8 | 3 | Let $z_1,z_2,z_3,\dots,z_{12}$ be the 12 zeroes of the polynomial $z^{12}-2^{36}$ . For each $j$ , let $w_j$ be one of $z_j$ or $i z_j$ . Then the maximum possible value of the real part of $\sum_{j=1}^{12} w_j$ can be written as $m+\sqrt{n}$ where $m$ and $n$ are positive integers. Find $m+n$ | Clearly, the roots are: $2^3*(\cos{\frac{k\pi}{12}}+i\sin{\frac{k\pi}{12}}), k\in{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}$
Now, realize for $z=a+bi$ $\operatorname{Re}(iz)=-b$ $\operatorname{Re}(z)=a$ $\operatorname{Re}(z)<\operatorname{Re}(iz)$ is true when $a<-b$
This means:
When $a>0$ $b<-a<0$
When $a<0$ $0<b<-a$
For ... | 784 |
6,108 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_8 | 4 | Let $z_1,z_2,z_3,\dots,z_{12}$ be the 12 zeroes of the polynomial $z^{12}-2^{36}$ . For each $j$ , let $w_j$ be one of $z_j$ or $i z_j$ . Then the maximum possible value of the real part of $\sum_{j=1}^{12} w_j$ can be written as $m+\sqrt{n}$ where $m$ and $n$ are positive integers. Find $m+n$ | Use De Moivre's Theorem to brute force all the roots out. Then choose the greater value of $\operatorname{Re}(z), \operatorname{Re}(iz)$ . After adding everything up, you get $\boxed{784}$ | 784 |
6,109 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_10 | 4 | circle with center $O$ has radius 25. Chord $\overline{AB}$ of length 30 and chord $\overline{CD}$ of length 14 intersect at point $P$ . The distance between the midpoints of the two chords is 12. The quantity $OP^2$ can be represented as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find th... | Define $M$ and $N$ as the midpoints of $AB$ and $CD$ , respectively. Because $\angle OMP = \angle ONP = 90^{\circ}$ , we have that $ONPM$ is a cyclic quadrilateral. Hence, $\angle PNM = \angle POM.$ Then, let these two angles be denoted as $\alpha$ .
Now, assume WLOG that $PD = x < 7$ and $PB = y < 15$ (We can do this... | 57 |
6,110 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_12 | 1 | Nine delegates, three each from three different countries, randomly select chairs at a round table that seats nine people. Let the probability that each delegate sits next to at least one delegate from another country be $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | We use complementary counting.
We can order the $9$ people around a circle in $\frac{9!}{9} = 8!$ ways. Now we count when there is at least one delegate surrounded by people from only his/her country.
Let the countries be $A,B,C$ . Suppose that the group $XXX$ (for some country $X$ ) appears. To account for circular ov... | 97 |
6,111 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_12 | 2 | Nine delegates, three each from three different countries, randomly select chairs at a round table that seats nine people. Let the probability that each delegate sits next to at least one delegate from another country be $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | $1680= \binom{9}{3} \binom{6}{3}$ is the total.
$540 = 3* 9* \binom{6}{3}$ is the case where one country has three people in a row. (Three is for selection of country and nine is for rotation of the seats.)
$108 = 3* 9* 4$ is the case where two countries has three people in a row. (Three and nine are for the same reaso... | 97 |
6,112 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_13 | 7 | Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP > CP$ . Let $O_{1}$ and $O_{2}$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB = 12$ and $\angle O_{1}PO_{2} = 120^{\circ}$ , then $AP = \sqrt{a} + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$ | Both $O_1$ and $O_2$ lie on the perpendicular bisector of $AB$
Claim: $O_1O_2=12$ and $O_1P=O_2P$
Proof. Translate $O_1$ and $P$ $12$ units down, and let their images be $O_1'$ and $P'$ , respectively. Note that $\triangle ABP\cong\triangle DCP'$ . Additionally, \[\angle CP'D = \angle BPA = 180^{\circ} - \angle BPC = 1... | 96 |
6,113 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_14 | 1 | There are $N$ permutations $(a_{1}, a_{2}, ... , a_{30})$ of $1, 2, \ldots, 30$ such that for $m \in \left\{{2, 3, 5}\right\}$ $m$ divides $a_{n+m} - a_{n}$ for all integers $n$ with $1 \leq n < n+m \leq 30$ . Find the remainder when $N$ is divided by $1000$ | Be wary of "position" versus "number" in the solution!
Each POSITION in the 30-position permutation is uniquely defined by an ordered triple $(i, j, k)$ . The $n$ th position is defined by this ordered triple where $i$ is $n \mod 2$ $j$ is $n \mod 3$ , and $k$ is $n \mod 5$ . There are 2 choices for $i$ , 3 for $j$ , a... | 440 |
6,114 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_14 | 2 | There are $N$ permutations $(a_{1}, a_{2}, ... , a_{30})$ of $1, 2, \ldots, 30$ such that for $m \in \left\{{2, 3, 5}\right\}$ $m$ divides $a_{n+m} - a_{n}$ for all integers $n$ with $1 \leq n < n+m \leq 30$ . Find the remainder when $N$ is divided by $1000$ | Note that $30=2\cdot 3\cdot 5$ . Since $\gcd(2, 3, 5)=1$ , by CRT, for each value $k=0\ldots 29$ modulo $30$ there exists a unique ordered triple of values $(a, b, c)$ such that $k\equiv a\pmod{2}$ $k\equiv b\pmod{3}$ , and $k\equiv c\pmod{5}$ . Therefore, we can independently assign the residues modulo $2, 3, 5$ , so ... | 440 |
6,115 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_14 | 3 | There are $N$ permutations $(a_{1}, a_{2}, ... , a_{30})$ of $1, 2, \ldots, 30$ such that for $m \in \left\{{2, 3, 5}\right\}$ $m$ divides $a_{n+m} - a_{n}$ for all integers $n$ with $1 \leq n < n+m \leq 30$ . Find the remainder when $N$ is divided by $1000$ | First let's look at the situation when $m$ is equal to 2. It isn't too difficult to see the given conditions are satisfied iff the sequences $S_1 = a_1,a_3,a_5,a_7...$ and $S_2 =a_2,a_4,a_6...$ each are assigned either $\equiv 0 \pmod 2$ or $\equiv 1 \pmod 2$ . Another way to say this is each element in the sequence $a... | 440 |
6,116 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_15 | 1 | Let $P(x) = x^2 - 3x - 9$ . A real number $x$ is chosen at random from the interval $5 \le x \le 15$ . The probability that $\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}$ is equal to $\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}$ , where $a$ $b$ $c$ $d$ , and $e$ are positive integers. Find $a + b + c + d + ... | Make the substitution $y=2x-3$ , so $P(x)=\frac{y^2-45}{4}.$ We're looking for solutions to \[\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=\sqrt{\frac{\lfloor{y\rfloor}^2-45}{4}}\] with the new bounds $y\in{[7,27]}$ . Since the left side is an integer, it must be that $\frac{\lfloor{y\rfloor}^2-45}{4}$ is a perfec... | 850 |
