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 |
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5,601 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_1 | 2 | Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$ | Similar to the previous solution, we plot the triangle and cut it in half. Then, we find the number of boundary points, which is $101+51+51-3$ , or just $200$ . Using Pick's theorem, we know that the area of the half-triangle, which is $2500$ , is just $I+100-1$ . That means that there are $2401$ interior points, plus ... | 600 |
5,602 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_1 | 3 | Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$ | Notice that for $x^2+ax+b$ to be true, for every $a$ $b$ will always be the product of the possibilities of how to add two integers to $a$ . For example, if $a=3$ $b$ will be the product of $(3,0)$ and $(2,1)$ , as those two sets are the only possibilities of adding two integers to $a$ . Note that order does not matter... | 600 |
5,603 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_1 | 4 | Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$ | We see the pattern $1, 2; 2, 3; 3, 4; ...$ . There are 50 pairs of $i, i+1$ in this pattern, and each pair sums to $2i+1$ . So the pattern condenses to $3, 5, 7, ...$ for 50 terms. This is just $1+3+5+...$ for 51 terms, minus $1$ , or $51^2-1=2601-1=2600\implies\boxed{600}$ | 600 |
5,604 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_1 | 5 | Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$ | The following link is the URL to the graph I drew showing the relationship between a-values and b-values http://artofproblemsolving.com/wiki/index.php?title=File:Screen_Shot_2018-04-30_at_8.15.00_PM.png#file
The pattern continues until $a=100$ , and in total, there are $49$ pairs of a-value with the same amount of b-va... | 600 |
5,605 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_1 | 6 | Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$ | Let's say that the quadratic $x^2 + ax + b$ can be factored into $(x+c)(x+d)$ where $c$ and $d$ are non-negative numbers. We can't have both of them zero because $a$ would not be within bounds. Also, $c+d \leq 100$ . Assume that $c < d$ $d$ can be written as $c + x$ where $x \geq 0$ . Therefore, $c + d = 2c + x \leq 10... | 600 |
5,606 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_1 | 7 | Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$ | By Vietas, the sum of the roots is $-a$ and the product is $b$ . Therefore, both roots are nonpositive. For each value of $a$ from $1$ to $100$ , the number of $b$ values is the number of ways to sum two numbers between $0$ and $a-1$ inclusive to $a$ . This is just $1 + 2 + 2 + 3 + 3 +... 50 + 50 + 51 = 2600$ . Thus, t... | 600 |
5,607 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_1 | 8 | Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$ | Similar to solution 1 we plot the triangle and half it. From dividing the triangle in half we are removing the other half of answers that are just flipped coordinates. We notice that we can measure the length of the longest side of the half triangle which is just from $0$ to $100$ , so the number of points on that line... | 600 |
5,608 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_2 | 1 | The number $n$ can be written in base $14$ as $\underline{a}\text{ }\underline{b}\text{ }\underline{c}$ , can be written in base $15$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{b}$ , and can be written in base $6$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }$ , ... | We have these equations: $196a+14b+c=225a+15c+b=222a+37c$ .
Taking the last two we get $3a+b=22c$ . Because $c \neq 0$ otherwise $a \ngtr 0$ , and $a \leq 5$ $c=1$
Then we know $3a+b=22$ .
Taking the first two equations we see that $29a+14c=13b$ . Combining the two gives $a=4, b=10, c=1$ . Then we see that $222 \times ... | 925 |
5,609 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_2 | 2 | The number $n$ can be written in base $14$ as $\underline{a}\text{ }\underline{b}\text{ }\underline{c}$ , can be written in base $15$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{b}$ , and can be written in base $6$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }$ , ... | We know that $196a+14b+c=225a+15c+b=222a+37c$ . Combining the first and third equations give that $196a+14b+c=222a+37c$ , or \[7b=13a+18c\] The second and third gives $222a+37c=225a+15c+b$ , or \[22c-3a=b\] \[154c-21a=7b=13a+18c\] \[4c=a\] We can have $a=4,8,12$ , but only $a=4$ falls within the possible digits of base... | 925 |
5,610 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_2 | 3 | The number $n$ can be written in base $14$ as $\underline{a}\text{ }\underline{b}\text{ }\underline{c}$ , can be written in base $15$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{b}$ , and can be written in base $6$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }$ , ... | We're given that $196a+14b+c=225a+15c+b=222a+37c.$ By taking the difference of the first $2$ equalities, we receive $29a+14c=13b.$ Taking $\pmod{13}$ , we receive $3a+c \equiv 0 \pmod{13}.$ We receive the following cases: $(a,c)=(4,1)$ or $(3,4).$ (Note that $(2,7)$ doesn't work since $a,c<6$ by third condition). We ca... | 925 |
5,611 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_3 | 2 | Kathy has $5$ red cards and $5$ green cards. She shuffles the $10$ cards and lays out $5$ of the cards in a row in a random order. She will be happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders RRGGG, GGGGR, or RRRRR will make Kathy happy... | Our probability will be $\dfrac{\text{number of "happy" configurations of cards}}{\text{total number of configurations of cards}}.$
First of all, we have $10$ choices for the first card, $9$ choices for the second card, $8$ choices for the third card, $7$ choices for the fourth card, and $6$ choices for the last card. ... | 157 |
5,612 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_3 | 3 | Kathy has $5$ red cards and $5$ green cards. She shuffles the $10$ cards and lays out $5$ of the cards in a row in a random order. She will be happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders RRGGG, GGGGR, or RRRRR will make Kathy happy... | As the problem states, some examples of valid are $RRGGG$ $GGGGR$ , and $RRRRR$ . Let's use each of these as more general cases.
