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int64 1
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5,801
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_14
| 2
|
Centered at each lattice point in the coordinate plane are a circle radius $\frac{1}{10}$ and a square with sides of length $\frac{1}{5}$ whose sides are parallel to the coordinate axes. The line segment from $(0,0)$ to $(1001, 429)$ intersects $m$ of the squares and $n$ of the circles. Find $m + n$
|
This is mostly a clarification to Solution 1, but let's take the diagram for the origin to $(7,3)$ . We have the origin circle and square intersected, then two squares, then the circle and square at $(7,3)$ . If we take the circle and square at the origin out of the diagram, we will be able to repeat the resulting segment (with its circles and squares) end to end from $(0,0)$ to $(1001,429)$ , which forms the line we need without overlapping. Since $143$ of these segments are needed to do this, and $3$ squares and $1$ circle are intersected with each, there are $143 \cdot (3+1) = 572$ squares and circles intersected. Adding the circle and square that are intersected at the origin back into the picture, we get that there are $572+2=\boxed{574}$ squares and circles intersected in total.
| 574
|
5,802
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_14
| 3
|
Centered at each lattice point in the coordinate plane are a circle radius $\frac{1}{10}$ and a square with sides of length $\frac{1}{5}$ whose sides are parallel to the coordinate axes. The line segment from $(0,0)$ to $(1001, 429)$ intersects $m$ of the squares and $n$ of the circles. Find $m + n$
|
This solution is a more systematic approach for finding when the line intersects the squares and circles. Because $1001 = 7*11*13$ and $429=3*11*13$ , the slope of our line is $\frac{3}{7}$ , and we only need to consider the line in the rectangle from the origin to $(7,3)$ , and we can iterate the line $11*13=143$ times. First, we consider how to figure out if the line intersects a square. Given a lattice point $(x_1, y_1)$ , we can think of representing a square centered at that lattice point as all points equal to $(x_1 \pm a, y_1 \pm b)$ s.t. $0 \leq a,b \leq \frac{1}{10}$ . If the line $y = \frac{3}{7}x$ intersects the square, then we must have $\frac{y_1 + b}{x_1 + a} = \frac{3}{7}$ . The line with the least slope that intersects the square intersects at the bottom right corner and the line with the greatest slope that intersects the square intersects at the top left corner; thus we must have that $\frac{3}{7}$ lies in between these slopes, or that $\frac{y_1-\frac{1}{10}}{x_1+\frac{1}{10}} \leq \frac{3}{7} \leq \frac{y_1+\frac{1}{10}}{x_1-\frac{1}{10}}$ . Simplifying, $3x_1 - 1 \leq 7y_1 \leq 3x_1 + 1$ . Because $y$ can only equal $0, 1, 2, 3$ , we just do casework based on the values of $y$ and find that the points $(2, 1)$ and $(5, 2)$ are intersected just at the corner of the square and $(0, 0), (7, 3)$ are intersected through the center of the square. However, we disregard one of $(0, 0)$ and $(7, 3)$ , WLOG $(0, 0)$ , since we just use it in our count for the next of the 143 segments. Therefore, in one of our "segments", 3 squares are intersected and 1 circle is intersected giving 4 total. Thus our answer is $143*4 = 572$ . HOWEVER, we cannot forget that we ignored $(0, 0)$ , which contributes another square and circle to our count, making the final answer $572 + 2 = \boxed{574}$
| 574
|
5,803
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15
| 1
|
Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$ . The three points $C$ $Y$ $D$ are collinear, $XC = 67$ $XY = 47$ , and $XD = 37$ . Find $AB^2$
|
Let $Z = XY \cap AB$ . By the radical axis theorem $AD, XY, BC$ are concurrent, say at $P$ . Moreover, $\triangle DXP \sim \triangle PXC$ by simple angle chasing. Let $y = PX, x = XZ$ . Then \[\frac{y}{37} = \frac{67}{y} \qquad \implies \qquad y^2 = 37 \cdot 67.\] Now, $AZ^2 = \tfrac 14 AB^2$ , and by power of a point, \begin{align*} x(y-x) &= \tfrac 14 AB^2, \quad \text{and} \\ x(47+x) &= \tfrac 14 AB^2 \end{align*} Solving, we get \[\tfrac 14 AB^2 = \tfrac 12 (y-47)\cdot \tfrac 12 (y+47) \qquad \implies\] \[\qquad AB ^ 2 = 37\cdot 67 - 47^2 = \boxed{270}\]
| 270
|
5,804
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15
| 2
|
Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$ . The three points $C$ $Y$ $D$ are collinear, $XC = 67$ $XY = 47$ , and $XD = 37$ . Find $AB^2$
|
By the Radical Axis Theorem $AD, XY, BC$ concur at point $E$
Let $AB$ and $EY$ intersect at $S$ . Note that because $AXDY$ and $CYXB$ are cyclic, by Miquel's Theorem $AXBE$ is cyclic as well. Thus \[\angle AEX = \angle ABX = \angle XCB = \angle XYB\] and \[\angle XEB = \angle XAB = \angle XDA = \angle XYA.\] Thus $AY \parallel EB$ and $YB \parallel EA$ , so $AEBY$ is a parallelogram. Hence $AS = SB$ and $SE = SY$ . But notice that $DXE$ and $EXC$ are similar by $AA$ Similarity, so $XE^2 = XD \cdot XC = 37 \cdot 67$ . But \[XE^2 - XY^2 = (XE + XY)(XE - XY) = EY \cdot 2XS = 2SY \cdot 2SX = 4SA^2 = AB^2.\] Hence $AB^2 = 37 \cdot 67 - 47^2 = \boxed{270}.$
| 270
|
5,805
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15
| 3
|
Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$ . The three points $C$ $Y$ $D$ are collinear, $XC = 67$ $XY = 47$ , and $XD = 37$ . Find $AB^2$
|
First, we note that as $\triangle XDY$ and $\triangle XYC$ have bases along the same line, $\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{DY}{YC}$ . We can also find the ratio of their areas using the circumradius area formula. If $R_1$ is the radius of $\omega_1$ and if $R_2$ is the radius of $\omega_2$ , then \[\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{(37\cdot 47\cdot DY)/(4R_1)}{(47\cdot 67\cdot YC)/(4R_2)}=\frac{37\cdot DY\cdot R_2}{67\cdot YC\cdot R_1}.\] Since we showed this to be $\frac{DY}{YC}$ , we see that $\frac{R_2}{R_1}=\frac{67}{37}$