6,117 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_15 | 2 | Let $P(x) = x^2 - 3x - 9$ . A real number $x$ is chosen at random from the interval $5 \le x \le 15$ . The probability that $\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}$ is equal to $\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}$ , where $a$ $b$ $c$ $d$ , and $e$ are positive integers. Find $a + b + c + d + ... | Note that all the "bounds" have to be less than the number+1, otherwise it wouldn't fit the answer format. Therefore, the answer is $\frac{3*3+\sqrt{9+4(4+9)}-10+\sqrt{9+4(16+9)}-12+\sqrt{9+4(144+9)}}{20} \implies \boxed{850}$ | 850 |
6,118 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_1 | 1 | Maya lists all the positive divisors of $2010^2$ . She then randomly selects two distinct divisors from this list. Let $p$ be the probability that exactly one of the selected divisors is a perfect square . The probability $p$ can be expressed in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive ... | $2010^2 = 2^2\cdot3^2\cdot5^2\cdot67^2$ . Thus there are $(2+1)^4$ divisors, $(1+1)^4$ of which are squares (the exponent of each prime factor must either be $0$ or $2$ ). Therefore the probability is \[\frac {2\cdot2^4\cdot(3^4 - 2^4)}{3^4(3^4 - 1)} = \frac {26}{81} \Longrightarrow 26+ 81 = \boxed{107}.\] | 107 |
6,119 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_1 | 2 | Maya lists all the positive divisors of $2010^2$ . She then randomly selects two distinct divisors from this list. Let $p$ be the probability that exactly one of the selected divisors is a perfect square . The probability $p$ can be expressed in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive ... | The prime factorization of $2010^2$ is $67^2\cdot3^2\cdot2^2\cdot5^2$ . Therefore, the number of divisors of $2010^2$ is $3^4$ or $81$ $16$ of which are perfect squares. The number of ways we can choose $1$ perfect square from the two distinct divisors is $\binom{16}{1}\binom{81-16}{1}$ . The total number of ways to pi... | 107 |
6,120 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_3 | 1 | Suppose that $y = \frac34x$ and $x^y = y^x$ . The quantity $x + y$ can be expressed as a rational number $\frac {r}{s}$ , where $r$ and $s$ are relatively prime positive integers. Find $r + s$ | Substitute $y = \frac34x$ into $x^y = y^x$ and solve. \[x^{\frac34x} = \left(\frac34x\right)^x\] \[x^{\frac34x} = \left(\frac34\right)^x \cdot x^x\] \[x^{-\frac14x} = \left(\frac34\right)^x\] \[x^{-\frac14} = \frac34\] \[x = \frac{256}{81}\] \[y = \frac34x = \frac{192}{81}\] \[x + y = \frac{448}{81}\] \[448 + 81 = \box... | 529 |
6,121 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_3 | 2 | Suppose that $y = \frac34x$ and $x^y = y^x$ . The quantity $x + y$ can be expressed as a rational number $\frac {r}{s}$ , where $r$ and $s$ are relatively prime positive integers. Find $r + s$ | We solve in general using $c$ instead of $3/4$ . Substituting $y = cx$ , we have:
Dividing by $x^x$ , we get $(x^x)^{c - 1} = c^x$
Taking the $x$ th root, $x^{c - 1} = c$ , or $x = c^{1/(c - 1)}$
In the case $c = \frac34$ $x = \Bigg(\frac34\Bigg)^{ - 4} = \Bigg(\frac43\Bigg)^4 = \frac {256}{81}$ $y = \frac {64}{27}$ $x... | 529 |
6,122 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_3 | 3 | Suppose that $y = \frac34x$ and $x^y = y^x$ . The quantity $x + y$ can be expressed as a rational number $\frac {r}{s}$ , where $r$ and $s$ are relatively prime positive integers. Find $r + s$ | Taking the logarithm base $x$ of both sides, we arrive with: \[y = \log_x y^x \Longrightarrow \frac{y}{x} = \log_{x} y = \log_x \left(\frac{3}{4}x\right) = \frac{3}{4}\] Now we proceed by the logarithm rule $\log(ab)=\log a + \log b$ . The equation becomes: \[\log_x \frac{3}{4} + \log_x x = \frac{3}{4}\] \[\Longleftrig... | 529 |
6,123 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_4 | 1 | Jackie and Phil have two fair coins and a third coin that comes up heads with probability $\frac47$ . Jackie flips the three coins, and then Phil flips the three coins. Let $\frac {m}{n}$ be the probability that Jackie gets the same number of heads as Phil, where $m$ and $n$ are relatively prime positive integers. Find... | This can be solved quickly and easily with generating functions
Let $x^n$ represent flipping $n$ heads.
The generating functions for these coins are $(1+x)$ $(1+x)$ ,and $(3+4x)$ in order.
The product is $3+10x+11x^2+4x^3$ . ( $ax^n$ means there are $a$ ways to get $n$ heads, eg there are $10$ ways to get $1$ head, and... | 515 |
6,124 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_5 | 1 | Positive integers $a$ $b$ $c$ , and $d$ satisfy $a > b > c > d$ $a + b + c + d = 2010$ , and $a^2 - b^2 + c^2 - d^2 = 2010$ . Find the number of possible values of $a$ | Using the difference of squares $2010 = (a^2 - b^2) + (c^2 - d^2) = (a + b)(a - b) + (c + d)(c - d) \ge a + b + c + d = 2010$ , where equality must hold so $b = a - 1$ and $d = c - 1$ . Then we see $a = 1004$ is maximal and $a = 504$ is minimal, so the answer is $\boxed{501}$ | 501 |
6,125 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_5 | 2 | Positive integers $a$ $b$ $c$ , and $d$ satisfy $a > b > c > d$ $a + b + c + d = 2010$ , and $a^2 - b^2 + c^2 - d^2 = 2010$ . Find the number of possible values of $a$ | Since $a+b$ must be greater than $1005$ , it follows that the only possible value for $a-b$ is $1$ (otherwise the quantity $a^2 - b^2$ would be greater than $2010$ ). Therefore the only possible ordered pairs for $(a,b)$ are $(504, 503)$ $(505, 504)$ , ... , $(1004, 1003)$ , so $a$ has $\boxed{501}$ possible values. | 501 |
6,126 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_6 | 1 | Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$ | Let $Q(x) = x^2 - 2x + 2$ $R(x) = 2x^2 - 4x + 3$ Completing the square , we have $Q(x) = (x-1)^2 + 1$ , and $R(x) = 2(x-1)^2 + 1$ , so it follows that $P(x) \ge Q(x) \ge 1$ for all $x$ (by the Trivial Inequality ).
Also, $1 = Q(1) \le P(1) \le R(1) = 1$ , so $P(1) = 1$ , and $P$ obtains its minimum at the point $(1,1)$... | 406 |
6,127 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_6 | 2 | Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$ | It can be seen that the function $P(x)$ must be in the form $P(x) = ax^2 - 2ax + c$ for some real $a$ and $c$ . This is because the derivative of $P(x)$ is $2ax - 2a$ , and a global minimum occurs only at $x = 1$ (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at $\frac{-b}{2a}$ ... | 406 |
6,128 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_6 | 3 | Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$ | Let $y = x^2 - 2x + 2$ ; note that $2y - 1 = 2x^2 - 4x + 3$ . Setting $y = 2y - 1$ , we find that equality holds when $y = 1$ and therefore when $x^2 - 2x + 2 = 1$ ; this is true iff $x = 1$ , so $P(1) = 1$
Let $Q(x) = P(x) - x$ ; clearly $Q(1) = 0$ , so we can write $Q(x) = (x - 1)Q'(x)$ , where $Q'(x)$ is some linear... | 406 |
6,129 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_6 | 4 | Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$ | Let $Q(x) = P(x) - (x^2-2x+2)$ , then $0\le Q(x) \le (x-1)^2$ (note this is derived from the given inequality chain). Therefore, $0\le Q(x+1) \le x^2 \Rightarrow Q(x+1) = Ax^2$ for some real value A.