Let $RRGGG$ be the case when there are 2 adjacents of one color, and 3 adjacents of the other color. This yields $4$ combinations ( $RRGGG$ $GGRRR$ $RRRGG$ , and $GGGRR$ ). The probability of... | 157 |
5,613 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_3 | 4 | Kathy has $5$ red cards and $5$ green cards. She shuffles the $10$ cards and lays out $5$ of the cards in a row in a random order. She will be happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders RRGGG, GGGGR, or RRRRR will make Kathy happy... | Kathy will draw the 10 cards in some order, but the restriction of having all greens in a row and all reds in a row only applies to the first 5 cards drawn. The total number of ways the 10 cards can be drawn is simply 10 choose 5 which is 252. Now we just count the number of possible successful configurations of the te... | 157 |
5,614 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_3 | 5 | Kathy has $5$ red cards and $5$ green cards. She shuffles the $10$ cards and lays out $5$ of the cards in a row in a random order. She will be happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders RRGGG, GGGGR, or RRRRR will make Kathy happy... | Assume without loss of generality that the first card laid out is red. Then the arrangements that satisfy Kathy’s requirements are RRRRR, RRRRG, RRRGG, RRGGG, and RGGGG. The probability that Kathy will lay out one of these
arrangements is \[\frac49\cdot\frac38\cdot\frac27\cdot\frac16\] \[\frac49\cdot\frac38\cdot\frac27... | 157 |
5,615 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_4 | 1 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Fi... | By the Law of Cosines on $\triangle ABC$ , we have: \[\cos(A) = \frac{10^2+10^2-12^2}{2*10*10} = \frac{7}{25}\] By the Law of Cosines on $\triangle ADE$ , then \[\frac{7}{25} = \frac{10-x}{2x} \iff x =\frac{250}{39}\] So, our answer is $250+39=\boxed{289}$ | 289 |
5,616 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_4 | 2 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Fi... | We draw the altitude from $B$ to $\overline{AC}$ to get point $F$ . We notice that the triangle's height from $A$ to $\overline{BC}$ is 8 because it is a $3-4-5$ Right Triangle. To find the length of $\overline{BF}$ , we let $h$ represent $\overline{BF}$ and set up an equation by finding two ways to express the area. T... | 289 |
5,617 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_4 | 3 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Fi... | We start by adding a few points to the diagram. Call $F$ the midpoint of $AE$ , and $G$ the midpoint of $BC$ . (Note that $DF$ and $AG$ are altitudes of their respective triangles). We also call $\angle BAC = \theta$ . Since triangle $ADE$ is isosceles, $\angle AED = \theta$ , and $\angle ADF = \angle EDF = 90 - \theta... | 289 |
5,618 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_4 | 4 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Fi... | As in the diagram, let $DE = x$ . Consider point $G$ on the diagram shown above. Our goal is to be able to perform Pythagorean Theorem on $DG, GC$ , and $DC$ . Let $GE = \frac{10-x}{2}$ . Therefore, it is trivial to see that $GC^2 = \Big(x + \frac{10-x}{2}\Big)^2$ (leave everything squared so that it cancels nicely at ... | 289 |
5,619 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_4 | 7 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Fi... | In isosceles triangle, draw the altitude from $D$ onto $\overline{AD}$ . Let the point of intersection be $X$ . Clearly, $AE=10-AD$ , and hence $AX=\frac{10-AD}{2}$
Now, we recognise that the perpendicular from $A$ onto $\overline{AD}$ gives us two $6$ $8$ $10$ triangles. So, we calculate $\sin \angle ABC=\frac{4}{5}$ ... | 289 |
5,620 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_4 | 8 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Fi... | We can have 2 Law of Cosines applied on $\angle A$ (one from $\triangle ADE$ and one from $\triangle ABC$ ),
$x^2=x^2+(10-x)^2-2(x)(10-x)\cdot \cos{A}$ and $12^2=10^2+10^2-2(10)(10)\cdot \cos{A}$
Solving for $\cos{A}$ in both equations, we get
$\cos{A} = \frac{(10-x)^2}{(2)(10-x)(x)}$ and $cos A = \frac{7}{25} \implies... | 289 |
5,621 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_4 | 9 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Fi... | Let $B=(0, 0)$ , and $C=(12, 0)$ . From there, we know that $A=(6, 8)$ , so line $AB$ is $y=\frac{4}{3}x$ . Hence, $D=(a, \frac{4}{3}a)$ for some $a$ , and $BD=\frac{5}{3}a$ so $AD=10-\frac{5}{3}a$ . Now, notice that by symmetry, $E=(6+a, 8-\frac{4}{3}a)$ , so $ED^2=6^2+(8-\frac{8}{3}a)^2$ . Because $AD=ED$ , we now ha... | 289 |
5,622 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_4 | 10 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Fi... | Let's label $\angle A = \theta$ and $\angle ECD = \alpha$ . Using isosceles triangle properties and the triangle angle sum equation, we get \[180-(180-2\theta+\alpha) + \frac{180-\theta}{2} + \left(\frac{180-\theta}{2} - \alpha\right) = 180.\] Solving, we find $\theta = 2 \alpha$
Relabelling our triangle, we get $\angl... | 289 |
5,623 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_4 | 11 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Fi... | We start by labelling a few angles (all of them in degrees). Let $\angle{BAC}=2\alpha = \angle{AED}, \angle{EDC}=\angle{ECD}=\alpha, \angle{DEC}=180-2\alpha, \angle{BDC}=3\alpha, \angle{DCB}=90-2\alpha, \angle{DBC}=90-\alpha$ . Also let $AD=a$ . By sine rule in $\triangle{ADE},$ we get $\frac{a}{\sin{2\alpha}}=\frac{10... | 289 |
5,624 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_5 | 1 | For each ordered pair of real numbers $(x,y)$ satisfying \[\log_2(2x+y) = \log_4(x^2+xy+7y^2)\] there is a real number $K$ such that \[\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).\] Find the product of all possible values of $K$ | Using the logarithmic property $\log_{a^n}b^n = \log_{a}b$ , we note that \[(2x+y)^2 = x^2+xy+7y^2\] That gives \[x^2+xy-2y^2=0\] upon simplification and division by $3$ . Factoring $x^2+xy-2y^2=0$ gives \[(x+2y)(x-y)=0\] Then, \[x=y \text{ or }x=-2y\] From the second equation, \[9x^2+6xy+y^2=3x^2+4xy+Ky^2\] If we take... | 189 |
5,625 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_5 | 2 | For each ordered pair of real numbers $(x,y)$ satisfying \[\log_2(2x+y) = \log_4(x^2+xy+7y^2)\] there is a real number $K$ such that \[\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).\] Find the product of all possible values of $K$ | Do as done in Solution 1 to get \[x^2+xy-2y^2=0\] \[\implies (\frac{x}{y})^2+\frac{x}{y}-2=0\] \[\implies \frac{x}{y}=\frac{-1\pm\sqrt{1+8}}{2}=1,-2\] Do as done in Solution 1 to get \[9x^2+6xy+y^2=3x^2+4xy+Ky^2\] \[\implies 6x^2+2xy+(1-K)y^2=0\] \[\implies 6(\frac{x}{y})^2+2\frac{x}{y}+(1-K)=0\] \[\implies \frac{x}{y}... | 189 |
5,626 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_6 | 1 | Let $N$ be the number of complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$ | Let $a=z^{120}$ . This simplifies the problem constraint to $a^6-a \in \mathbb{R}$ . This is true if $\text{Im}(a^6)=\text{Im}(a)$ . Let $\theta$ be the angle $a$ makes with the positive x-axis. Note that there is exactly one $a$ for each angle $0\le\theta<2\pi$ . We are given $\sin\theta = \sin{6\theta}$ . Note that $... | 440 |
5,627 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_6 | 2 | Let $N$ be the number of complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$ | The constraint mentioned in the problem is equivalent to the requirement that the imaginary part is equal to $0$ . Since $|z|=1$ , let $z=\cos \theta + i\sin \theta$ , then we can write the imaginary part of $\Im(z^{6!}-z^{5!})=\Im(z^{720}-z^{120})=\sin\left(720\theta\right)-\sin\left(120\theta\right)=0$ . Using the su... | 440 |
5,628 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_6 | 3 | Let $N$ be the number of complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$ | As mentioned in solution one, for the difference of two complex numbers to be real, their imaginary parts must be equal. We use exponential form of complex numbers. Let $z = e^{i \theta}$ . We have two cases to consider. Either $z^{6!} = z^{5!}$ , or $z^{6!}$ and $z^{5!}$ are reflections across the imaginary axis.