We extend $AD$ and $BC$ to meet at point $P$ , and we extend $AB$ and $CD$ to meet at point $Q$ as shown below. [asy] size(200); import olympiad; real R1=45,R2=67*R1/37; real m1=sqrt(R1^2-23.5^2); real m2=sqrt(R2^2-23.5^2); pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5); draw(circle(o1,R1)); draw(circle(o2,R2)); pair q=(-R1/(R2-R1)*o2.x,0); pair a=tangent(q,o1,R1,2); pair b=tangent(q,o2,R2,2); pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q))[0]; pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q))[1]; pair p=extension(a,d,b,c); dot(q^^a^^b^^x^^y^^c^^d^^p); draw(q--b^^q--c); draw(p--d^^p--c^^x--y); draw(a--y^^b--y); draw(d--x--c); label("$A$",a,NW,fontsize(8)); label("$B$",b,NE,fontsize(8)); label("$C$",c,SE,fontsize(8)); label("$D$",d,SW,fontsize(8)); label("$X$",x,2*WNW,fontsize(8)); label("$Y$",y,3*S,fontsize(8)); label("$P$",p,N,fontsize(8)); label("$Q$",q,W,fontsize(8)); [/asy] As $ABCD$ is cyclic, we know that $\angle BCD=180-\angle DAB=\angle BAP$ . But then as $AB$ is tangent to $\omega_2$ at $B$ , we see that $\angle BCD=\angle ABY$ . Therefore, $\angle ABY=\angle BAP$ , and $BY\parallel PD$ . A similar argument shows $AY\parallel PC$ . These parallel lines show $\triangle PDC\sim\triangle ADY\sim\triangle BYC$ . Also, we showed that $\frac{R_2}{R_1}=\frac{67}{37}$ , so the ratio of similarity between $\triangle ADY$ and $\triangle BYC$ is $\frac{37}{67}$ , or rather \[\frac{AD}{BY}=\frac{DY}{YC}=\frac{YA}{CB}=\frac{37}{67}.\] We can now use the parallel lines to find more similar triangles. As $\triangle AQD\sim \triangle BQY$ , we know that \[\frac{QA}{QB}=\frac{QD}{QY}=\frac{AD}{BY}=\frac{37}{67}.\] Setting $QA=37x$ , we see that $QB=67x$ , hence $AB=30x$ , and the problem simplifies to finding $30^2x^2$ . Setting $QD=37^2y$ , we also see that $QY=37\cdot 67y$ , hence $DY=37\cdot 30y$ . Also, as $\triangle AQY\sim \triangle BQC$ , we find that \[\frac{QY}{QC}=\frac{YA}{CB}=\frac{37}{67}.\] As $QY=37\cdot 67y$ , we see that $QC=67^2y$ , hence $YC=67\cdot30y$
Applying Power of a Point to point $Q$ with respect to $\omega_2$ , we find \[67^2x^2=37\cdot 67^3 y^2,\] or $x^2=37\cdot 67 y^2$ . We wish to find $AB^2=30^2x^2=30^2\cdot 37\cdot 67y^2$
Applying Stewart's Theorem to $\triangle XDC$ , we find \[37^2\cdot (67\cdot 30y)+67^2\cdot(37\cdot 30y)=(67\cdot 30y)\cdot (37\cdot 30y)\cdot (104\cdot 30y)+47^2\cdot (104\cdot 30y).\] We can cancel $30\cdot 104\cdot y$ from both sides, finding $37\cdot 67=30^2\cdot 67\cdot 37y^2+47^2$ . Therefore, \[AB^2=30^2\cdot 37\cdot 67y^2=37\cdot 67-47^2=\boxed{270}.\]
| 270
|
5,806
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15
| 4
|
Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$ . The three points $C$ $Y$ $D$ are collinear, $XC = 67$ $XY = 47$ , and $XD = 37$ . Find $AB^2$
|
[asy] size(9cm); import olympiad; real R1=45,R2=67*R1/37; real m1=sqrt(R1^2-23.5^2); real m2=sqrt(R2^2-23.5^2); pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5); draw(circle(o1,R1)); draw(circle(o2,R2)); pair q=(-R1/(R2-R1)*o2.x,0); pair a=tangent(q,o1,R1,2); pair b=tangent(q,o2,R2,2); pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q))[0]; pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q))[1]; dot(a^^b^^x^^y^^c^^d); draw(x--y); draw(a--y^^b--y); draw(d--x--c); draw(a--b--c--d--cycle); draw(x--a^^x--b); label("$A$",a,NW,fontsize(9)); label("$B$",b,NE,fontsize(9)); label("$C$",c,SE,fontsize(9)); label("$D$",d,SW,fontsize(9)); label("$X$",x,2*N,fontsize(9)); label("$Y$",y,3*S,fontsize(9)); [/asy] First of all, since quadrilaterals $ADYX$ and $XYCB$ are cyclic, we can let $\angle DAX = \angle XYC = \theta$ , and $\angle XYD = \angle CBX = 180 - \theta$ , due to the properties of cyclic quadrilaterals. In addition, let $\angle BAX = x$ and $\angle ABX = y$ . Thus, $\angle ADX = \angle AYX = x$ and $\angle XYB = \angle XCB = y$ . Then, since quadrilateral $ABCD$ is cyclic as well, we have the following sums: \[\theta + x +\angle XCY + y = 180^{\circ}\] \[180^{\circ} - \theta + y + \angle XDY + x = 180^{\circ}\] Cancelling out $180^{\circ}$ in the second equation and isolating $\theta$ yields $\theta = y + \angle XDY + x$ . Substituting $\theta$ back into the first equation, we obtain \[2x + 2y + \angle XCY + \angle XDY = 180^{\circ}\] Since \[x + y +\angle XAY + \angle XCY + \angle DAY = 180^{\circ}\] \[x + y + \angle XDY + \angle XCY + \angle DAY = 180^{\circ}\] we can then imply that $\angle DAY = x + y$ . Similarly, $\angle YBC = x + y$ . So then $\angle DXY = \angle YXC = x + y$ , so since we know that $XY$ bisects $\angle DXC$ , we can solve for $DY$ and $YC$ with Stewart’s Theorem. Let $DY = 37n$ and $YC = 67n$ . Then \[37n \cdot 67n \cdot 104n + 47^2 \cdot 104n = 37^2 \cdot 67n + 67^2 \cdot 37n\] \[37n \cdot 67n + 47^2 = 37 \cdot 67\] \[n^2 = \frac{270}{2479}\] Now, since $\angle AYX = x$ and $\angle BYX = y$ $\angle AYB = x + y$ . From there, let $\angle AYD = \alpha$ and $\angle BYC = \beta$ . From angle chasing we can derive that $\angle YDX = \angle YAX = \beta - x$ and $\angle YCX = \angle YBX = \alpha - y$ . From there, since $\angle ADX = x$ , it is quite clear that $\angle ADY = \beta$ , and $\angle YAB = \beta$ can be found similarly. From there, since $\angle ADY = \angle YAB = \angle BYC = \beta$ and $\angle DAY = \angle AYB = \angle YBC = x + y$ , we have $AA$ similarity between $\triangle DAY$ $\triangle AYB$ , and $\triangle YBC$ . Therefore the length of $AY$ is the geometric mean of the lengths of $DA$ and $YB$ (from $\triangle DAY \sim \triangle AYB$ ). However, $\triangle DAY \sim \triangle AYB \sim \triangle YBC$ yields the proportion $\frac{AD}{DY} = \frac{YA}{AB} = \frac{BY}{YC}$ ; hence, the length of $AB$ is the geometric mean of the lengths of $DY$ and $YC$ .
We can now simply use arithmetic to calculate $AB^2$ \[AB^2 = DY \cdot YC\] \[AB^2 = 37 \cdot 67 \cdot \frac{270}{2479}\] \[AB^2 = \boxed{270}\]
| 270
|
5,807
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15
| 5
|
Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$ . The three points $C$ $Y$ $D$ are collinear, $XC = 67$ $XY = 47$ , and $XD = 37$ . Find $AB^2$
|
Let $E = DA \cap CB$ . By Radical Axes, $E$ lies on $XY$ . Note that $EAXB$ is cyclic as $X$ is the Miquel point of $\triangle EDC$ in this configuration.