$Q(11) = 10^2A \Rightarrow P(11)-(11^2-22+2)=100A \Rightarrow 80=100A \Rightarrow A=\frac{4}{5}$
$Q(16)=15^2A=180 \Righ... | 406 |
6,130 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_6 | 5 | Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$ | Let $P(x) = ax^2 + bx + c$ . Plugging in $x = 1$ to the expressions on both sides of the inequality, we see that $a + b + c = 1$ . We see from the problem statement that $121a + 11b + c = 181$ . Since we know the vertex of $P(x)$ lies at $x = 1$ , by symmetry we get $81a -9b + c = 181$ as well. Since we now have three ... | 406 |
6,131 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_6 | 6 | Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$ | Similar to Solution 5, let $P(x) = ax^2 + bx + c$ . Note that $(1,1)$ is a vertex of the polynomial. Additionally, this means that $b = -2a$ (since $\frac{-b}{2a}$ is the minimum $x$ point). Thus, we have $P(x) = ax^2 - 2ax + c$ . Therefore $a - 2a + c = 1$ . Moreover, $99a + c = 181$ . And so our polynomial is $\frac{... | 406 |
6,132 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_6 | 7 | Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$ | Very similar to Solution 6, start by noticing that $P(x)$ is between two polynomials, we try to set them equal to find a point where the two polynomials intersect, meaning that $P(x)$ would also have to intersect that point (it must be between the two graphs). Setting $x^2 - 2x + 2 = 2x^2 - 4x + 3$ , we find that $x = ... | 406 |
6,133 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_8 | 1 | For a real number $a$ , let $\lfloor a \rfloor$ denote the greatest integer less than or equal to $a$ . Let $\mathcal{R}$ denote the region in the coordinate plane consisting of points $(x,y)$ such that $\lfloor x \rfloor ^2 + \lfloor y \rfloor ^2 = 25$ . The region $\mathcal{R}$ is completely contained in a disk of ra... | The desired region consists of 12 boxes, whose lower-left corners are integers solutions of $x^2 + y^2 = 25$ , namely $(\pm5,0), (0,\pm5), (\pm3,\pm4), (\pm4,\pm3).$ Since the points themselves are symmetric about $(0,0)$ , the boxes are symmetric about $\left(\frac12,\frac12\right)$ . The distance from $\left(\frac12,... | 132 |
6,134 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_9 | 1 | Let $(a,b,c)$ be the real solution of the system of equations $x^3 - xyz = 2$ $y^3 - xyz = 6$ $z^3 - xyz = 20$ . The greatest possible value of $a^3 + b^3 + c^3$ can be written in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | This is almost the same as solution 1. Note $a^3 + b^3 + c^3 = 28 + 3abc$ . Next, let $k = a^3$ . Note that $b = \sqrt [3]{k + 4}$ and $c = \sqrt [3]{k + 18}$ , so we have $28 + 3\sqrt [3]{k(k+4)(k+18)} = 28+3abc=a^3+b^3+c^3=3k+22$ . Move 28 over, divide both sides by 3, then cube to get $k^3-6k^2+12k-8 = k^3+22k^2+18k... | 158 |
6,135 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_10 | 2 | Let $N$ be the number of ways to write $2010$ in the form $2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$ , where the $a_i$ 's are integers, and $0 \le a_i \le 99$ . An example of such a representation is $1\cdot 10^3 + 3\cdot 10^2 + 67\cdot 10^1 + 40\cdot 10^0$ . Find $N$ | Note that $a_0 \equiv 2010\ (\textrm{mod}\ 10)$ and $a_1 \equiv 2010 - a_0\ (\textrm{mod}\ 100)$ . It's easy to see that exactly 10 values in $0 \leq a_0 \leq 99$ that satisfy our first congruence. Similarly, there are 10 possible values of $a_1$ for each choice of $a_0$ . Thus, there are $10 \times 10 = 100$ possible... | 202 |
6,136 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_10 | 3 | Let $N$ be the number of ways to write $2010$ in the form $2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$ , where the $a_i$ 's are integers, and $0 \le a_i \le 99$ . An example of such a representation is $1\cdot 10^3 + 3\cdot 10^2 + 67\cdot 10^1 + 40\cdot 10^0$ . Find $N$ | We immediately see that $a_3$ can only be $0$ $1$ or $2$ . We also note that the maximum possible value for $10a_1 + a_0$ is $990 + 99 = 1089$ . We then split into cases:
Case 1: $a_3 = 0$ .
We try to find possible values of $a_2$ . We plug in $a_3 = 0$ and $10a_1 + a_0 = 1089$ to our initial equation, which gives us $... | 202 |
6,137 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_10 | 4 | Let $N$ be the number of ways to write $2010$ in the form $2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$ , where the $a_i$ 's are integers, and $0 \le a_i \le 99$ . An example of such a representation is $1\cdot 10^3 + 3\cdot 10^2 + 67\cdot 10^1 + 40\cdot 10^0$ . Find $N$ | We will represent the problem using generating functions. Consider the generating function \[f(x) = (1+x^{1000}+x^{2000}+\cdots+x^{99000})(1+x^{100}+x^{200}+\cdots+x^{9900})(1+x^{10}+x^{20}+\cdots+x^{990})(1+x+x^2+\cdots+x^{99})\] where the first factor represents $a_3$ , the second factor $a_2$ , and so forth. We want... | 202 |
6,138 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_10 | 5 | Let $N$ be the number of ways to write $2010$ in the form $2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$ , where the $a_i$ 's are integers, and $0 \le a_i \le 99$ . An example of such a representation is $1\cdot 10^3 + 3\cdot 10^2 + 67\cdot 10^1 + 40\cdot 10^0$ . Find $N$ | First note that $a_3$ has to be a single-digit number( $0$ $1$ , or $2$ to be exact), and that $a_1$ has to be a two-digit multiple of ten.
Then, $a_3$ $a_2$ $a_1$ and $a_0$ can be represented as follows: \begin{align*} a_3 = a \\ a_2 = 10b+c \\ a_1= 10d+e \\ a_0 = 10f \end{align*} , where $a$ $b$ $c$ $d$ $e$ , and $f$... | 202 |
6,139 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_11 | 1 | Let $\mathcal{R}$ be the region consisting of the set of points in the coordinate plane that satisfy both $|8 - x| + y \le 10$ and $3y - x \ge 15$ . When $\mathcal{R}$ is revolved around the line whose equation is $3y - x = 15$ , the volume of the resulting solid is $\frac {m\pi}{n\sqrt {p}}$ , where $m$ $n$ , and $p$ ... | The inequalities are equivalent to $y \ge x/3 + 5, y \le 10 - |x - 8|$ . We can set them equal to find the two points of intersection, $x/3 + 5 = 10 - |x - 8| \Longrightarrow |x - 8| = 5 - x/3$ . This implies that one of $x - 8, 8 - x = 5 - x/3$ , from which we find that $(x,y) = \left(\frac 92, \frac {13}2\right), \le... | 365 |
6,140 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_12 | 1 | Let $m \ge 3$ be an integer and let $S = \{3,4,5,\ldots,m\}$ . Find the smallest value of $m$ such that for every partition of $S$ into two subsets, at least one of the subsets contains integers $a$ $b$ , and $c$ (not necessarily distinct) such that $ab = c$
Note : a partition of $S$ is a pair of sets $A$ $B$ such that... | We claim that $243$ is the minimal value of $m$ . Let the two partitioned sets be $A$ and $B$ ; we will try to partition $3, 9, 27, 81,$ and $243$ such that the $ab=c$ condition is not satisfied. Without loss of generality , we place $3$ in $A$ . Then $9$ must be placed in $B$ , so $81$ must be placed in $A$ , and $27$... | 243 |
6,141 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_14 | 1 | For each positive integer $n$ , let $f(n) = \sum_{k = 1}^{100} \lfloor \log_{10} (kn) \rfloor$ . Find the largest value of $n$ for which $f(n) \le 300$
Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ | Observe that $f$ is strictly increasing in $n$ . We realize that we need $100$ terms to add up to around $300$ , so we need some sequence of $2$ s, $3$ s, and then $4$ s.