If $... | 440 |
5,629 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_6 | 4 | Let $N$ be the number of complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$ | Because $|z| = 1,$ we know that $z\overline{z} = 1^2 = 1.$ Hence $\overline{z} = \frac 1 {z}.$ Because $z^{6!}-z^{5!}$ is real, it is equal to its complex conjugate. Hence $z^{6!}-z^{5!} = \overline{z^{6!}}-\overline{z^{5!}}.$ Substituting the expression we that we derived earlier, we get $z^{720}-z^{120} = \frac 1{z^{... | 440 |
5,630 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_9 | 1 | Find the number of four-element subsets of $\{1,2,3,4,\dots, 20\}$ with the property that two distinct elements of a subset have a sum of $16$ , and two distinct elements of a subset have a sum of $24$ . For example, $\{3,5,13,19\}$ and $\{6,10,20,18\}$ are two such subsets. | This problem is tricky because it is the capital of a few "bashy" calculations. Nevertheless, the process is straightforward. Call the set $\{a, b, c, d\}$
Note that there are only two cases: 1 where $a + b = 16$ and $c + d = 24$ or 2 where $a + b = 16$ and $a + c = 24$ . Also note that there is no overlap between the ... | 210 |
5,631 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_9 | 2 | Find the number of four-element subsets of $\{1,2,3,4,\dots, 20\}$ with the property that two distinct elements of a subset have a sum of $16$ , and two distinct elements of a subset have a sum of $24$ . For example, $\{3,5,13,19\}$ and $\{6,10,20,18\}$ are two such subsets. | Let's say our four elements in our subset are $a,b,c,d$ . We have two cases. Note that the order of the elements / the element letters themselves don't matter since they are all on equal grounds at the start.
$\textrm{Case } 1 \textrm{:}$ $a+b = 16$ and $c+d = 24$
List out possibilities for $a+b$ $(\text{i.e. } 1+15, 2... | 210 |
5,632 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_11 | 1 | Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$ | Note that the given condition is equivalent to $3^n \equiv 1 \pmod{143^2}$ and $143=11\cdot 13$ . Because $\gcd(11^2, 13^2) = 1$ , the desired condition is equivalent to $3^n \equiv 1 \pmod{121}$ and $3^n \equiv 1 \pmod{169}$
If $3^n \equiv 1 \pmod{121}$ , one can see the sequence $1, 3, 9, 27, 81, 1, 3, 9...$ so $5|n$... | 195 |
5,633 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_11 | 2 | Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$ | Note that Euler's Totient Theorem would not necessarily lead to the smallest $n$ and that in this case that $n$ is greater than $1000$
We wish to find the least $n$ such that $3^n \equiv 1 \pmod{143^2}$ . This factors as $143^2=11^{2}*13^{2}$ . Because $\gcd(121, 169) = 1$ , we can simply find the least $n$ such that $... | 195 |
5,634 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_11 | 3 | Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$ | Listing out the powers of $3$ , modulo $169$ and modulo $121$ , we have: \[\begin{array}{c|c|c} n & 3^n\mod{169} & 3^n\mod{121}\\ \hline 0 & 1 & 1\\ 1 & 3 & 3\\ 2 & 9 & 9\\ 3 & 27 & 27\\ 4 & 81 & 81\\ 5 & 74 & 1\\ 6 & 53\\ 7 & 159\\ 8 & 139\\ 9 & 79\\ 10 & 68\\ 11 & 35\\ 12 & 105\\ 13 & 146\\ 14 & 100\\ 15 & 131\\ 16 &... | 195 |
5,635 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_11 | 4 | Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$ | We have that \[3^n \equiv 1 \pmod{143^2}.\] Now, $3^{110} \equiv 1 \pmod{11^2}$ so by the Fundamental Theorem of Orders, $\text{ord}_{11^2}(3)|110$ and with some bashing, we get that it is $5$ . Similarly, we get that $\text{ord}_{13^2}(3)=39$ . Now, $\text{lcm}(39,5)=\boxed{195}$ which is our desired solution. | 195 |
5,636 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_11 | 6 | Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$ | We can see that $3^n-1 = 143^2*x$ , which means that $v_{11}(3^n-1) \geq 2$ $v_{13}(3^n-1) \geq 2$ $v_{11}(3^n-1) = v_{11}(242) + v_{11}(\frac{n}{5})$ $v_{13}(3^n-1) = v_{13}(26) + v_{13}(\frac{n}{3})$ by the Lifting the Exponent lemma. From the first equation we gather that 5 divides n, while from the second equation ... | 195 |
5,637 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_11 | 7 | Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$ | Note that the problem is basically asking for the least positive integer $n$ such that $11^2 \cdot 13^2 | 3^n - 1.$ It is easy to see that $n = \text{lcm } (a, b),$ where $a$ is the least positive integer satisfying $11^2 | 3^a - 1$ and $b$ the least positive integer satisfying $13^2 | 3^b - 1$ . Luckily, finding $a$ i... | 195 |
5,638 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_11 | 8 | Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$ | The requested positive integer is the least value of $n>0$ such that $3^n\equiv 1\pmod{143^2}.$ Note that $143=11\cdot 13.$ The least power of $3$ that is congruent to $1$ modulo $11^2$ is $3^5=243=2\cdot 11^2+1.$ It follows that. $3^n\equiv 1\pmod {11^2}$ if and only if $n=5j$ for some positive integer $j$
The least p... | 195 |
5,639 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_11 | 9 | Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$ | We first note that we wish to find $3^n \equiv 1 \pmod{11^2}$ and $3^n \equiv 1 \pmod{13^2}.$ Not thinking of anything else, we try a few numbers for the first condition to get that $5 \mid n.$ For the second condition, upon testing up to 729, we find that it doesn't yield a solution readily, so we use Lifting the Expo... | 195 |
5,640 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | 1 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. ... | The question asks us for the probability that a randomly chosen subset of the set of the first 18 positive integers has the property that the sum of its elements is divisible by 3. Note that the total number of subsets is $2^{18}$ because each element can either be in or not in the subset. To find the probability, we w... | 1 |
5,641 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | 2 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. ... | Consider the elements of $U$ modulo $3.$
Ignore the $0$ 's because we're gonna multiply $\binom{6}{0}+..+\binom{6}{6}=2^6$ at the end. Let $a$ be the $1's$ and $b$ be the $2's$ . The key here is that $2 \equiv -1 \pmod{3}$ so the difference between the number of $a$ and $b$ is a multiple of $3$