Claim. $\triangle DXE \sim \triangle EXC$ Proof. We angle chase. \[\measuredangle XEC = \measuredangle XEB = \measuredangle XAB = \measuredangle XDA = \measuredangle XDE\] and \[\measuredangle XCE = \measuredangle XCB = \measuredangle XBA = \measuredangle XEA = \measuredangle XED. \square\]
Let $F = EX \cap AB$ . Note \[FA^2 = FX \cdot FY = FB^2\] and \[EF \cdot FX = AF \cdot FB = FA^2 = FX \cdot FY \implies EF = FY\] By our claim, \[\frac{DX}{XE} = \frac{EX}{XC} \implies EX^2 = DX \cdot XC = 67 \cdot 37 \implies FY = \frac{EY}{2} = \frac{EX+XY}{2} = \frac{\sqrt{67 \cdot 37}+47}{2}\] and \[FX=FY-47=\frac{\sqrt{67 \cdot 37}-47}{2}\] Finally, \[AB^2 = (2 \cdot FA)^2 = 4 \cdot FX \cdot FY = 4 \cdot \frac{(67 \cdot 37) - 47^2}{4} = \boxed{270}. \blacksquare\] ~Mathscienceclass
| 270
|
5,808
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15
| 6
|
Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$ . The three points $C$ $Y$ $D$ are collinear, $XC = 67$ $XY = 47$ , and $XD = 37$ . Find $AB^2$
|
$AB^2 = 4 AM^2 =2x(2x+ 2 XY) =(XP - XY) (XP + XY) = XP^2 - XY^2 = XC \cdot XD - XY^2 = 67 \cdot 37 - 47^2 = \boxed{270}.$
| 270
|
5,809
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15
| 7
|
Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$ . The three points $C$ $Y$ $D$ are collinear, $XC = 67$ $XY = 47$ , and $XD = 37$ . Find $AB^2$
|
Extend $\overline{AD}$ and $\overline{BC}$ to meet at point $P$ . Let $M$ be the midpoint of segment $AB$ . Then by radical axis on $(ADY)$ $(BCY)$ and $(ABCD)$ $P$ lies on $XY$ . By the bisector lemma, $M$ lies on $XY$ . It is well-known that $P$ $A$ $X$ , and $B$ are concyclic. By Power of a point on $M$ with respect to $(PAXB)$ and $(ADY)$ \[|\text{Pow}(M, (PAXB))| = MX \cdot MP = MA^2 = |\text{Pow}(M, (ADY))| = MX \cdot MY,\] so $MP=MY$ . Thus $AB$ and $PY$ bisect each other, so $PAYB$ is a parallelogram. This implies that \[\angle DAY = \angle YBC,\] so by the inscribed angle theorem $\overline{XY}$ bisects $\angle DXC$
Claim: $AB^2 = DY \cdot YC$
Proof. Define the linear function $f(\bullet) := \text{Pow}(\bullet, (ADY)) - \text{Pow}(\bullet, (ABCD))$ . Since $\overline{BY}$ is parallel to the radical axis $\overline{AD}$ of $(ADY)$ and $(ABCD)$ by our previous parallelism, $f(B)=f(Y)$ . Note that $f(B)=AB^2$ while $f(Y)=DY \cdot YC$ , so we conclude. $\square$
By Stewart's theorem on $\triangle DXC$ $DY \cdot YC=37 \cdot 67 - 47^2 = 270$ , so $AB^2=\boxed{270}$
| 270
|
5,810
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_1
| 1
|
Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially.
|
Let $r$ be the common ratio, where $r>1$ . We then have $ar-9-(a-5)=a(r-1)-4=ar^{2}-25-(ar-9)=ar(r-1)-16$ . We now have, letting, subtracting the 2 equations, $ar^{2}+-2ar+a=12$ , so we have $3ar=432,$ or $ar=144$ , which is how much Betty had. Now we have $144+\dfrac{144}{r}+144r=444$ , or $144(r+\dfrac{1}{r})=300$ , or $r+\dfrac{1}{r}=\dfrac{25}{12}$ , which solving for $r$ gives $r=\dfrac{4}{3}$ , since $r>1$ , so Alex had $\dfrac{3}{4} \cdot 144=\boxed{108}$ peanuts.
| 108
|
5,811
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_1
| 2
|
Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially.
|
Let $a$ be Alex's peanuts and $k$ the common ratio. Then we have $a(k^2+k+1)=444$ . Adding $k$ to both sides and factoring,
\[\frac{444}{a}+k=(k+1)^2\]
For the common difference, $ak=5-(a-5)=ak^2-25-(ak-9)$ . Simplifying, $k^2-2k+1=\frac{12}{a}$ . Factoring, \[(k-1)^2=\frac{12}{a}\]
\[(k+1)^2-(k-1)^2=4k \implies 4k=\frac{444-12}{a}+k \implies k=\frac{144}{a}\]
Substitute $k$ in the second equation to get $(\frac{144-a}{a})^2=\frac{12}{a}$ . Expanding and applying the quadratic formula, \[a=150\pm\frac{\sqrt{300^2-4(144^2)}}{2}\] Taking out $4^2\cdot3^2$ from under the radical leaves \[a=150\pm6\sqrt{625-576}=108, 192\] Since Alex's peanut number was the lowest of the trio, and $3*192>444$ , Alex initially had $\boxed{108}$ peanuts.
| 108
|
5,812
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_1
| 3
|
Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially.
|
Let the initial numbers of peanuts Alex, Betty and Charlie had be $a$ $b$ , and $c$ respectively.
Let the final numbers of peanuts, after eating, be $a'$ $b'$ , and $c'$
We are given that $a + b + c = 444$ . Since a total of $5 + 9 + 25 = 39$ peanuts are eaten, we must have $a' + b' + c' = 444 - 39 = 405$ .
Since $a'$ $b'$ , and $c'$ form an arithmetic progression, we have that $a' = b' - x$ and $c' = b' + x$ for some integer $x$ .
Substituting yields $3b' = 405$ and so $b' = 135$ . Since Betty ate $9$ peanuts, it follows that $b = b' + 9 = 144$
Since $a$ $b$ , and $c$ form a geometric progression, we have that $a = \frac{b}{r}$ and $c = br$ .
Multiplying yields $ac = b^2 = 144^2$ .
Since $a + c = 444 - b = 300$ , it follows that $a = 150 - \lambda$ and $c = 150 + \lambda$ for some integer $\lambda$ .
Substituting yields $(150-\lambda)(150+\lambda) = 144^2$ , which expands and rearranges to $\lambda^2 = 150^2-144^2 = 42^2$ .
Since $\lambda > 0$ , we must have $\lambda = 42$ , and so $a = 150 - \lambda = \boxed{108}$
| 108
|
5,813
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_1
| 4
|
Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially.
|
Bashing is not difficult. All we have to consider is the first equation. We can write it as $x*(1+r+r^2) = 444$ . The variable $x$ must be an integer, and after trying all the factors of $444$ , it's clear that $r$ is a fraction smaller than $10$ . When calculating the coefficient of $x$ , we must consider that the fraction produced will very likely have a numerator that divides $444$ . Trying a couple will make it easier to find the fraction, and soon you will find that $\frac{4}{3}$ gives a numerator of $37$ , a rather specific factor of $444$ . Solving for the rest will give you an integer value of $\boxed{108}$ . This is by no means a good solution, but it may be faster in a competition if you don't want to mess with several other equations. This is purely up to different individuals.
| 108
|
5,814
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_1
| 5
|
Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially.
|
Let $b$ be the finish number of Betty's peanuts. Then \[3b = 444-(5 + 9 + 25) = 405 = 3 \cdot 135 \implies b = 135, b+ 9 = 144.\]
Let $k > 1$ be the common ratio. Then \[\frac{144}{k} + 144 + k \cdot 144 = 444 \implies \frac{144}{k} + k \cdot 144 = 300\implies \frac{12}{k} + k \cdot 12 = 25\implies k = \frac{4}{3} \implies \frac {144\cdot 3}{4} = \boxed{108}.\]
| 108
|
5,815
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_2
| 1
|
There is a $40\%$ chance of rain on Saturday and a $30\%$ chance of rain on Sunday. However, it is twice as likely to rain on Sunday if it rains on Saturday than if it does not rain on Saturday. The probability that it rains at least one day this weekend is $\frac{a}{b}$ , where $a$ and $b$ are relatively prime positive integers. Find $a+b$
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Let $x$ be the probability that it rains on Sunday given that it doesn't rain on Saturday. We then have $\dfrac{4}{5}x+\dfrac{2}{5}2x = \dfrac{3}{10} \implies \dfrac{7}{5}x=\dfrac{3}{10}$ $\implies x=\dfrac{3}{14}$ . Therefore, the probability that it doesn't rain on either day is $\left(1-\dfrac{3}{14}\right)\left(\dfrac{3}{5}\right)=\dfrac{33}{70}$ . Therefore, the probability that rains on at least one of the days is $1-\dfrac{33}{70}=\dfrac{37}{70}$ , so adding up the $2$ numbers, we have $37+70=\boxed{107}$
| 107
|
5,816
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_3
| 1
|
Let $x,y,$ and $z$ be real numbers satisfying the system \begin{align*} \log_2(xyz-3+\log_5 x)&=5,\\ \log_3(xyz-3+\log_5 y)&=4,\\ \log_4(xyz-3+\log_5 z)&=4.\\ \end{align*} Find the value of $|\log_5 x|+|\log_5 y|+|\log_5 z|$
|
First, we get rid of logs by taking powers: $xyz-3+\log_5 x=2^{5}=32$ $xyz-3+\log_5 y=3^{4}=81$ , and $(xyz-3+\log_5 z)=4^{4}=256$ . Adding all the equations up and using the $\log {xy}=\log {x}+\log{y}$ property, we have $3xyz+\log_5{xyz} = 378$ , so we have $xyz=125$ . Solving for $x,y,z$ by substituting $125$ for $xyz$ in each equation, we get $\log_5 x=-90, \log_5 y=-41, \log_5 z=134$ , so adding all the absolute values we have $90+41+134=\boxed{265}$
| 265
|
5,817
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_4
| 1
|
An $a \times b \times c$ rectangular box is built from $a \cdot b \cdot c$ unit cubes. Each unit cube is colored red, green, or yellow. Each of the $a$ layers of size $1 \times b \times c$ parallel to the $(b \times c)$ faces of the box contains exactly $9$ red cubes, exactly $12$ green cubes, and some yellow cubes. Each of the $b$ layers of size $a \times 1 \times c$ parallel to the $(a \times c)$ faces of the box contains exactly $20$ green cubes, exactly $25$ yellow cubes, and some red cubes. Find the smallest possible volume of the box.