It follows that $n \approx 100$ (alternatively, use binary search to get to this, with $n\le 1000$ ). Manually checking shows that $f(109) = 300$ and... | 109 |
6,142 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_14 | 2 | For each positive integer $n$ , let $f(n) = \sum_{k = 1}^{100} \lfloor \log_{10} (kn) \rfloor$ . Find the largest value of $n$ for which $f(n) \le 300$
Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ | Because we want the value for which $f(n)=300$ , the average value of the 100 terms of the sequence should be around $3$ . For the value of $\lfloor \log_{10} (kn) \rfloor$ to be $3$ $1000 \le kn < 10000$ . We want kn to be around the middle of that range, and for k to be in the middle of 0 and 100, let $k=50$ , so $50... | 109 |
6,143 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_14 | 3 | For each positive integer $n$ , let $f(n) = \sum_{k = 1}^{100} \lfloor \log_{10} (kn) \rfloor$ . Find the largest value of $n$ for which $f(n) \le 300$
Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ | For any $n$ where the sum is close to $300$ , all the terms in the sum must be equal to $2$ $3$ or $4$ . Let $M$ be the number of terms less than or equal to $3$ and $N$ be the number of terms equal to $2$ (also counted in $M$ ). With this definition of $M$ and $N$ the total will be $400 - M - N \le 300$ , from which $... | 109 |
6,144 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_15 | 6 | In $\triangle{ABC}$ with $AB = 12$ $BC = 13$ , and $AC = 15$ , let $M$ be a point on $\overline{AC}$ such that the incircles of $\triangle{ABM}$ and $\triangle{BCM}$ have equal radii . Then $\frac{AM}{CM} = \frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ | Let $BM = d, AM = x, CM = 15 - x$ . Observe that we have the equation by the incircle formula: \[\frac{[ABM]}{12 + AM + MB} = \frac{[CBM]}{13 + CM + MB} \implies \frac{AM}{CM} = \frac{12 + MB}{13 + MB} \implies \frac{x}{15 - x} = \frac{12 + d}{13 + d}.\] Now let $X$ be the point of tangency between the incircle of $\tr... | 45 |
6,145 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_2 | 1 | A point $P$ is chosen at random in the interior of a unit square $S$ . Let $d(P)$ denote the distance from $P$ to the closest side of $S$ . The probability that $\frac{1}{5}\le d(P)\le\frac{1}{3}$ is equal to $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Any point outside the square with side length $\frac{1}{3}$ that has the same center and orientation as the unit square and inside the square with side length $\frac{3}{5}$ that has the same center and orientation as the unit square has $\frac{1}{5}\le d(P)\le\frac{1}{3}$
Since the area of the unit square is $1$ , the ... | 281 |
6,146 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_2 | 2 | A point $P$ is chosen at random in the interior of a unit square $S$ . Let $d(P)$ denote the distance from $P$ to the closest side of $S$ . The probability that $\frac{1}{5}\le d(P)\le\frac{1}{3}$ is equal to $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | First, let's figure out $d(P) \geq \frac{1}{3}$ which is \[\left(\frac{3}{5}\right)^2=\frac{9}{25}.\] Then, $d(P) \geq \frac{1}{5}$ is a square inside $d(P) \geq \frac{1}{3}$ , so \[\left(\frac{1}{3}\right)^2=\frac{1}{9}.\] Therefore, the probability that $\frac{1}{5}\le d(P)\le\frac{1}{3}$ is \[\frac{9}{25}-\frac{1}{9... | 281 |
6,147 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_3 | 1 | Let $K$ be the product of all factors $(b-a)$ (not necessarily distinct) where $a$ and $b$ are integers satisfying $1\le a < b \le 20$ . Find the greatest positive integer $n$ such that $2^n$ divides $K$ | In general, there are $20-n$ pairs of integers $(a, b)$ that differ by $n$ because we can let $b$ be any integer from $n+1$ to $20$ and set $a$ equal to $b-n$ . Thus, the product is $(1^{19})(2^{18})\cdots(19^1)$ (or alternatively, $19! \cdot 18! \cdots 1!$ .)
When we count the number of factors of $2$ , we have 4 grou... | 150 |
6,148 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_7 | 1 | Let $P(z)=z^3+az^2+bz+c$ , where a, b, and c are real. There exists a complex number $w$ such that the three roots of $P(z)$ are $w+3i$ $w+9i$ , and $2w-4$ , where $i^2=-1$ . Find $|a+b+c|$ | Set $w=x+yi$ , so $x_1 = x+(y+3)i$ $x_2 = x+(y+9)i$ $x_3 = 2x-4+2yi$
Since $a,b,c\in{R}$ , the imaginary part of $a,b,c$ must be $0$
Start with a, since it's the easiest one to do: $y+3+y+9+2y=0, y=-3$
and therefore: $x_1 = x$ $x_2 = x+6i$ $x_3 = 2x-4-6i$
Now, do the part where the imaginary part of c is 0 since it's t... | 136 |
6,149 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_7 | 2 | Let $P(z)=z^3+az^2+bz+c$ , where a, b, and c are real. There exists a complex number $w$ such that the three roots of $P(z)$ are $w+3i$ $w+9i$ , and $2w-4$ , where $i^2=-1$ . Find $|a+b+c|$ | Note that at least one of $w+3i$ $w+9i$ , or $2w-4$ is real by complex conjugate roots. We now separate into casework based on which one.
Let $w=x+yi$ , where $x$ and $y$ are reals.
Case 1: $w+3i$ is real. This implies that $x+yi+3i$ is real, so by setting the imaginary part equal to zero we get $y=-3$ , so $w=x-3i$ . ... | 136 |
6,150 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_8 | 1 | Let $N$ be the number of ordered pairs of nonempty sets $\mathcal{A}$ and $\mathcal{B}$ that have the following properties:
Find $N$ | Let us partition the set $\{1,2,\cdots,12\}$ into $n$ numbers in $A$ and $12-n$ numbers in $B$
Since $n$ must be in $B$ and $12-n$ must be in $A$ $n\ne6$ , we cannot partition into two sets of 6 because $6$ needs to end up somewhere, $n\ne 0$ or $12$ either).