1. Counted twice because... | 683 |
5,642 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | 3 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. ... | Elements $0 \pmod{3}$ can either be included or excluded, for a total of $2^6$ . Then, like the previous solution, let $a$ be the number of elements $1 \pmod{3}$ and $b$ be the number of elements $2 \pmod{3}$ . Then, $a + 2b \equiv 0 \pmod{3} \implies a - b \equiv 0 \pmod{3}$ . Since $0 \le a, b \le 6$ , there are 3 ca... | 683 |
5,643 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | 4 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. ... | Note that in general, the answer will be around $1/3$ of $2^{18}$ . We will use this to our advantage. Partition $U$ into disjoint subsets $\{1,2,3\}, \{4,5,6\}, \cdots, \{16,17,18\}$ . Within each of these subsets, there are 3 possible remainders $\mod 3$ depending on what elements we choose to include into $T$ : (Usi... | 683 |
5,644 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | 5 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. ... | Rewrite the set after mod 3 as above
1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0
All 0s can be omitted
Case 1
No 1 No 2 $1$
Case 2 $222$ $20$
Case 3 $222222$ $1$
Case 4 $12$ $6*6=36$
Case 5 $12222$ $6*15=90$
Case 6 $1122$ $15*15=225$
Case 7 $1122222$ $15*6=90$
Case 8 $111$ $20$
Case 9 $111222$ $20*20=400$
Case 10 $111222222$ $... | 683 |
5,645 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | 6 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. ... | Notice that six numbers are $0\pmod3$ , six are $1\pmod3$ , and six are $2\pmod3$ . Having numbers $0\pmod3$ will not change the remainder when $s(T)$ is divided by $3$ , so we can choose any number of these in our subset. We ignore these for now. The number of numbers that are $1\pmod3$ , minus the number of number... | 683 |
5,646 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | 7 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. ... | We use generating functions. Each element of $U$ has two choices that occur with equal probability--either it is in the set or out of the set. Therefore, given $n\in U$ , the probability generating function is \[\frac{1}{2}+\frac{1}{2}x^n.\] Therefore, in the generating function \[\frac{1}{2^{18}}(1+x)(1+x^2)(1+x^3)\cd... | 683 |
5,647 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | 8 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. ... | Define a set that "works" to be a set for which the sum of the terms is $0$ mod $3$ . The given set mod $3$ is \[\{1,2,0,1,2,0,1,2,0,1,2,0,1,2,0,1,2,0\}.\] Let $w(N)$ be the number of subsets of the first $N$ terms of this set that "work."
Consider just the first $3$ terms: \[\{1,2,0\}.\] There are $2^3=8$ total subset... | 683 |
5,648 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | 9 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. ... | Try smaller cases and find a pattern. Using similar casework as in Solution 1, we can easily find the desired probability if $U$ is of a small size.
If $U = \{ 1,2,0\} \pmod 3$ , then $4$ out of $8$ subsets work, for a probability of $\tfrac12$
If $U = \{ 1,2,0,1,2,0\} \pmod 3$ , then $24$ out of $64$ subsets work, for... | 683 |
5,649 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | 10 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. ... | The total number of subsets is simply $2^{18}.$ Now we need to find the number of subsets that have a sum divisible by $3.$
Ignore the 6 numbers in the list that are divisible by 3. We look only at the number of subsets of $\{1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17\}$ then multiply by $2^6$ at the end. This is because... | 683 |
5,650 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | 11 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. ... | The total number of subsets is $\sum_{i=0}^{18}\tbinom{18}{i}=2^{18}$ . If $s(T)\equiv 0\bmod{3}$ , the sum of the elements divides 3. We can rewrite the set as 6 0s, 6 1s, and 6 2s. We can ignore the zeros for now, since they won't influence the sum so we focus on each configuration of the 6 1s and 6 2s such that the ... | 683 |
5,651 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_13 | 1 | Let $\triangle ABC$ have side lengths $AB=30$ $BC=32$ , and $AC=34$ . Point $X$ lies in the interior of $\overline{BC}$ , and points $I_1$ and $I_2$ are the incenters of $\triangle ABX$ and $\triangle ACX$ , respectively. Find the minimum possible area of $\triangle AI_1I_2$ as $X$ varies along $\overline{BC}$ | First note that \[\angle I_1AI_2 = \angle I_1AX + \angle XAI_2 = \frac{\angle BAX}2 + \frac{\angle CAX}2 = \frac{\angle A}2\] is a constant not depending on $X$ , so by $[AI_1I_2] = \tfrac12(AI_1)(AI_2)\sin\angle I_1AI_2$ it suffices to minimize $(AI_1)(AI_2)$ . Let $a = BC$ $b = AC$ $c = AB$ , and $\alpha = \angle AX... | 126 |
5,652 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_13 | 2 | Let $\triangle ABC$ have side lengths $AB=30$ $BC=32$ , and $AC=34$ . Point $X$ lies in the interior of $\overline{BC}$ , and points $I_1$ and $I_2$ are the incenters of $\triangle ABX$ and $\triangle ACX$ , respectively. Find the minimum possible area of $\triangle AI_1I_2$ as $X$ varies along $\overline{BC}$ | It's clear that $\angle I_{1}AI_{2}=\frac{1}{2}\angle BAX+\frac{1}{2}\angle CAX=\frac{1}{2}\angle A$ . Thus \begin{align*} AI_{1}I_{2}&=\frac{1}{2}\cdot AI_{1}\cdot AI_{2}\cdot\sin\angle I_{1}AI_{2} \\ &=\frac{1}{2}\cdot AI_{1}\cdot AI_{2}\cdot\sin\left(\frac{1}{2}\angle A\right).\end{align*} By the Law of Sines on $\t... | 126 |
5,653 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_13 | 3 | Let $\triangle ABC$ have side lengths $AB=30$ $BC=32$ , and $AC=34$ . Point $X$ lies in the interior of $\overline{BC}$ , and points $I_1$ and $I_2$ are the incenters of $\triangle ABX$ and $\triangle ACX$ , respectively. Find the minimum possible area of $\triangle AI_1I_2$ as $X$ varies along $\overline{BC}$ | First, instead of using angles to find $[AI_1I_2]$ , let's try to find the area of other, simpler figures, and subtract that from $[ABC]$ . However, to do this, we need to be able to figure out the length of the inradii, and so, we need to find $AX$
To minimize $[AI_1I_2]$ , intuitively, we should try to minimize the l... | 126 |