|
By counting the number of green cubes $2$ different ways, we have $12a=20b$ , or $a=\dfrac{5}{3} b$ . Notice that there are only $3$ possible colors for unit cubes, so for each of the $1 \times b \times c$ layers, there are $bc-21$ yellow cubes, and similarly there are $ac-45$ red cubes in each of the $1 \times a \times c$ layers. Therefore, we have $a(bc-21)=25b$ and $b(ac-45)=9a$ . We check a few small values of $a,b$ and solve for $c$ , checking $(a,b)=(5,3)$ gives $c=12$ with a volume of $180$ $(a,b)=(10,6)$ gives $c=6$ with a volume of $360$ , and $(a,b)=(15,9)$ gives $c=4$ , with a volume of $540$ . Any higher $(a,b)$ will $ab>180$ , so therefore, the minimum volume is $\boxed{180}$
| 180
|
5,818
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_4
| 2
|
An $a \times b \times c$ rectangular box is built from $a \cdot b \cdot c$ unit cubes. Each unit cube is colored red, green, or yellow. Each of the $a$ layers of size $1 \times b \times c$ parallel to the $(b \times c)$ faces of the box contains exactly $9$ red cubes, exactly $12$ green cubes, and some yellow cubes. Each of the $b$ layers of size $a \times 1 \times c$ parallel to the $(a \times c)$ faces of the box contains exactly $20$ green cubes, exactly $25$ yellow cubes, and some red cubes. Find the smallest possible volume of the box.
|
The total number of green cubes is given by $12a=20b\Longrightarrow a=\frac{5}{3}b$
Let $r$ be the number of red cubes on each one of the $b$ layers then the total number of red cubes is $9a=br$ . Substitute $a=\frac{5}{3}b$ gives $r=15$
Repeating the procedure on the number of yellow cubes $y$ on each of the $a$ layers gives $y=15$
Therefore $bc=9+12+15=36$ and $ac=15+20+25=60$ . Multiplying yields $abc^2=2160$
Since $abc^2$ is fixed, $abc$ is minimized when $c$ is maximized, which occurs when $a$ $b$ are minimized (since each of $ac$ $bc$ is fixed). Thus $(a,b,c)=(3,5,12)\Longrightarrow abc=\boxed{180}$
| 180
|
5,819
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_5
| 1
|
Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$ . Find $p$
|
Do note that by counting the area in 2 ways, the first altitude is $x = \frac{ab}{c}$ . By similar triangles, the common ratio is $\rho = \frac{a}{c}$ for each height, so by the geometric series formula, we have \begin{align} 6p=\frac{x}{1-\rho} = \frac{ab}{c-a}. \end{align} Writing $p=a+b+c$ and clearing denominators, we get \[13a=6p .\] Thus $p=13q$ $a=6q$ , and $b+c=7q$ , i.e. $c=7q-b$ . Plugging these into $(1)$ , we get $78q(q-b)=6bq$ , i.e., $14b=13q$ . Thus $q=14r$ and $p=182r$ $b=13r$ $a=84r$ $c=85r$ . Taking $r=1$ (since $a,b,c$ are relatively prime) we get $p=\boxed{182}$
| 182
|
5,820
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_5
| 2
|
Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$ . Find $p$
|
Note that by counting the area in 2 ways, the first altitude is $\dfrac{ab}{c}$ . By similar triangles, the common ratio is $\dfrac{a}{c}$ for reach height, so by the geometric series formula, we have $6p=\dfrac{\dfrac{ab}{c}}{1-\dfrac{a}{c}}$ . Multiplying by the denominator and expanding, the equation becomes $\dfrac{ab}{c}=6a+6b+6c-\dfrac{6a^2}{c}-\dfrac{6ab}{c}-6a$ . Cancelling $6a$ and multiplying by $c$ yields $ab=6bc+6c^2-6a^2-6ab$ , so $7ab = 6bc+6b^2$ and $7a=6b+6c$ . Checking for Pythagorean triples gives $13,84,$ and $85$ , so $p=13+84+85=\boxed{182}$
| 182
|
5,821
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_5
| 3
|
Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$ . Find $p$
|
We start by splitting the sum of all $C_{n-2}C_{n-1}$ into two parts: those where $n-2$ is odd and those where $n-2$ is even.
First consider the sum of the lengths of the segments for which $n-2$ is odd for each $n\geq2$ . The perimeters of these triangles can be expressed using $p$ and ratios that result because of similar triangles. Considering triangles where $n-2$ is odd, we find that the perimeter for each such $n$ is $p\left(\frac{C_{n-1}C_{n}}{C_{0}B}\right)$ . Thus,
$p\sum_{n=1}^{\infty}\frac{C_{2n-1}C_{2n}}{C_{0}B}=6p+C_{0}B$
Simplifying,
$\sum_{n=1}^{\infty}C_{2n-1}C_{2n}=6C_{0}B + \frac{(C_{0}B)^2}{p}=C_{0}B\left(6+\frac{C_{0}B}{p}\right)$ . (1)
Continuing with a similar process for the sum of the lengths of the segments for which $n-2$ is even,
$p\sum_{n=1}^{\infty}\frac{C_{2n-2}C_{2n-1}}{C_{0}B}=6p+C_{0}A+AB=7p-C_{0}B$
Simplifying,
$\sum_{n=1}^{\infty}C_{2n-2}C_{2n-1}=C_{0}B\left(7-\frac{C_{0}B}{p}\right)$ . (2)
Adding (1) and (2) together, we find that
$6p=13C_{0}B \Rightarrow p=\frac{13C_{0}B}{6}=C_{0}B+C_{0}A+AB \Rightarrow \frac{7C_{0}B}{6}=C_{0}A+AB \Rightarrow 7C_{0}B=6C_{0}A + 6AB$
Setting $a=C_{0}B$ $b=C_{0}A$ , and $c=AB$ , we can now proceed as in Shaddoll's solution, and our answer is $p=13+84+85=\boxed{182}$
| 182
|
5,822
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_5
| 4
|
Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$ . Find $p$
|
[asy] size(10cm); // Setup pair A, B; pair C0, C1, C2, C3, C4, C5, C6, C7, C8; A = (5, 0); B = (0, 3); C0 = (0, 0); C1 = foot(C0, A, B); C2 = foot(C1, C0, B); C3 = foot(C2, C1, B); C4 = foot(C3, C2, B); C5 = foot(C4, C3, B); C6 = foot(C5, C4, B); C7 = foot(C6, C5, B); C8 = foot(C7, C6, B); // Labels label("$A$", A, SE); label("$B$", B, NW); label("$C_0$", C0, SW); label("$C_1$", C1, NE); label("$C_2$", C2, W); label("$C_3$", C3, NE); label("$C_4$", C4, W); label("$a$", (B+C0)/2, W); label("$b$", (A+C0)/2, S); label("$c$", (A+B)/2, NE); // Drawings draw(A--B--C0--cycle); draw(C0--C1--C2, red); draw(C2--C3--C4--C5--C6--C7--C8); draw(C0--C2, blue); [/asy]
Let $a = BC_0$ $b = AC_0$ , and $c = AB$ .
Note that the total length of the red segments in the figure above is equal to the length of the blue segment times $\frac{a+c}{b}$
The desired sum is equal to the total length of the infinite path $C_0 C_1 C_2 C_3 \cdots$ , shown in red in the figure below.