We have $\dbinom{10}{n-1}$ ways of picking the numbers to be... | 772 |
6,151 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_8 | 2 | Let $N$ be the number of ordered pairs of nonempty sets $\mathcal{A}$ and $\mathcal{B}$ that have the following properties:
Find $N$ | Regardless of the size $n$ of $A$ (ignoring the case when $n = 6$ ), $n$ must not be in $A$ and $12 - n$ must be in $A$
There are $10$ remaining elements whose placements have yet to be determined. Note that the actual value of $n$ does not matter; there is always $1$ necessary element, $1$ forbidden element, and $10$ ... | 772 |
6,152 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_8 | 3 | Let $N$ be the number of ordered pairs of nonempty sets $\mathcal{A}$ and $\mathcal{B}$ that have the following properties:
Find $N$ | The total number of possible subsets is $\sum_{i=1}^{11}\dbinom{12}{i}$ , which is $2^{12}-2$ . Note that picking a subset from the set leaves the rest of the set to be in the other subset. We exclude $i=0$ and $i=12$ since they leave a set empty. We proceed with complementary counting and casework:
We apply the Princi... | 772 |
6,153 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_10 | 1 | Find the number of second-degree polynomials $f(x)$ with integer coefficients and integer zeros for which $f(0)=2010$ | Let $f(x) = a(x-r)(x-s)$ . Then $ars=2010=2\cdot3\cdot5\cdot67$ . First consider the case where $r$ and $s$ (and thus $a$ ) are positive. There are $3^4 = 81$ ways to split up the prime factors between $a$ $r$ , and $s$ . However, $r$ and $s$ are indistinguishable. In one case, $(a,r,s) = (2010,1,1)$ , we have $r=s$ . ... | 163 |
6,154 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_10 | 3 | Find the number of second-degree polynomials $f(x)$ with integer coefficients and integer zeros for which $f(0)=2010$ | The equation can be written in the form of $k(x-a)(x-b)$ , where $|k|$ is a divisor of $2010$ . Factor $2010=2\cdot 3\cdot 5\cdot 67$ . We divide into few cases to study.
Firstly, if $|k|=1$ , the equation can be $-(x-a)(x+b)$ or $(x-a)(x-b)$ or $(x+a)(x+b)$ , there are $2^4+2^4=32$ ways
If $|k|\in \{2,3,5,67\}$ . Take... | 163 |
6,155 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_11 | 1 | Define a T-grid to be a $3\times3$ matrix which satisfies the following two properties:
Find the number of distinct T-grids | The T-grid can be considered as a tic-tac-toe board: five $1$ 's (or X's) and four $0$ 's (or O's).
There are only $\dbinom{9}{5} = 126$ ways to fill the board with five $1$ 's and four $0$ 's. Now we just need to subtract the number of bad grids. Bad grids are ones with more than one person winning, or where someone h... | 68 |
6,156 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_11 | 2 | Define a T-grid to be a $3\times3$ matrix which satisfies the following two properties:
Find the number of distinct T-grids | We can use generating functions to compute the case that no row or column is completely filled with $1$ 's. Given a row, let $a$ $b$ $c$ be the events that the first, second, third respective squares are $1$ 's. Then the generating function representing the possible events that exclude a row of $1,1,1$ or $0,0,0$ from ... | 68 |
6,157 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_12 | 1 | Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is $8: 7$ . Find the minimum possible value of their common perimeter. | Let $s$ be the semiperimeter of the two triangles. Also, let the base of the longer triangle be $16x$ and the base of the shorter triangle be $14x$ for some arbitrary factor $x$ . Then, the dimensions of the two triangles must be $s-8x,s-8x,16x$ and $s-7x,s-7x,14x$ . By Heron's Formula, we have
Since $15$ and $338$ are... | 676 |
6,158 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_12 | 2 | Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is $8: 7$ . Find the minimum possible value of their common perimeter. | Let the first triangle have sides $16n,a,a$ , so the second has sides $14n,a+n,a+n$ . The height of the first triangle is $\frac{7}{8}$ the height of the second triangle. Therefore, we have \[a^2-64n^2=\frac{49}{64}((a+n)^2-49n^2).\] Multiplying this, we get \[64a^2-4096n^2=49a^2+98an-2352n^2,\] which simplifies to \[1... | 676 |
6,159 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_13 | 1 | The $52$ cards in a deck are numbered $1, 2, \cdots, 52$ . Alex, Blair, Corey, and Dylan each picks a card from the deck without replacement and with each card being equally likely to be picked, The two persons with lower numbered cards form a team, and the two persons with higher numbered cards form another team. Let ... | Once the two cards are drawn, there are $\dbinom{50}{2} = 1225$ ways for the other two people to draw. Alex and Dylan are the team with higher numbers if Blair and Corey both draw below $a$ , which occurs in $\dbinom{a-1}{2}$ ways. Alex and Dylan are the team with lower numbers if Blair and Corey both draw above $a+9$ ... | 263 |
6,160 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_15 | 1 | In triangle $ABC$ $AC = 13$ $BC = 14$ , and $AB=15$ . Points $M$ and $D$ lie on $AC$ with $AM=MC$ and $\angle ABD = \angle DBC$ . Points $N$ and $E$ lie on $AB$ with $AN=NB$ and $\angle ACE = \angle ECB$ . Let $P$ be the point, other than $A$ , of intersection of the circumcircles of $\triangle AMN$ and $\triangle ADE... | Define the function $f:\mathbb{R}^{2}\rightarrow\mathbb{R}$ by \[f(X)=\text{Pow}_{(AMN)}(X)-\text{Pow}_{(ADE)}(X)\] for points $X$ in the plane. Then $f$ is linear, so $\frac{BQ}{CQ}=\frac{f(B)-f(Q)}{f(Q)-f(C)}$ . But $f(Q)=0$ since $Q$ lies on the radical axis of $(AMN)$ $(ADE)$ thus \[\frac{BQ}{CQ}=-\frac{f(B)}{f(C)}... | 218 |
6,161 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_15 | 2 | In triangle $ABC$ $AC = 13$ $BC = 14$ , and $AB=15$ . Points $M$ and $D$ lie on $AC$ with $AM=MC$ and $\angle ABD = \angle DBC$ . Points $N$ and $E$ lie on $AB$ with $AN=NB$ and $\angle ACE = \angle ECB$ . Let $P$ be the point, other than $A$ , of intersection of the circumcircles of $\triangle AMN$ and $\triangle ADE... | Let $Y = MN \cap AQ$ $\frac {BQ}{QC} = \frac {NY}{MY}$ since $\triangle AMN \sim \triangle ACB$ . Since quadrilateral $AMPN$ is cyclic, $\triangle MYA \sim \triangle PYN$ and $\triangle MYP \sim \triangle AYN$ , yielding $\frac {YM}{YA} = \frac {MP}{AN}$ and $\frac {YA}{YN} = \frac {AM}{PN}$ . Multiplying these togethe... | 218 |
6,162 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_1 | 1 | Call a $3$ -digit number geometric if it has $3$ distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers. | Assume that the largest geometric number starts with a $9$ . We know that the common ratio must be a rational of the form $k/3$ for some integer $k$ , because a whole number should be attained for the 3rd term as well. When $k = 1$ , the number is $931$ . When $k = 2$ , the number is $964$ . When $k = 3$ , we get $999$... | 840 |