5,654 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_14 | 1 | Let $SP_1P_2P_3EP_4P_5$ be a heptagon. A frog starts jumping at vertex $S$ . From any vertex of the heptagon except $E$ , the frog may jump to either of the two adjacent vertices. When it reaches vertex $E$ , the frog stops and stays there. Find the number of distinct sequences of jumps of no more than $12$ jumps that ... | This is easily solved by recursion/dynamic programming. To simplify the problem somewhat, let us imagine the seven vertices on a line $E \leftrightarrow P_4 \leftrightarrow P_5 \leftrightarrow S \leftrightarrow P_1 \leftrightarrow P_2 \leftrightarrow P_3 \leftrightarrow E$ . We can count the number of left/right (L/R) ... | 351 |
5,655 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_14 | 2 | Let $SP_1P_2P_3EP_4P_5$ be a heptagon. A frog starts jumping at vertex $S$ . From any vertex of the heptagon except $E$ , the frog may jump to either of the two adjacent vertices. When it reaches vertex $E$ , the frog stops and stays there. Find the number of distinct sequences of jumps of no more than $12$ jumps that ... | Let $E_n$ denotes the number of sequences with length $n$ that ends at $E$ . Define similarly for the other vertices. We seek for a recursive formula for $E_n$ \begin{align*} E_n&=P_{3_{n-1}}+P_{4_{n-1}} \\ &=P_{2_{n-2}}+P_{5_{n-2}} \\ &=P_{1_{n-3}}+P_{3_{n-3}}+S_{n-3}+P_{4_{n-3}} \\ &=(P_{3_{n-3}}+P_{4_{n-3}})+S_{n-3}... | 351 |
5,656 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_1 | 1 | Points $A$ $B$ , and $C$ lie in that order along a straight path where the distance from $A$ to $C$ is $1800$ meters. Ina runs twice as fast as Eve, and Paul runs twice as fast as Ina. The three runners start running at the same time with Ina starting at $A$ and running toward $C$ , Paul starting at $B$ and running tow... | We know that in the same amount of time given, Ina will run twice the distance of Eve, and Paul would run quadruple the distance of Eve. Let's consider the time it takes for Paul to meet Eve: Paul would've run 4 times the distance of Eve, which we can denote as $d$ . Thus, the distance between $B$ and $C$ is $4d+d=5d$ ... | 800 |
5,657 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_1 | 2 | Points $A$ $B$ , and $C$ lie in that order along a straight path where the distance from $A$ to $C$ is $1800$ meters. Ina runs twice as fast as Eve, and Paul runs twice as fast as Ina. The three runners start running at the same time with Ina starting at $A$ and running toward $C$ , Paul starting at $B$ and running tow... | Let $x$ be the distance from $A$ to $B$ . Then the distance from $B$ to $C$ is $1800-x$ . Since Eve is the slowest, we can call her speed $v$ , so that Ina's speed is $2v$ and Paul's speed is $4v$
For Paul and Eve to meet, they must cover a total distance of $1800-x$ which takes them a time of $\frac{1800-x}{4v+v}$ . P... | 800 |
5,658 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_2 | 1 | Let $a_{0} = 2$ $a_{1} = 5$ , and $a_{2} = 8$ , and for $n > 2$ define $a_{n}$ recursively to be the remainder when $4$ $a_{n-1}$ $+$ $a_{n-2}$ $+$ $a_{n-3}$ ) is divided by $11$ . Find $a_{2018} \cdot a_{2020} \cdot a_{2022}$ | When given a sequence problem, one good thing to do is to check if the sequence repeats itself or if there is a pattern.
After computing more values of the sequence, it can be observed that the sequence repeats itself every 10 terms starting at $a_{0}$
$a_{0} = 2$ $a_{1} = 5$ $a_{2} = 8$ $a_{3} = 5$ $a_{4} = 6$ $a_{5} ... | 112 |
5,659 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_2 | 2 | Let $a_{0} = 2$ $a_{1} = 5$ , and $a_{2} = 8$ , and for $n > 2$ define $a_{n}$ recursively to be the remainder when $4$ $a_{n-1}$ $+$ $a_{n-2}$ $+$ $a_{n-3}$ ) is divided by $11$ . Find $a_{2018} \cdot a_{2020} \cdot a_{2022}$ | Notice that the characteristic polynomial of this is $x^3-4x^2-4x-4\equiv 0\pmod{11}$
Then since $x\equiv1$ is a root, using Vieta's formula, the other two roots $r,s$ satisfy $r+s\equiv3$ and $rs\equiv4$
Let $r=7+d$ and $s=7-d$
We have $49-d^2\equiv4$ so $d\equiv1$ . We found that the three roots of the characteristic... | 112 |
5,660 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_3 | 1 | Find the sum of all positive integers $b < 1000$ such that the base- $b$ integer $36_{b}$ is a perfect square and the base- $b$ integer $27_{b}$ is a perfect cube. | The first step is to convert $36_{b}$ and $27_{b}$ into base-10 numbers. Then, we can write \[36_{b} = 3b + 6\] and \[27_{b} = 2b + 7\] . It should also be noted that $8 \leq b < 1000$
Because there are less perfect cubes than perfect squares for the restriction we are given on $b$ , it is best to list out all the perf... | 371 |
5,661 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_3 | 3 | Find the sum of all positive integers $b < 1000$ such that the base- $b$ integer $36_{b}$ is a perfect square and the base- $b$ integer $27_{b}$ is a perfect cube. | As shown above, let \[3b+6 = n^2\] \[2b+7 = m^3\] such that \[6b+12=2n^2\] \[6b+21=3m^3\]
Subtracting the equations we have \[3m^3-2n^2=9 \implies 3m^3-2n^2-9=0.\]
We know that $m$ and $n$ both have to be integers, because then the base wouldn't be an integer. Furthermore, any integer solution $m$ must divide $9$ by th... | 371 |
5,662 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_5 | 4 | Suppose that $x$ $y$ , and $z$ are complex numbers such that $xy = -80 - 320i$ $yz = 60$ , and $zx = -96 + 24i$ , where $i$ $=$ $\sqrt{-1}$ . Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$ . Find $a^2 + b^2$ | Dividing the first equation by the second equation given, we find that $\frac{xy}{yz}=\frac{x}{z}=\frac{-80-320i}{60}=-\frac{4}{3}-\frac{16}{3}i \implies x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)$ . Substituting this into the third equation, we get $z^2=\frac{-96+24i}{-\frac{4}{3}-\frac{16}{3}i}=3\cdot \frac{-24+6i}{-... | 74 |
5,663 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_5 | 7 | Suppose that $x$ $y$ , and $z$ are complex numbers such that $xy = -80 - 320i$ $yz = 60$ , and $zx = -96 + 24i$ , where $i$ $=$ $\sqrt{-1}$ . Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$ . Find $a^2 + b^2$ | We can turn the expression $x+y+z$ into $\sqrt{x^2+y^2+z^2+2xy+2yz+2xz}$ , and this would allow us to plug in the values after some computations.