Since each of the triangles $\triangle C_0 C_1 C_2, \triangle C_2 C_3 C_4, \dots$ on the left are similar, it follows that the total length of the red segments in the figure below is equal to the length of the blue segment times $\frac{a+c}{b}$ .
In other words, we have that $a\left(\frac{a+c}{b}\right) = 6p$
Guessing and checking Pythagorean triples reveals that $a = 84$ $b=13$ $c = 85$ , and $p = a + b + c = \boxed{182}$ satisfies this equation.
| 182
|
5,823
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_5
| 5
|
Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$ . Find $p$
|
This solution proceeds from $\frac{\frac{ab}{c}}{1-\frac{a}{c}} = \frac{\frac{ab}{c}}{\frac{c-a}{c}} = \frac{ab}{c-a} = 6(a+b+c)$ . Note the general from for a primitive pythagorean triple, $m^2-n^2, 2mn, m^2+n^2$ and after substitution, letting $a = m^2-n^2, b = 2mn, c = m^2+n^2$ into the previous equation simplifies down very nicely into $m = 13n$ . Thus $a = 168n^2, b = 26n^2, c = 170n^2$ . Since we know all three side lengths are relatively prime, we must divide each by 2 and let n = 1 giving $a = 84, b = 13, c = 85$ yielding $p = a + b + c = \boxed{182}$
| 182
|
5,824
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_5
| 6
|
Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$ . Find $p$
|
For this problem, first notice that its an infinite geometric series of $6(a+b+c)=\frac{ab}{c-b}$ if $c$ is the hypotenuse. WLOG $a<b$ , we can generalize a pythagorean triple of $x^2-y^2, 2xy, x^2+y^2$ . Let $b=2xy$ , then this generalization gives $6(a+b+c)(c-b)=ab$ \[(x^2-y^2)2xy=6(2x^2+2xy)(x-y)^2\] \[(x+y)xy=6(x^2+xy)(x-y)\] \[xy=6x(x-y)\] \[7xy=6x^2\] \[y=\frac{6}{7}x\]
Now this is just clear. Let $x=7m$ and $y=6m$ for $m$ to be a positive integer, the pythagorean triple is $13-84-85$ which yields $\boxed{182}$
| 182
|
5,825
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_6
| 1
|
For polynomial $P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}$ , define $Q(x)=P(x)P(x^{3})P(x^{5})P(x^{7})P(x^{9})=\sum_{i=0}^{50} a_ix^{i}$ .
Then $\sum_{i=0}^{50} |a_i|=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
Note that all the coefficients of odd-powered terms is an odd number of odd degree terms multiplied together, and all coefficients of even-powered terms have an even number of odd degree terms multiplied together. Since every odd degree term is negative, and every even degree term is positive, the sum is just equal to $Q(-1)=P(-1)^{5}=\left( \dfrac{3}{2}\right)^{5}=\dfrac{243}{32}$ , so the desired answer is $243+32=\boxed{275}$
| 275
|
5,826
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_6
| 2
|
For polynomial $P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}$ , define $Q(x)=P(x)P(x^{3})P(x^{5})P(x^{7})P(x^{9})=\sum_{i=0}^{50} a_ix^{i}$ .
Then $\sum_{i=0}^{50} |a_i|=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
We are looking for the sum of the absolute values of the coefficients of $Q(x)$ . By defining $P'(x) = 1 + \frac{1}{3}x+\frac{1}{6}x^2$ , and defining $Q'(x) = P'(x)P'(x^3)P'(x^5)P'(x^7)P'(x^9)$ , we have made it so that all coefficients in $Q'(x)$ are just the positive/absolute values of the coefficients of $Q(x)$ . .
To find the sum of the absolute values of the coefficients of $Q(x)$ , we can just take the sum of the coefficients of $Q'(x)$ . This sum is equal to \[Q'(1) = P'(1)P'(1)P'(1)P'(1)P'(1) = \left(1+\frac{1}{3}+\frac{1}{6}\right)^5 = \frac{243}{32},\]
so our answer is $243+32 = \boxed{275}$
| 275
|
5,827
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_6
| 3
|
For polynomial $P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}$ , define $Q(x)=P(x)P(x^{3})P(x^{5})P(x^{7})P(x^{9})=\sum_{i=0}^{50} a_ix^{i}$ .
Then $\sum_{i=0}^{50} |a_i|=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
Multiply $P(x)P(x^3)$ and notice that the odd degree terms have a negative coefficient. Observing that this is probably true for all polynomials like this (including $P(x)P(x^3)P(x^5)P(x^7)P(x^9)$ ), we plug in $-1$ to get $\frac{243}{32} \implies \boxed{275}$
| 275
|
5,828
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_7
| 1
|
Squares $ABCD$ and $EFGH$ have a common center and $\overline{AB} || \overline{EF}$ . The area of $ABCD$ is 2016, and the area of $EFGH$ is a smaller positive integer. Square $IJKL$ is constructed so that each of its vertices lies on a side of $ABCD$ and each vertex of $EFGH$ lies on a side of $IJKL$ . Find the difference between the largest and smallest positive integer values for the area of $IJKL$
|
Letting $AI=a$ and $IB=b$ , we have \[IJ^{2}=a^{2}+b^{2} \geq 1008\] by AM-GM inequality . Also, since $EFGH||ABCD$ , the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and $2$ adjacent vertices of a smaller square are similar. Therefore, the areas form a geometric progression, so since \[2016=12^{2} \cdot 14\] we have the maximum area is \[2016 \cdot \dfrac{11}{12} = 1848\] (the areas of the squares from largest to smallest are $12^{2} \cdot 14, 11 \cdot 12 \cdot 14, 11^{2} \cdot 14$ forming a geometric progression).
The minimum area is $1008$ (every square is half the area of the square whose sides its vertices touch), so the desired answer is \[1848-1008=\boxed{840}\]
| 840
|
5,829
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_8
| 1
|
Find the number of sets $\{a,b,c\}$ of three distinct positive integers with the property that the product of $a,b,$ and $c$ is equal to the product of $11,21,31,41,51,61$
|
Note that the prime factorization of the product is $3^{2}\cdot 7 \cdot 11 \cdot 17 \cdot 31 \cdot 41 \cdot 61$ . Ignoring overcounting, by stars and bars there are $6$ ways to choose how to distribute the factors of $3$ , and $3$ ways to distribute the factors of the other primes, so we have $3^{6} \cdot 6$ ways. However, some sets have $2$ numbers that are the same, namely the ones in the form $1,1,x$ and $3,3,x$ , which are each counted $3$ times, and each other set is counted $6$ times, so the desired answer is $\dfrac{729 \cdot 6-6}{6} = \boxed{728}$
| 728
|
5,830
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_8
| 2
|
Find the number of sets $\{a,b,c\}$ of three distinct positive integers with the property that the product of $a,b,$ and $c$ is equal to the product of $11,21,31,41,51,61$
|
Again, notice that the prime factors of the product are $3, 3, 7, 11, 17, 31, 41, 61$ . In this problem, we are asked to partition this set of distinct(ish) factors into three smaller indistinct sets. To do this, we can use Stirling numbers of the second kind, but Stirling numbers of the second kind would assume no empty parts, which isn't what we want. However, this is easy to fix. Denote Stirling numbers of the second kind with $S(n,k)$ . We may start at the situation when all three sets are nonempty. Then, the number of partitions is simply $S(8,3)$ . However, we are overcounting, since every time the two threes are in different sets, (unless they are both individually in their own sets), each one is double counted. To fix this, we can see that they were in different sets $S(8, 3) - S(7,3)$ times by complementary counting, but one of these times they were in their own individual sets, so the total overcount is $\dfrac{S(8,3) - S(7,3) - 1}{2}$ . Similarly, we can do the cases for if two of the sets are nonempty and one of the sets are nonempty, (but in those last two cases the threes cannot be individually in their own sets). We then find that the "answer" is given by: \[S(8,3) - \dfrac{S(8,3) - S(7,3) - 1}{2} + S(8,2) - \dfrac{S(8,2) - S(7,2)}{2}+ S(8,1) - \dfrac{S(8,1) - S(7,1)}{2}\] \[= \dfrac{S(8,3)+S(8,2) + S(8,1) + S(7,3) + S(7,2) + S(7,1) + 1}{2}\] Drawing out the Stirling number triangle and evaluating yields $730$ for the last expression. However, throughout the entire solution, we ignored the fact that the three numbers $a$ $b$ , and $c$ needed to be distinct. However, this is easy to fix, since there are only two sets in which they are not, namely $(1,1,x)$ and $(3,3,x)$ . Thus, the actual answer is $730 - 2 = \boxed{728}$