6,163 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_1 | 2 | Call a $3$ -digit number geometric if it has $3$ distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers. | Consider the three-digit number $abc$ . If its digits form a geometric progression, we must have that ${a \over b} = {b \over c}$ , that is, $b^2 = ac$
The minimum and maximum geometric numbers occur when $a$ is minimized and maximized, respectively. The minimum occurs when $a = 1$ ; letting $b = 2$ and $c = 4$ achieve... | 840 |
6,164 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_1 | 3 | Call a $3$ -digit number geometric if it has $3$ distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers. | The smallest geometric number is $124$ because $123$ and any number containing a zero does not work. $964$ is the largest geometric number because the middle digit cannot be 8 or 7. Subtracting the numbers gives $\boxed{840}.$ | 840 |
6,165 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_2 | 1 | There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that
\[\frac {z}{z + n} = 4i.\]
Find $n$ | Let $z = a + 164i$
Then \[\frac {a + 164i}{a + 164i + n} = 4i\] and \[a + 164i = \left (4i \right ) \left (a + n + 164i \right ) = 4i \left (a + n \right ) - 656.\]
By comparing coefficients, equating the real terms on the leftmost and rightmost side of the equation,
we conclude that \[a = -656.\]
By equating the imagi... | 697 |
6,166 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_2 | 2 | There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that
\[\frac {z}{z + n} = 4i.\]
Find $n$ | \[\frac {z}{z+n}=4i\]
\[1-\frac {n}{z+n}=4i\]
\[1-4i=\frac {n}{z+n}\]
\[\frac {1}{1-4i}=\frac {z+n}{n}\]
\[\frac {1+4i}{17}=\frac {z}{n}+1\]
Since their imaginary part has to be equal,
\[\frac {4i}{17}=\frac {164i}{n}\]
\[n=\frac {(164)(17)}{4}=697\]
\[n = \boxed{697}.\] | 697 |
6,167 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_2 | 3 | There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that
\[\frac {z}{z + n} = 4i.\]
Find $n$ | Below is an image of the complex plane. Let $\operatorname{Im}(z)$ denote the imaginary part of a complex number $z$ [asy] unitsize(1cm); xaxis("Re",Arrows); yaxis("Im",Arrows); real f(real x) {return 164;} pair F(real x) {return (x,f(x));} draw(graph(f,-700,100),red,Arrows); label("Im$(z)=164$",F(-330),N); pair... | 697 |
6,168 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_2 | 4 | There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that
\[\frac {z}{z + n} = 4i.\]
Find $n$ | Taking the reciprocal of our equation gives us $1 + \frac{n}{z} = \frac{1}{4i}.$ Therefore, \[\frac{n}{z} = \frac{1-4i}{4i} = \frac{17}{-16+4i}.\] Since $z$ has an imaginary part of $164$ , we must multiply both sides of our RHS fraction by $\frac{164}{4} = 41$ so that its denominator's imaginary part matches the LHS f... | 697 |
6,169 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_4 | 1 | In parallelogram $ABCD$ , point $M$ is on $\overline{AB}$ so that $\frac {AM}{AB} = \frac {17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac {AN}{AD} = \frac {17}{2009}$ . Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$ . Find $\frac {AC}{AP}$ | One of the ways to solve this problem is to make this parallelogram a straight line.
So the whole length of the line is $APC$ $AMC$ or $ANC$ ), and $ABC$ is $1000x+2009x=3009x.$
$AP$ $AM$ or $AN$ ) is $17x.$
So the answer is $3009x/17x = \boxed{177}$ | 177 |
6,170 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_4 | 2 | In parallelogram $ABCD$ , point $M$ is on $\overline{AB}$ so that $\frac {AM}{AB} = \frac {17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac {AN}{AD} = \frac {17}{2009}$ . Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$ . Find $\frac {AC}{AP}$ | Draw a diagram with all the given points and lines involved. Construct parallel lines $\overline{DF_2F_1}$ and $\overline{BB_1B_2}$ to $\overline{MN}$ , where for the lines the endpoints are on $\overline{AM}$ and $\overline{AN}$ , respectively, and each point refers to an intersection. Also, draw the median of quadril... | 177 |
6,171 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_4 | 3 | In parallelogram $ABCD$ , point $M$ is on $\overline{AB}$ so that $\frac {AM}{AB} = \frac {17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac {AN}{AD} = \frac {17}{2009}$ . Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$ . Find $\frac {AC}{AP}$ | Assume, for the ease of computation, that $AM=AN=17$ $AB=1000$ , and $AD=2009$ . Now, let line $MN$ intersect line $CD$ at point $X$ and let $Y$ be a point such that $XY\parallel AD$ and $AY\parallel DX$ . As a result, $ADXY$ is a parallelogram. By construction, $\triangle MAN\sim \triangle MYX$ so \[\frac{MY}{MA}=\... | 177 |
6,172 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_4 | 4 | In parallelogram $ABCD$ , point $M$ is on $\overline{AB}$ so that $\frac {AM}{AB} = \frac {17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac {AN}{AD} = \frac {17}{2009}$ . Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$ . Find $\frac {AC}{AP}$ | Assign $A = (0,0)$ . Since there are no constraints in the problem against this, assume $ABCD$ to be a rectangle with dimensions $1000 \times 2009.$ Now, we can assign \[A=(0,0)\] \[B=(1000, 0)\] \[C=(1000,-2009)\] \[D=(0,-2009).\]
Then, since $\frac{AM}{AB} = \frac{17}{1000}$ and $AB = 1000$ , we can place $M$ at $(17... | 177 |
6,173 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_6 | 1 | How many positive integers $N$ less than $1000$ are there such that the equation $x^{\lfloor x\rfloor} = N$ has a solution for $x$ | First, $x$ must be less than $5$ , since otherwise $x^{\lfloor x\rfloor}$ would be at least $3125$ which is greater than $1000$
Because ${\lfloor x\rfloor}$ must be an integer, let’s do case work based on ${\lfloor x\rfloor}$
For ${\lfloor x\rfloor}=0$ $N=1$ as long as $x \neq 0$ . This gives us $1$ value of $N$
For ${... | 412 |
6,174 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_6 | 2 | How many positive integers $N$ less than $1000$ are there such that the equation $x^{\lfloor x\rfloor} = N$ has a solution for $x$ | For a positive integer $k$ , we find the number of positive integers $N$ such that $x^{\lfloor x\rfloor}=N$ has a solution with ${\lfloor x\rfloor}=k$ . Then $x=\sqrt[k]{N}$ , and because $k \le x < k+1$ , we have $k^k \le x^k < (k+1)^k$ , and because $(k+1)^k$ is an integer, we get $k^k \le x^k \le (k+1)^k-1$ . The nu... | 412 |
6,175 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_7 | 2 | The sequence $(a_n)$ satisfies $a_1 = 1$ and $5^{(a_{n + 1} - a_n)} - 1 = \frac {1}{n + \frac {2}{3}}$ for $n \geq 1$ . Let $k$ be the least integer greater than $1$ for which $a_k$ is an integer. Find $k$ | We notice that by multiplying the equation from an arbitrary $a_n$ all the way to $a_1$ , we get: \[5^{a_n-a_1}=\dfrac{n+\tfrac23}{1+\tfrac23}\] This simplifies to \[5^{a_n}=3n+2.\] We can now test powers of $5$
$5$ - that gives us $n=1$ , which is useless.