Based off of the given products, we have
$xy^2z=60(-80-320i)$
$xyz^2=60(-96+24i)$
$x^2yz=(-96+24i)(-80-320i)$
Dividing by the given products, we have
$y^2=\frac{60(-80-320i)}... | 74 |
5,664 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_8 | 1 | A frog is positioned at the origin of the coordinate plane. From the point $(x, y)$ , the frog can jump to any of the points $(x + 1, y)$ $(x + 2, y)$ $(x, y + 1)$ , or $(x, y + 2)$ . Find the number of distinct sequences of jumps in which the frog begins at $(0, 0)$ and ends at $(4, 4)$ | We solve this problem by working backwards. Notice, the only points the frog can be on to jump to $(4,4)$ in one move are $(2,4),(3,4),(4,2),$ and $(4,3)$ . This applies to any other point, thus we can work our way from $(0,0)$ to $(4,4)$ , recording down the number of ways to get to each point recursively.
$(0,0): 1$
... | 556 |
5,665 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_8 | 2 | A frog is positioned at the origin of the coordinate plane. From the point $(x, y)$ , the frog can jump to any of the points $(x + 1, y)$ $(x + 2, y)$ $(x, y + 1)$ , or $(x, y + 2)$ . Find the number of distinct sequences of jumps in which the frog begins at $(0, 0)$ and ends at $(4, 4)$ | We'll refer to the moves $(x + 1, y)$ $(x + 2, y)$ $(x, y + 1)$ , and $(x, y + 2)$ as $R_1$ $R_2$ $U_1$ , and $U_2$ , respectively. Then the possible sequences of moves that will take the frog from $(0,0)$ to $(4,4)$ are all the permutations of $U_1U_1U_1U_1R_1R_1R_1R_1$ $U_2U_1U_1R_1R_1R_1R_1$ $U_1U_1U_1U_1R_2R_1R_1$ ... | 556 |
5,666 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_9 | 1 | Octagon $ABCDEFGH$ with side lengths $AB = CD = EF = GH = 10$ and $BC = DE = FG = HA = 11$ is formed by removing 6-8-10 triangles from the corners of a $23$ $\times$ $27$ rectangle with side $\overline{AH}$ on a short side of the rectangle, as shown. Let $J$ be the midpoint of $\overline{AH}$ , and partition the octago... | We represent octagon $ABCDEFGH$ in the coordinate plane with the upper left corner of the rectangle being the origin. Then it follows that $A=(0,-6), B=(8, 0), C=(19,0), D=(27, -6), E=(27, -17), F=(19, -23), G=(8, -23), H=(0, -17), J=(0, -\frac{23}{2})$ . Recall that the centroid is $\frac{1}{3}$ way up each median in ... | 184 |
5,667 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_10 | 1 | Find the number of functions $f(x)$ from $\{1, 2, 3, 4, 5\}$ to $\{1, 2, 3, 4, 5\}$ that satisfy $f(f(x)) = f(f(f(x)))$ for all $x$ in $\{1, 2, 3, 4, 5\}$ | We perform casework on the number of fixed points (the number of points where $f(x) = x$ ). To better visualize this, use the grid from Solution 1.
Case 1: 5 fixed points
Case 2: 4 fixed points
Case 3: 3 fixed points
Case 4: 2 fixed points
Case 5: 1 fixed point
Therefore, the answer is $1+20+150+380+205 = \boxed{756}$ | 756 |
5,668 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_11 | 1 | Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$ , at least one of the first $k$ terms of the permutation is greater than $k$ | If the first number is $6$ , then there are no restrictions. There are $5!$ , or $120$ ways to place the other $5$ numbers.
If the first number is $5$ $6$ can go in four places, and there are $4!$ ways to place the other $4$ numbers. $4 \cdot 4! = 96$ ways.
If the first number is $4$ , ....
4 6 _ _ _ _ $\implies$ 24 wa... | 461 |
5,669 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_11 | 2 | Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$ , at least one of the first $k$ terms of the permutation is greater than $k$ | If $6$ is the first number, then there are no restrictions. There are $5!$ , or $120$ ways to place the other $5$ numbers.
If $6$ is the second number, then the first number can be $2, 3, 4,$ or $5$ , and there are $4!$ ways to place the other $4$ numbers. $4 \cdot 4! = 96$ ways.
If $6$ is the third number, then we can... | 461 |
5,670 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_11 | 3 | Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$ , at least one of the first $k$ terms of the permutation is greater than $k$ | Note the condition in the problem is equivalent to the following condition: for each $k$ with $1 \le k \le 5$ , the first $k$ terms is not a permutation $(1, 2, \ldots, k)$ (since it would mean there must be some integer $x$ in the first $k$ terms such that $x \not \in \{1, 2, \ldots, k\} \implies x > k$ ). Then, let $... | 461 |
5,671 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_11 | 4 | Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$ , at least one of the first $k$ terms of the permutation is greater than $k$ | Let $A_i$ be the set of permutations such that there is no number greater than $i$ in the first $i$ places. Note that $\bigcap^{k}_{i=0}{A_{b_i}}=\prod^k_{i=1}{(b_i-b_{i-1})!}$ for all $1\le b_0 < b_1\cdots < b_k \le 5$ and that the set of restricted permutations is $A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5$
We will com... | 461 |
5,672 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_11 | 5 | Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$ , at least one of the first $k$ terms of the permutation is greater than $k$ | Define the function $f(p,q)$ as the amount of permutations with maximum digit $q$ and string length $p$ that satisfy the condition within bounds. For example, $f(4,5)$ would be the amount of ways to make a string with length $4$ with the highest digit being $5$ . We wish to obtain $f(6,6)=f(5,6)$
To generate recursion,... | 461 |
5,673 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_11 | 6 | Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$ , at least one of the first $k$ terms of the permutation is greater than $k$ | We can also solve this problem by counting the number of permutations that do NOT satisfy the given conditions; namely, these permutations must satisfy the condition that none of the first $k$ terms is greater than $k$ for $1\leq$ $k$ $\leq$ $5$ . We can further simplify this method by approaching it through casework o... | 461 |
5,674 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_12 | 1 | Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$ | For reference, $2\sqrt{65} \approx 16$ , so $\overline{AD}$ is the longest of the four sides of $ABCD$ . Let $h_1$ be the length of the altitude from $B$ to $\overline{AC}$ , and let $h_2$ be the length of the altitude from $D$ to $\overline{AC}$ . Then, the triangle area equation becomes
\[\frac{h_1}{2}AP + \frac{h_2}... | 112 |
5,675 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_12 | 2 | Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$ | Denote $\angle APB$ by $\alpha$ . Then $\sin(\angle APB)=\sin \alpha = \sin(\angle APD)$ .