| 728
|
5,831
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_9
| 1
|
The sequences of positive integers $1,a_2, a_3,...$ and $1,b_2, b_3,...$ are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let $c_n=a_n+b_n$ . There is an integer $k$ such that $c_{k-1}=100$ and $c_{k+1}=1000$ . Find $c_k$
|
Since all the terms of the sequences are integers, and 100 isn't very big, we should just try out the possibilities for $b_2$ . When we get to $b_2=9$ and $a_2=91$ , we have $a_4=271$ and $b_4=729$ , which works, therefore, the answer is $b_3+a_3=81+181=\boxed{262}$
| 262
|
5,832
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_9
| 2
|
The sequences of positive integers $1,a_2, a_3,...$ and $1,b_2, b_3,...$ are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let $c_n=a_n+b_n$ . There is an integer $k$ such that $c_{k-1}=100$ and $c_{k+1}=1000$ . Find $c_k$
|
Using the same reasoning ( $100$ isn't very big), we can guess which terms will work. The first case is $k=3$ , so we assume the second and fourth terms of $c$ are $100$ and $1000$ . We let $r$ be the common ratio of the geometric sequence and write the arithmetic relationships in terms of $r$
The common difference is $100-r - 1$ , and so we can equate: $2(99-r)+100-r=1000-r^3$ . Moving all the terms to one side and the constants to the other yields
$r^3-3r = 702$ , or $r(r^2-3) = 702$ . Simply listing out the factors of $702$ shows that the only factor $3$ less than a square that works is $78$ . Thus $r=9$ and we solve from there to get $\boxed{262}$
| 262
|
5,833
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_10
| 1
|
Triangle $ABC$ is inscribed in circle $\omega$ . Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$ . Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$ ), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$ , then $ST=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
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[asy] import cse5; pathpen = black; pointpen = black; pointfontsize = 9; size(8cm); pair A = origin, B = (13,0), P = (4,0), Q = (7,0), T = B + 5 dir(220), C = IP(circumcircle(A,B,T),Line(T,Q,-0.1,10)), S = IP(circumcircle(A,B,C),Line(C,P,-0.1,10)); Drawing(A--B--C--cycle); D(circumcircle(A,B,C),rgb(0,0.6,1)); DrawPathArray(C--S^^C--T,rgb(1,0.4,0.1)); DrawPathArray(A--S^^B--T,rgb(0,0.4,0)); D(S--T,rgb(1,0.2,0.4)); D("A",A,dir(215)); D("B",B,dir(330)); D("P",P,dir(240)); D("Q",Q,dir(240)); D("T",T,dir(290)); D("C",C,dir(120)); D("S",S,dir(250)); MP("4",(A+P)/2,dir(90)); MP("3",(P+Q)/2,dir(90)); MP("6",(Q+B)/2,dir(90)); MP("5",(B+T)/2,dir(140)); MP("7",(A+S)/2,dir(40)); [/asy] Let $\angle ACP=\alpha$ $\angle PCQ=\beta$ , and $\angle QCB=\gamma$ . Note that since $\triangle ACQ\sim\triangle TBQ$ we have $\tfrac{AC}{CQ}=\tfrac56$ , so by the Ratio Lemma \[\dfrac{AP}{PQ}=\dfrac{AC}{CQ}\cdot\dfrac{\sin\alpha}{\sin\beta}\quad\implies\quad \dfrac{\sin\alpha}{\sin\beta}=\dfrac{24}{15}.\] Similarly, we can deduce $\tfrac{PC}{CB}=\tfrac47$ and hence $\tfrac{\sin\beta}{\sin\gamma}=\tfrac{21}{24}$
Now Law of Sines on $\triangle ACS$ $\triangle SCT$ , and $\triangle TCB$ yields \[\dfrac{AS}{\sin\alpha}=\dfrac{ST}{\sin\beta}=\dfrac{TB}{\sin\gamma}.\] Hence \[\dfrac{ST^2}{\sin^2\beta}=\dfrac{TB\cdot AS}{\sin\alpha\sin\gamma},\] so \[TS^2=TB\cdot AS\left(\dfrac{\sin\beta}{\sin\alpha}\dfrac{\sin\beta}{\sin\gamma}\right)=\dfrac{15\cdot 21}{24^2}\cdot 5\cdot 7=\dfrac{35^2}{8^2}.\] Hence $ST=\tfrac{35}8$ and the requested answer is $35+8=\boxed{43}$
| 43
|
5,834
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_10
| 3
|
Triangle $ABC$ is inscribed in circle $\omega$ . Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$ . Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$ ), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$ , then $ST=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
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By Ptolemy's Theorem applied to quadrilateral $ASTB$ , we find \[5\cdot 7+13\cdot ST=AT\cdot BS.\] Therefore, in order to find $ST$ , it suffices to find $AT\cdot BS$ . We do this using similar triangles, which can be found by using Power of a Point theorem.
As $\triangle APS\sim \triangle CPB$ , we find \[\frac{4}{PC}=\frac{7}{BC}.\] Therefore, $\frac{BC}{PC}=\frac{7}{4}$
As $\triangle BQT\sim\triangle CQA$ , we find \[\frac{6}{CQ}=\frac{5}{AC}.\] Therefore, $\frac{AC}{CQ}=\frac{5}{6}$
As $\triangle ATQ\sim\triangle CBQ$ , we find \[\frac{AT}{BC}=\frac{7}{CQ}.\] Therefore, $AT=\frac{7\cdot BC}{CQ}$
As $\triangle BPS\sim \triangle CPA$ , we find \[\frac{9}{PC}=\frac{BS}{AC}.\] Therefore, $BS=\frac{9\cdot AC}{PC}$ . Thus we find \[AT\cdot BS=\left(\frac{7\cdot BC}{CQ}\right)\left(\frac{9\cdot AC}{PC}\right).\] But now we can substitute in our previously found values for $\frac{BC}{PC}$ and $\frac{AC}{CQ}$ , finding \[AT\cdot BS=63\cdot \frac{7}{4}\cdot \frac{5}{6}=\frac{21\cdot 35}{8}.\] Substituting this into our original expression from Ptolemy's Theorem, we find \begin{align*}35+13ST&=\frac{21\cdot 35}{8}\\13ST&=\frac{13\cdot 35}{8}\\ST&=\frac{35}{8}.\end{align*} Thus the answer is $\boxed{43}$
| 43
|
5,835
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_10
| 5
|
Triangle $ABC$ is inscribed in circle $\omega$ . Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$ . Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$ ), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$ , then $ST=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
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By Ptolemy's Theorem applied to quadrilateral $ASTB$ , we find \[AS\cdot BT+AB\cdot ST=AT\cdot BS.\] Projecting through $C$ we have \[\frac{AQ \cdot PB}{PQ \cdot AB} = (A,Q; P,B)\stackrel{C}{=}(A,T; S,B)=\frac{AT \cdot BS}{ST \cdot AB}.\] Therefore \[AT \cdot BS = \frac {AQ \cdot PB}{PQ} \times ST \implies\] \[\left(\frac {AQ \cdot PB}{PQ} - AB\right)\times ST = AS \cdot BT \implies\] \[ST = \frac {AS \cdot BT \cdot PQ}{AQ \cdot PB – AB \cdot PQ}\] \[ST = \frac {7\cdot 5 \cdot 3}{7\cdot 9 – 13 \cdot 3 } = \frac {35}{8} \implies 35 + 8 = \boxed{43}.\]
| 43
|
5,836
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_11
| 1
|
For positive integers $N$ and $k$ , define $N$ to be $k$ -nice if there exists a positive integer $a$ such that $a^{k}$ has exactly $N$ positive divisors. Find the number of positive integers less than $1000$ that are neither $7$ -nice nor $8$ -nice.