$25$ - that gives a non-integer $n$
$125$ - that gives $n=\box... | 41 |
6,176 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_8 | 2 | Let $S = \{2^0,2^1,2^2,\ldots,2^{10}\}$ . Consider all possible positive differences of pairs of elements of $S$ . Let $N$ be the sum of all of these differences. Find the remainder when $N$ is divided by $1000$ | When computing $N$ , the number $2^x$ will be added $x$ times (for terms $2^x-2^0$ $2^x-2^1$ , ..., $2^x - 2^{x-1}$ ), and subtracted $10-x$ times. Hence $N$ can be computed as $N=10\cdot 2^{10} + 8\cdot 2^9 + 6\cdot 2^8 + \cdots - 8\cdot 2^1 - 10\cdot 2^0$ . Evaluating $N \bmod {1000}$ yields:
\begin{align*} N & = 10... | 398 |
6,177 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_8 | 3 | Let $S = \{2^0,2^1,2^2,\ldots,2^{10}\}$ . Consider all possible positive differences of pairs of elements of $S$ . Let $N$ be the sum of all of these differences. Find the remainder when $N$ is divided by $1000$ | This solution can be generalized to apply when $10$ is replaced by other positive integers.
Extending from Solution 2, we get that the sum $N$ of all possible differences of pairs of elements in $S$ when $S = \{2^0,2^1,2^2,\ldots,2^{n}\}$ is equal to $\sum_{x=0}^{n} (2x-n) 2^x$ . Let $A = \sum_{x=0}^{n} x2^x$ $B=\sum_{... | 398 |
6,178 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_9 | 1 | A game show offers a contestant three prizes A, B and C, each of which is worth a whole number of dollars from $$ 1$ to $$ 9999$ inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A, B, C. As a hint, the digits of the three prices are given. On a particular day, the dig... | [Clarification: You are supposed to find the number of all possible tuples of prices, $(A, B, C)$ , that could have been on that day.]
Since we have three numbers, consider the number of ways we can put these three numbers together in a string of 7 digits. For example, if $A=113, B=13, C=31$ , then the string is
\[1131... | 420 |
6,179 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_10 | 1 | The Annual Interplanetary Mathematics Examination (AIME) is written by a committee of five Martians, five Venusians, and five Earthlings. At meetings, committee members sit at a round table with chairs numbered from $1$ to $15$ in clockwise order. Committee rules state that a Martian must occupy chair $1$ and an Eart... | Since the 5 members of each planet committee are distinct we get that the number of arrangement of sittings is in the form $N*(5!)^3$ because for each $M, V, E$ sequence we have $5!$ arrangements within the Ms, Vs, and Es.
Pretend the table only seats $3$ "people", with $1$ "person" from each planet. Counting clockwise... | 346 |
6,180 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_11 | 1 | Consider the set of all triangles $OPQ$ where $O$ is the origin and $P$ and $Q$ are distinct points in the plane with nonnegative integer coordinates $(x,y)$ such that $41x + y = 2009$ . Find the number of such distinct triangles whose area is a positive integer. | Let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$
We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem).
$\det \left(\begin{array}{c} P \\ Q\end{array}\right)=\det \left(\begin{array}{cc}x_1 &y_1... | 600 |
6,181 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_11 | 2 | Consider the set of all triangles $OPQ$ where $O$ is the origin and $P$ and $Q$ are distinct points in the plane with nonnegative integer coordinates $(x,y)$ such that $41x + y = 2009$ . Find the number of such distinct triangles whose area is a positive integer. | As in the solution above, let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$
If the coordinates of $P$ and $Q$ have nonnegative integer coordinates, $P$ and $Q$ must be lattice points either
We can calculate the y-intercept of the line $41x+y=2009$ to be $(0,2009)$ and the x-int... | 600 |
6,182 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_12 | 5 | In right $\triangle ABC$ with hypotenuse $\overline{AB}$ $AC = 12$ $BC = 35$ , and $\overline{CD}$ is the altitude to $\overline{AB}$ . Let $\omega$ be the circle having $\overline{CD}$ as a diameter. Let $I$ be a point outside $\triangle ABC$ such that $\overline{AI}$ and $\overline{BI}$ are both tangent to circle $\o... | This solution is not a real solution and is solving the problem with a ruler and compass.
Draw $AC = 4.8, BC = 14, AB = 14.8$ . Then, drawing the tangents and intersecting them, we get that $IA$ is around $6.55$ and $IB$ is around $18.1$ . We then find the ratio to be around $\frac{39.45}{14.8}$ . Using long division, ... | 11 |
6,183 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_12 | 6 | In right $\triangle ABC$ with hypotenuse $\overline{AB}$ $AC = 12$ $BC = 35$ , and $\overline{CD}$ is the altitude to $\overline{AB}$ . Let $\omega$ be the circle having $\overline{CD}$ as a diameter. Let $I$ be a point outside $\triangle ABC$ such that $\overline{AI}$ and $\overline{BI}$ are both tangent to circle $\o... | Denoting three tangents has length $h_1,h_2,h_3$ while $h_1,h_3$ lies on $AB$ with $h_1>h_3$ .The area of $ABC$ is $1/2*12*35 = 1/2*37*CD$ , so $CD=\frac{420}{37}$ and the inradius of $\triangle ABI$ is $r=\frac{210}{37}$ .As we know that the diameter of the circle is the height of $\triangle ACB$ from $C$ to $AB$ . As... | 11 |
6,184 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_13 | 3 | The terms of the sequence $(a_i)$ defined by $a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}$ for $n \ge 1$ are positive integers. Find the minimum possible value of $a_1 + a_2$ | Essentially you see that it must be an integer for infinite numbers, which doesn't quite seem probable. The most logical explanation is that the sequence repeats, and the numbers in the sequence that repeat are integers. We list out some terms. \begin{align*} a_{1} &= a \\ a_{2} &= b \\ a_{3} &=\frac{a+2009}{1+b} \\ a_... | 90 |
6,185 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_14 | 1 | For $t = 1, 2, 3, 4$ , define $S_t = \sum_{i = 1}^{350}a_i^t$ , where $a_i \in \{1,2,3,4\}$ . If $S_1 = 513$ and $S_4 = 4745$ , find the minimum possible value for $S_2$ | Because the order of the $a$ 's doesn't matter, we simply need to find the number of $1$ $2$ $3$ s and $4$ s that minimize $S_2$ . So let $w, x, y,$ and $z$ represent the number of $1$ s, $2$ s, $3$ s, and $4$ s respectively. Then we can write three equations based on these variables. Since there are a total of $350$ $... | 905 |
6,186 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_15 | 1 | In triangle $ABC$ $AB = 10$ $BC = 14$ , and $CA = 16$ . Let $D$ be a point in the interior of $\overline{BC}$ . Let points $I_B$ and $I_C$ denote the incenters of triangles $ABD$ and $ACD$ , respectively. The circumcircles of triangles $BI_BD$ and $CI_CD$ meet at distinct points $P$ and $D$ . The maximum possible area ... | First, by the Law of Cosines , we have \[\cos BAC = \frac {16^2 + 10^2 - 14^2}{2\cdot 10 \cdot 16} = \frac {256+100-196}{320} = \frac {1}{2},\] so $\angle BAC = 60^\circ$
Let $O_1$ and $O_2$ be the circumcenters of triangles $BI_BD$ and $CI_CD$ , respectively. We first compute \[\angle BO_1D = \angle BO_1I_B + \angle ... | 150 |