Using the formula for the area of a triangle, we get \[\frac{1}{2} (AP\cdot BP+ CP\cdot DP)\sin\alpha=\frac{1}{2}(AP\cdot DP+ CP\cdot BP)\sin\alpha ,\] so \[(AP-CP)(BP-DP)=0\] Hence $AP=CP$ (note that $BP=DP$ makes no difference ... | 112 |
5,676 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_12 | 3 | Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$ | As in all other solutions, we can first find that either $AP=CP$ or $BP=DP$ , but it's an AIME problem, we can take $AP=CP$ , and assume the other choice will lead to the same result (which is true).
From $AP=CP$ , we have $[DAP]=[DCP]$ , and $[BAP]=[BCP] \implies [ABD] = [CBD]$ , therefore, \begin{align} \nonumber \fr... | 112 |
5,677 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_12 | 4 | Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$ | Either $PA=PC$ or $PD=PB$ . Let $PD=PB=s$ . Applying Stewart's Theorem on $\triangle ABD$ and $\triangle BCD$ , dividing by $2s$ and rearranging, \[\tag{1}CP^2+s^2=148\] \[\tag{2}AP^2+s^2=180\] Applying Stewart on $\triangle CAB$ and $\triangle CAD$ \[\tag{3} 5CP^2=3AP^2\] Substituting equations 1 and 2 into 3 and rear... | 112 |
5,678 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_12 | 5 | Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$ | Now we prove P is the midpoint of $BD$ . Denote the height from $B$ to $AC$ as $h_1$ , height from $D$ to $AC$ as $h_2$ .According to the problem, $AP* h_1 +CP* h_2 =CP* h_1 +AP* h_2$ implies $h_1 (AP-CP)= h_2 (AP-CP), h_1 = h_2$ . Then according to basic congruent triangles we get $BP=DP$ Firstly, denote that $CP=a,BP... | 112 |
5,679 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_12 | 6 | Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$ | $BP = PD$ as in another solutions.
Let $D'$ be the reflection of $D$ across $C$ .
Let points $E, E',$ and $H$ be the foot of perpendiculars on $AC$ from $D,D'$ , and $B$ respectively. \begin{align*} &\qquad AB = CD = CD', \quad \textrm{and} \quad BH = DE = D'E' \\ \Rightarrow &\qquad \triangle ABH = \triangle CDE = \... | 112 |
5,680 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_12 | 7 | Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$ | Use your favorite method to get that $P$ is the midpoint of one of the two diagonals (suppose it's the midpoint of $\overline{AC}$ ). From here, let $x=AP=PC, y=BP, z=PD, a=\cos\theta$ where $\theta$ is the angle that the diagonals make. Then we have a system of four equations:
\begin{align*} x^2+y^2+2xya &= 100 \\ z^2... | 112 |
5,681 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_12 | 8 | Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$ | Note that $\angle APB = \angle CPD = 180-\angle APD = 180-\angle BPC.$ (All angles are in degrees)
Since $\sin(\theta)=\sin(180-\theta),$ we can use sine area formula to get the following(after some simplifying steps): \begin{align*} BP \times AP + CP \times DP = BP \times PC + AP \times PD. \end{align*} For convenienc... | 112 |
5,682 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_12 | 9 | Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$ | Use any method to derive that $P$ is the midpoint of $A$ and $C$ . Now, either construct a parallelogram with as shown in the diagram to the right, or use law of sines. We will take the parallelogram route here, but the same thing can be done with law of sines on triangles $\triangle \textnormal{ABP}$ and $\triangle \t... | 112 |
5,683 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_13 | 1 | Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $P_n$ be the probability of getting consecutive $1,2,3$ rolls in $n$ rolls and not rolling $1,2,3$ prior to the nth roll.
Let $x = P_3+P_5+...=1-(P_4+P_6+..)$ . Following Solution 2, one can see that \[P_{n+1}=P_{n}-\frac{P_{n-2}}{6^3}\] for all positive integers $n \ge 5$ . Summing for $n=5,7,...$ gives \[(1-x)-\f... | 647 |
5,684 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_13 | 2 | Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $P_{odd}=\frac{m}{n}$ , with the subscript indicating an odd number of rolls. Then $P_{even}=1-\frac{m}{n}$
The ratio of $\frac{P_{odd}}{P_{even}}$ is just $\frac{P_{odd}}{1-P_{odd}}$
We see that $P_{odd}$ is the sum of $P_3$ $P_5$ $P_7$ ,... , while $P_{even}$ is the sum of $P_4$ $P_6$ $P_8$ ,... .
$P_3$ , the pro... | 647 |
5,685 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_13 | 3 | Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Call the probability you win on a certain toss $f_n$ , where $n$ is the toss number.
Obviously, since the sequence has length 3, $f_1=0$ and $f_2=0$ .
Additionally, $f_3=\left(\frac{1}{6}\right)^3$ . We can call this value $x$ , to keep our further equations looking clean.
We can now write our general form for $f$ as $... | 647 |
5,686 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_13 | 4 | Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $S(n)$ be the number of strings of length $n$ containing the digits $1$ through $6$ that do not contain the string $123$ . Then we have $S(n) = 6 \cdot S(n-1) - S(n-3)$ because we can add any digit to end of a string with length $n-1$ but we have to subtract all the strings that end in $123$ . We rewrite this as \b... | 647 |
5,687 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_13 | 5 | Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $A=\frac{1}{6} \begin{bmatrix} 5 & 1 & 0 & 0 \\ 4 & 1 & 1 & 0 \\ 4 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$ $A$ is a transition matrix for the prefix of 1-2-3 matched so far. The state corresponding to a complete match has no outgoing probability mass. The probability that we roll the dice exactly $k$ times i... | 647 |
5,688 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_13 | 6 | Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Consider it as a contest of Odd and Even. Let $P_o$ and $P_e$ be probability that Odd and Even wins, respectively. If we consider every 3 rolls as an atomic action, then we can have a simple solution. If the rolls is 1-2-3, Odd wins; otherwise, Odd and Even switch the odds of winning. Therefore, we have \[P_o = \frac{1... | 647 |
5,689 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_13 | 7 | Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $P_o$ $P_e$ be the winning probabilities respectively. We call Odd "in position" when a new sequence of 1-2-3 starts at odd position, and likewise, call Even is "in position" when a new sequence starts at even position.