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We claim that an integer $N$ is only $k$ -nice if and only if $N \equiv 1 \pmod k$ . By the number of divisors formula, the number of divisors of $\prod_{i=1}^n p_i^{a_i}$ is $\prod_{i=1}^n (a_i+1)$ . Since all the $a_i$ s are divisible by $k$ in a perfect $k$ power, the only if part of the claim follows. To show that all numbers $N \equiv 1 \pmod k$ are $k$ -nice, write $N=bk+1$ . Note that $2^{kb}$ has the desired number of factors and is a perfect kth power. By PIE, the number of positive integers less than $1000$ that are either $1 \pmod 7$ or $1\pmod 8$ is $142+125-17=250$ , so the desired answer is $999-250=\boxed{749}$
| 749
|
5,837
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12
| 1
|
The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy]
|
Choose a section to start coloring. Assume, WLOG, that this section is color $1$ . We proceed coloring clockwise around the ring. Let $f(n,C)$ be the number of ways to color the first $n$ sections (proceeding clockwise) such that the last section has color $C$ . In general (except for when we complete the coloring), we see that \[f(n,C_i)=\sum_{j\ne i} f(n-1,C_j),\] i.e., $f(n,C_i)$ is equal to the number of colorings of $n-1$ sections that end in any color other than $C_i$ . Using this, we can compute the values of $f(n,C)$ in the following table.
$\begin{tabular}{c|c|c|c|c } \multicolumn{1}{c}{}&\multicolumn{4}{c}{\(C\)}\\ \(n\)&1 & 2 & 3& 4 \\ \hline 1& 1 & 0 & 0 & 0\\ 2 & 0 & 1 & 1 & 1 \\ 3& 3 & 2 & 2 & 2 \\ 4 & 6 & 7 & 7 & 7 \\ 5 & 21 & 20 & 20 & 20\\ 6& 0& 61 & 61 & 61\\ \end{tabular}$
Note that $f(6, 1)=0$ because then $2$ adjacent sections are both color $1$ . We multiply this by $4$ to account for the fact that the initial section can be any color. Thus the desired answer is $(61+61+61) \cdot 4 = \boxed{732}$
| 732
|
5,838
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12
| 2
|
The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy]
|
We use complementary counting. There are $4^6$ total colorings of the ring without restriction. To count the complement, we wish to count the number of colorings in which at least one set of adjacent sections are the same color. There are six possible sets of adjacent sections that could be the same color (think of these as borders). Call these $B_1,B_2,\dots,B_6$ . Let $\mathcal{A}_1, \mathcal{A}_2,\dots,\mathcal{A}_6$ be the sets of colorings of the ring where the sections on both sides of $B_1,B_2,\dots,B_6$ are the same color. We wish to determine $|\mathcal{A}_1\cup\mathcal{A}_2\cup\cdots\cup\mathcal{A}_6|$ . Note that all of these cases are symmetric, and in general, $|\mathcal{A}_i|=4^5$ . There are $6$ such sets $\mathcal{A}_i$ . Also, $|\mathcal{A}_i\cup\mathcal{A}_j|=4^4$ , because we can only change colors at borders, so if we have two borders along which we cannot change colors, then there are four borders along which we have a choice of color. There are $\binom{6}{2}$ such pairs $\mathcal{A}_i\cup\mathcal{A}_j$ . Similarly, $|\mathcal{A}_i\cup \mathcal{A}_j\cup \mathcal{A}_k|=4^3$ , with $\binom{6}{3}$ such triples, and we see that the pattern will continue for 4-tuples and 5-tuples. For 6-tuples, however, these cases occur when there are no changes of color along the borders, i.e., each section has the same color. Clearly, there are four such possibilities.
Therefore, by PIE, \[|\mathcal{A}_1\cup\mathcal{A}_2\cup\cdots\cup\mathcal{A}_6|=\binom{6}{1}\cdot 4^5-\binom{6}{2}\cdot 4^4+\binom{6}{3}\cdot 4^3-\binom{6}{4}\cdot 4^2+\binom{6}{5}\cdot 4^1-4.\] We wish to find the complement of this, or \[4^6-\left(\binom{6}{1}\cdot 4^5-\binom{6}{2}\cdot 4^4+\binom{6}{3}\cdot 4^3-\binom{6}{4}\cdot 4^2+\binom{6}{5}\cdot 4^1-4\right).\] By the Binomial Theorem, this is $(4-1)^6+3=\boxed{732}$
| 732
|
5,839
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12
| 3
|
The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy]
|
We use generating functions. Suppose that the colors are $0,1,2,3$ . Then as we proceed around a valid coloring of the ring in the clockwise direction, we know that between two adjacent sections with colors $s_i$ and $s_{i+1}$ , there exists a number $d_i\in\{1,2,3\}$ such that $s_{i+1}\equiv s_i+d_i\pmod{4}$ . Therefore, we can represent each border between sections by the generating function $(x+x^2+x^3)$ , where $x,x^2,x^3$ correspond to increasing the color number by $1,2,3\pmod4$ , respectively. Thus the generating function that represents going through all six borders is $A(x)=(x+x^2+x^3)^6$ , where the coefficient of $x^n$ represents the total number of colorings where the color numbers are increased by a total of $n$ as we proceed around the ring. But if we go through all six borders, we must return to the original section, which is already colored. Therefore, we wish to find the sum of the coefficients of $x^n$ in $A(x)$ with $n\equiv 0\pmod4$
Now we note that if $P(x)=x^n$ , then \[P(x)+P(ix)+P(-x)+P(-ix)=\begin{cases}4x^n&\text{if } n\equiv0\pmod{4}\\0&\text{otherwise}.\end{cases}\] Therefore, the sum of the coefficients of $A(x)$ with powers congruent to $0\pmod 4$ is \[\frac{A(1)+A(i)+A(-1)+A(-i)}{4}=\frac{3^6+(-1)^6+(-1)^6+(-1)^6}{4}=\frac{732}{4}.\] We multiply this by $4$ to account for the initial choice of color, so our answer is $\boxed{732}$
| 732
|
5,840
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12
| 4
|
The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy]
|
Let $f(n)$ be the number of valid ways to color a ring with $n$ sections (which we call an $n$ -ring), so the answer is given by $f(6)$ . For $n=2$ , we compute $f(n)=4\cdot3=12$ . For $n \ge 3$ , we can count the number of valid colorings as follows: choose one of the sections arbitrarily, which we may color in $4$ ways. Moving clockwise around the ring, there are $3$ ways to color each of the $n-1$ other sections. Therefore, we have $4 \cdot 3^{n-1}$ colorings of an $n$ -ring.
However, note that the first and last sections could be the same color under this method. To count these invalid colorings, we see that by "merging" the first and last sections into one, we get a valid coloring of an $(n-1)$ -ring. That is, there are $f(n-1)$ colorings of an $n$ -ring in which the first and last sections have the same color. Thus, $f(n) = 4 \cdot 3^{n-1} - f(n-1)$ for all $n \ge 3$
To compute the requested value $f(6)$ , we repeatedly apply this formula: \begin{align*} f(6)&=4\cdot3^5-f(5)\\&=4\cdot3^5-4\cdot3^4+f(4)\\&=4\cdot3^5-4\cdot3^4+4\cdot3^3-f(3)\\&=4\cdot3^5-4\cdot3^4+4\cdot3^3-4\cdot3^2+f(2)\\&=4(3^5-3^4+3^3-3^2+3)\\&=4\cdot3\cdot\frac{3^5+1}{3+1}\\&=\boxed{732} (Solution by MSTang.)
| 732
|
5,841
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12
| 5
|
The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy]
|
Label the sections 1, 2, 3, 4, 5, 6 clockwise. We do casework on the colors of sections 1, 3, 5.
Case 1: the colors of the three sections are the same.
In this case, each of sections 2, 4, 6 can be one of 3 colors, so this case yields $4 \times 3^3 = 108$ ways.
Case 2: two of sections 1, 3, 5 are the same color.
Note that there are 3 ways for which two of the three sections have the same color, and $4 \times 3 = 12$ ways to determine their colors. After this, the section between the two with the same color can be one of 3 colors and the other two can be one of 2 colors. This case yields $3 \times (4 \times 3) \times (3 \times 2 \times 2) = 432$ ways.