6,187 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_15 | 2 | In triangle $ABC$ $AB = 10$ $BC = 14$ , and $CA = 16$ . Let $D$ be a point in the interior of $\overline{BC}$ . Let points $I_B$ and $I_C$ denote the incenters of triangles $ABD$ and $ACD$ , respectively. The circumcircles of triangles $BI_BD$ and $CI_CD$ meet at distinct points $P$ and $D$ . The maximum possible area ... | From Law of Cosines on $\triangle{ABC}$ \[\cos{A}=\frac{16^2+10^2-14^2}{2\cdot 10\cdot 16}=\frac{1}{2}\implies\angle{A}=60^\circ.\] Now, \[\angle{CI_CD}+\angle{BI_BD}=180^\circ+\frac{\angle{A}}{2}=210^\circ.\] Since $CI_CDP$ and $BI_BDP$ are cyclic quadrilaterals, it follows that \[\angle{BPC}=\angle{CPD}+\angle{DPB}=(... | 150 |
6,188 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_15 | 3 | In triangle $ABC$ $AB = 10$ $BC = 14$ , and $CA = 16$ . Let $D$ be a point in the interior of $\overline{BC}$ . Let points $I_B$ and $I_C$ denote the incenters of triangles $ABD$ and $ACD$ , respectively. The circumcircles of triangles $BI_BD$ and $CI_CD$ meet at distinct points $P$ and $D$ . The maximum possible area ... | Proceed as in Solution 2 until you find $\angle CPB = 150$ . The locus of points $P$ that give $\angle CPB = 150$ is a fixed arc from $B$ to $C$ $P$ will move along this arc as $D$ moves along $BC$ ) and we want to maximise the area of [ $\triangle BPC$ ]. This means we want $P$ to be farthest distance away from $BC$ a... | 150 |
6,189 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_1 | 1 | Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When ... | Let $x$ be the amount of paint in a strip. The total amount of paint is $130+164+188=482$ . After painting the wall, the total amount of paint is $482-4x$ . Because the total amount in each color is the same after finishing, we have
\[\frac{482-4x}{3} = 130-x\] \[482-4x=390-3x\] \[x=92\]
Our answer is $482-4\cdot 92 = ... | 114 |
6,190 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_1 | 2 | Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When ... | Let the stripes be $b, r, w,$ and $p$ , respectively. Let the red part of the pink be $\frac{r_p}{p}$ and the white part be $\frac{w_p}{p}$ for $\frac{r_p+w_p}{p}=p$
Since the stripes are of equal size, we have $b=r=w=p$ . Since the amounts of paint end equal, we have $130-b=164-r-\frac{r_p}{p}=188-w-\frac{w_p}{p}$ . T... | 114 |
6,191 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_1 | 3 | Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When ... | After the pink stripe is drawn, all three colors will be used equally so the pink stripe must bring the amount of red and white paint down to $130$ ounces each. Say $a$ is the fraction of the pink paint that is red paint and $x$ is the size of each stripe. Then equations can be written: $ax = 164 - 130 = 34$ and $(1-a)... | 114 |
6,192 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_1 | 4 | Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When ... | We know that all the stripes are of equal size. We can then say that $r$ is the amount of paint per stripe. Then $130 - r$ will be the amount of blue paint left. Now for the other two stripes. The amount of white paint left after the white stripe and the amount of red paint left after the blue stripe are $188 - r$ and ... | 114 |
6,193 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_1 | 5 | Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When ... | Let the amount of paint each stripe painted used be $x$ . Also, let the amount of paint of each color left be $y$ . 1 stripe is drawn with the blue paint, and 3 stripes are drawn with the red and white paints. Add together the amount of red and white paint, $164 + 188 = 352$ and obtain the following equations : $352 - ... | 114 |
6,194 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_1 | 6 | Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When ... | Let $x$ be the number of ounces of paint needed for a single stripe. We know that in the end, the total amount of red and white paint will equal double the amount of blue paint.
After painting, the amount of red and white paint remaining is equal to $164+188-2x$ , and then minus another $x$ for the pink stripe. The amo... | 114 |
6,195 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_1 | 7 | Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When ... | Just like in solution 1, we note that all colors will be used equally, except for the pink stripe. This must bring red and white down to $130$ each, so $34$ red and $58$ white are used, making for a total of $92$ for the pink stripe. Thus, the other stripes also use $92$ . The answer is $130+164+188-4(92)=\boxed{114}$ | 114 |
6,196 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_2 | 1 | Suppose that $a$ $b$ , and $c$ are positive real numbers such that $a^{\log_3 7} = 27$ $b^{\log_7 11} = 49$ , and $c^{\log_{11}25} = \sqrt{11}$ . Find \[a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}.\] | First, we have: \[x^{(\log_y z)^2} = x^{\left( (\log_y z)^2 \right) } = x^{(\log_y z) \cdot (\log_y z) } = \left( x^{\log_y z} \right)^{\log_y z}\]
Now, let $x=y^w$ , then we have: \[x^{\log_y z} = \left( y^w \right)^{\log_y z} = y^{w\log_y z} = y^{\log_y (z^w)} = z^w\]
This is all we need to evaluate the given for... | 469 |
6,197 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_2 | 2 | Suppose that $a$ $b$ , and $c$ are positive real numbers such that $a^{\log_3 7} = 27$ $b^{\log_7 11} = 49$ , and $c^{\log_{11}25} = \sqrt{11}$ . Find \[a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}.\] | We know from the first three equations that $\log_a27 = \log_37$ $\log_b49 = \log_711$ , and $\log_c\sqrt{11} = \log_{11}25$ . Substituting, we find
\[a^{(\log_a27)(\log_37)} + b^{(\log_b49)(\log_711)} + c^{(\log_c\sqrt {11})(\log_{11}25)}.\]
We know that $x^{\log_xy} =y$ , so we find
\[27^{\log_37} + 49^{\log_711} + \... | 469 |
6,198 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_3 | 1 | In rectangle $ABCD$ $AB=100$ . Let $E$ be the midpoint of $\overline{AD}$ . Given that line $AC$ and line $BE$ are perpendicular, find the greatest integer less than $AD$ | From the problem, $AB=100$ and triangle $FBA$ is a right triangle. As $ABCD$ is a rectangle, triangles $BCA$ , and $ABE$ are also right triangles. By $AA$ $\triangle FBA \sim \triangle BCA$ , and $\triangle FBA \sim \triangle ABE$ , so $\triangle ABE \sim \triangle BCA$ . This gives $\frac {AE}{AB}= \frac {AB}{BC}$ $AE... | 141 |
6,199 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_3 | 2 | In rectangle $ABCD$ $AB=100$ . Let $E$ be the midpoint of $\overline{AD}$ . Given that line $AC$ and line $BE$ are perpendicular, find the greatest integer less than $AD$ | Let $x$ be the ratio of $BC$ to $AB$ . On the coordinate plane, plot $A=(0,0)$ $B=(100,0)$ $C=(100,100x)$ , and $D=(0,100x)$ . Then $E=(0,50x)$ . Furthermore, the slope of $\overline{AC}$ is $x$ and the slope of $\overline{BE}$ is $-x/2$ . They are perpendicular, so they multiply to $-1$ , that is, \[x\cdot-\frac{x}{2}... | 141 |
6,200 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_3 | 3 | In rectangle $ABCD$ $AB=100$ . Let $E$ be the midpoint of $\overline{AD}$ . Given that line $AC$ and line $BE$ are perpendicular, find the greatest integer less than $AD$ | Draw $CX$ and $EX$ to form a parallelogram $AEXC$ . Since $EX \parallel AC$ $\angle BEX=90^\circ$ by the problem statement, so $\triangle BEX$ is right.
Letting $AE=y$ , we have $BE=\sqrt{100^2+y^2}$ and $AC=EX=\sqrt{100^2+(2y)^2}$ . Since $CX=EA$ $\sqrt{100^2+y^2}^2+\sqrt{100^2+(2y)^2}=(3y)^2$ . Solving this, we have... | 141 |
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