Now, consider the situation when the first roll is $1.$ The conditional probability for Odd or E... | 647 |
5,690 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_14 | 1 | The incircle $\omega$ of triangle $ABC$ is tangent to $\overline{BC}$ at $X$ . Let $Y \neq X$ be the other intersection of $\overline{AX}$ with $\omega$ . Points $P$ and $Q$ lie on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $\overline{PQ}$ is tangent to $\omega$ at $Y$ . Assume that $AP = 3$ $PB = 4$ $... | Let the sides $\overline{AB}$ and $\overline{AC}$ be tangent to $\omega$ at $Z$ and $W$ , respectively. Let $\alpha = \angle BAX$ and $\beta = \angle AXC$ . Because $\overline{PQ}$ and $\overline{BC}$ are both tangent to $\omega$ and $\angle YXC$ and $\angle QYX$ subtend the same arc of $\omega$ , it follows that $\ang... | 227 |
5,691 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_14 | 3 | The incircle $\omega$ of triangle $ABC$ is tangent to $\overline{BC}$ at $X$ . Let $Y \neq X$ be the other intersection of $\overline{AX}$ with $\omega$ . Points $P$ and $Q$ lie on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $\overline{PQ}$ is tangent to $\omega$ at $Y$ . Assume that $AP = 3$ $PB = 4$ $... | Let the center of the incircle of $\triangle ABC$ be $O$ . Link $OY$ and $OX$ . Then we have $\angle OYP=\angle OXB=90^{\circ}$
$\because$ $OY=OX$
$\therefore$ $\angle OYX=\angle OXY$
$\therefore$ $\angle PYX=\angle YXB$
$\therefore$ $\sin \angle PYX=\sin \angle YXB=\sin \angle YXC=\sin \angle PYA$
Let the incircle of ... | 227 |
5,692 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_15 | 1 | Find the number of functions $f$ from $\{0, 1, 2, 3, 4, 5, 6\}$ to the integers such that $f(0) = 0$ $f(6) = 12$ , and \[|x - y| \leq |f(x) - f(y)| \leq 3|x - y|\] for all $x$ and $y$ in $\{0, 1, 2, 3, 4, 5, 6\}$ | Because $f(n)$ and $f(n+1)$ can differ by at most 3 for $n=0,1,2,3,4,5$ $f$ can decrease at most once. This is because if $f$ decreases $2$ times (or more) then we get a contradiction: $12 = f(6) \le 3\cdot 4 + (-1)\cdot 2 = 10$
If $f$ never decreases, then $f(n+1)-f(n)\in \{1,2,3\}$ for all $n$ . Let $a, b$ , and $c$ ... | 185 |
5,693 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_1 | 1 | Fifteen distinct points are designated on $\triangle ABC$ : the 3 vertices $A$ $B$ , and $C$ $3$ other points on side $\overline{AB}$ $4$ other points on side $\overline{BC}$ ; and $5$ other points on side $\overline{CA}$ . Find the number of triangles with positive area whose vertices are among these $15$ points. | Every triangle is uniquely determined by 3 points. There are $\binom{15}{3}=455$ ways to choose 3 points, but that counts the degenerate triangles formed by choosing three points on a line. There are $\binom{5}{3}$ invalid cases on segment $AB$ $\binom{6}{3}$ invalid cases on segment $BC$ , and $\binom{7}{3}$ invalid c... | 390 |
5,694 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_4 | 1 | A pyramid has a triangular base with side lengths $20$ $20$ , and $24$ . The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length $25$ . The volume of the pyramid is $m\sqrt{n}$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square... | Let the triangular base be $\triangle ABC$ , with $\overline {AB} = 24$ . We find that the altitude to side $\overline {AB}$ is $16$ , so the area of $\triangle ABC$ is $(24*16)/2 = 192$
Let the fourth vertex of the tetrahedron be $P$ , and let the midpoint of $\overline {AB}$ be $M$ . Since $P$ is equidistant from $A$... | 803 |
5,695 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_4 | 2 | A pyramid has a triangular base with side lengths $20$ $20$ , and $24$ . The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length $25$ . The volume of the pyramid is $m\sqrt{n}$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square... | We can place a three dimensional coordinate system on this pyramid. WLOG assume the vertex across from the line that has length $24$ is at the origin, or $(0, 0, 0)$ . Then, the two other vertices can be $(-12, -16, 0)$ and $(12, -16, 0)$ . Let the fourth vertex have coordinates of $(x, y, z)$ . We have the following $... | 803 |
5,696 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_4 | 3 | A pyramid has a triangular base with side lengths $20$ $20$ , and $24$ . The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length $25$ . The volume of the pyramid is $m\sqrt{n}$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square... | Label the four vertices of the tetrahedron and the midpoint of $\overline {AB}$ , and notice that the area of the base of the tetrahedron, $\triangle ABC$ , equals $192$ , according to Solution 1.
Notice that the altitude of $\triangle CPM$ from $\overline {CM}$ to point $P$ is the height of the tetrahedron. Side $\ove... | 803 |
5,697 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_4 | 4 | A pyramid has a triangular base with side lengths $20$ $20$ , and $24$ . The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length $25$ . The volume of the pyramid is $m\sqrt{n}$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square... | Notation is shown on diagram. \[AM = MB = c = 12, AC = BC = b = 20,\] \[DA = DB = DC = a = 25.\] \[CM = x + y = \sqrt{b^2-c^2} = 16,\] \[x^2 - y^2 = CD^2 – DM^2 = CD^2 – (BD^2 – BM^2) = c^2 = 144,\] \[x – y = \frac{x^2 – y^2}{x+y} = \frac {c^2} {16} = 9,\] \[x = \frac {16 + 9}{2} = \frac {a}{2},\] \[h = \sqrt{a^2 -\fra... | 803 |
5,698 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_5 | 1 | A rational number written in base eight is $\underline{ab} . \underline{cd}$ , where all digits are nonzero. The same number in base twelve is $\underline{bb} . \underline{ba}$ . Find the base-ten number $\underline{abc}$ | First, note that the first two digits will always be a positive number. We will start with base twelve because of its repetition. List all the positive numbers in base eight that have equal tens and ones digits in base 12.
$11_{12}=15_8$
$22_{12}=32_8$
$33_{12}=47_8$
$44_{12}=64_8$
$55_{12}=101_8$
We stop because we on... | 321 |
5,699 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_5 | 2 | A rational number written in base eight is $\underline{ab} . \underline{cd}$ , where all digits are nonzero. The same number in base twelve is $\underline{bb} . \underline{ba}$ . Find the base-ten number $\underline{abc}$ | The parts before and after the decimal points must be equal. Therefore $8a + b = 12b + b$ and $c/8 + d/64 = b/12 + a/144$ . Simplifying the first equation gives $a = (3/2)b$ . Plugging this into the second equation gives $3b/32 = c/8 + d/64$ . Multiplying both sides by 64 gives $6b = 8c + d$ $a$ and $b$ are both digi... | 321 |
5,700 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_5 | 3 | A rational number written in base eight is $\underline{ab} . \underline{cd}$ , where all digits are nonzero. The same number in base twelve is $\underline{bb} . \underline{ba}$ . Find the base-ten number $\underline{abc}$ | Converting to base $10$ we get
$4604a+72c+9d=6960b$
Since $72c$ and $9d$ are much smaller than the other two terms, dividing by $100$ and approximating we get
$46a=70b$
Writing out the first few values of $a$ and $b$ , the first possible tuple is
$a=3, b=2, c=1, d=4$
and the second possible tuple is
$a=6, b=4, c=3, d=0... | 321 |
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