Case 3: all three sections of 1, 3, 5 are of different colors.
Clearly, there are $4 \times 3 \times 2 = 24$ choices for which three colors to use, and there are 2 ways to choose the colors of each of sections 2, 4, 6. Thus, this case gives $4 \times 3 \times 2 \times 2^3 = 192$ ways.
In total, there are $108 + 432 + 192 = \boxed{732}$ valid colorings.
| 732
|
5,842
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12
| 6
|
The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy]
|
We will take a recursive approach to this problem. We can start by writing out the number of colorings for a circle with $1, 2,$ and $3$ compartments, which are $4, 12,$ and $24.$ Now we will try to find a recursive formula, $C(n)$ , for a circle with an arbitrary number of compartments $n.$ We will do this by focusing on the $n-1$ section in the circle. This section can either be the same color as the first compartment, or it can be a different color as the first compartment. We will focus on each case separately.
Case 1:
If they are the same color, we can say there are $C(n-2)$ to fill the first $n-1$ compartments. The $nth$ compartment must be different from the first and second to last compartments, which are the same color. Hence this case adds $3*C(n-2)$ to our recursive formula.
Case 2:
If they are different colors, we can say that there are $C(n-1)$ to fill the first $n-1$ compartments, and for the the $nth$ compartment, there are $2$ ways to color it because the $n-1$ and $1$ compartments are different colors. Hence this case adds $2*C(n-1).$
So our recursive formula, $C(n)$ , is $3*C(n-2) + 2*C(n-1).$ Using the initial values we calculated, we can evaluate this recursive formula up to $n=6.$ When $n=6,$ we get $\boxed{732}$ valid colorings.
| 732
|
5,843
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12
| 7
|
The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy]
|
WLOG, color the top left section $1$ and the top right section $2$ . Then the left section can be colored $2$ $3$ , or $4$ , and the right section can be colored $1$ $3$ , or $4$ . There are $3 \cdot 3 = 9$ ways to color the left and right sections. We split this up into two cases.
Case 1: The left and right sections are of the same color. There are $2$ ways this can happen: either they both are $3$ or they both are $4$ . We have $3$ colors to choose for the bottom left, and $2$ remaining colors to choose for the bottom right, for a total of $2 \cdot 3 \cdot 2$ cases.
Case 2: The left and right sections are of different colors. There are $9 - 2 = 7$ ways this can happen. Assume the left is $3$ and the right is $4$ . Then the bottom left can be $1$ $2$ , or $4$ , and the bottom right can be $1$ $2$ , or $3$ . However the bottom sections cannot both be $1$ or both be $2$ , so there are $3 \cdot 3 - 2 = 7$ ways to color the bottom sections, for a total for $7 \cdot 7 = 49$ colorings.
Since there were $4 \cdot 3 = 12$ ways to color the top sections, the answer is $12 \cdot (12 + 49) = \boxed{732}$
| 732
|
5,844
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12
| 9
|
The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy]
|
Let's number the regions $1,2,\dots 6$ . Suppose we color regions $1,2,3$ . Then, how many ways are there to color $4,5,6$
Note: the numbers are numbered as shown:
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); label("1",(-3.5,0)); label("2",(-1.6,3)); label("3",(1.6,3)); label("4",(3.5,0)); label("5",(1.6,-3)); label("6",(-1.6,-3)); [/asy]
$\textbf{Case 1:}$ The colors of $1,2,3$ are $BAB$ , in that order.
Then the colors of $6,5,4$ can be $AXA$ $AXC$ $CXA$ $CXC$ , or $CXD$ in that order, where $X$ is any color not equal to its surroundings. Then there are $4$ choices for $A$ $3$ choices for $B$ (it cannot be $A$ ), $2$ choices for $C$ , and $1$ for $D$ , the last color. So, summing up, we have \[4*3(1*3+2*2+2*2+2*3+2*1*2)=12*21=252\] colorings.
$\textbf{Case 2:}$ The colors of $1,2,3$ are $BAC$ , in that order.
Again, we list out the possible arrangements of $6,5,4$ $AXA$ $AXB$ $CXA$ $AXD$ $DXA$ $CXB$ $CXD$ $DXB$ , or $DXD$ . (Easily simplified; listed here for clarity.) Then there are $4$ choices for $A$ as usual, $3$ choices for $B$ , and so on. Hence we have \[4*3*2(1*3+1*2+1*2+1*2+1*2+1*2+1*2+1*2+1*3)=24*20=480\] colorings in this case.
Adding up, we have $252+480=\boxed{732}$ as our answer.
| 732
|
5,845
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12
| 11
|
The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy]
|
Let's label the regions as $1,2,3,4,5,6$ in that order. We start with region $1$ . There are no restrictions on the color of region $1$ so it can be any of the four colors. We know move on the region $2$ . It can be any color but color used for region $1$ , giving us $3$ choices. Section $3$ is where it gets a bit complicated; we will have to do casework based on whether the color of region $3$ is that of region $1$ or not.
If we have the color of region $3$ being different from that of region $1$ (in which we color do so in $2$ ways), then we have need for another casework at region $4$ . If the color of region $4$ is different from that of region $1$ (which can be achieved in $2$ ways), then we have yet another casework split.
If the color of region $5$ is different from that of region $1$ (which can be done in $2$ ways, then we would have a total of $2$ possible colorings for region $6$ (for it cannot be the color of regions $1$ nor $5$ ). Moving on to the case where the color of section $5$ is the same as that of section $1$ (which can be done in $1$ way), we will have $3$ ways (region $6$ cannot be that color of both region $1$ and $5$ ). Thus, if the color of region $4$ is different of that of region $1$ , then we have $2\cdot 2 + 3 = 7$ ways.
If the color of region $4$ is the same as that of region $1$ (which can be done in one way), then the color of region $5$ have to be different from that of sector $1$ $3$ ways). That means there will be $2$ choices for the color of region $6$ . So if the color of region $4$ is the same as region $1$ , then we will have $3\cdot 2 = 6$ . That means if the color of section $3$ is different from that of region $1$ , then there are $2\cdot 7 + 6 = 20$
Now, moving on to the case where section $3$ has the same color of region $1$ . That means section $4$ will have to be a different color of that of region $1$ $3$ ways). So, that means we have region $5$ to be either the same color or different color as region $1$ . If it is different (which can be done in $2$ ways), then there will be $2$ possibilities on the color of sector $6$ . If it is same (which can be done in $1$ way), then there are $3$ ways to color region $6$ . So, if section $3$ has the same color as section $1$ , then we have $3(2\cdot 2 + 3) = 3(7) = 21$
Now, in overall we will have $4\cdot 3 (2(20)+21) = 12(61) = \boxed{732}$
| 732
|
5,846
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12
| 12
|
The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy]
|
Let the top-right segment be segment $1,$ and remaining segments are numbered $2,3,4,5,6$ in clockwise order.
We have $4$ choices for segment $1,$ and $3$ choices for segments $2,3,4,5.$ For segment $6,$ we wish to find the expected value of the number of choices for segment $6$ 's color, which depends on whether segment $5$ is red. If we let $P(n)$ denote the probability of segment $P$ being the same color as segment $1$ (for simplicity, denote segment $1$ 's color as red), we get the following recurrence relation: \[P(n)=\dfrac{1-P(n-1)}{3}.\] This is because that you cannot have two reds in a row (hence the $1-P(n-1)$ ) and if segment $n-1$ is not red, there are three possible colors, one of which is red (hence the divide by $3$ ). Using the obvious fact that $P(1)=1$ by the definition of $P(n),$ we find that \[P(5)=\dfrac{7}{27}.\] If segment $5$ is red, then there are three possible colors for segment $6.$ If it is not, there are $2$ possible choices for segment $6.$ This means the expected number of color choices for segment $6$ is \[\dfrac{7}{27}\cdot3+\dfrac{20}{27}\cdot2,\] and the total number of colorings of the ring is \[4\cdot3\cdot3\cdot3\cdot3\cdot\left(\dfrac{7}{27}\cdot3+\dfrac{20}{27}\cdot2\right)=\boxed{732}.\]
| 732
|
5,847
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_13
| 1
|
Beatrix is going to place six rooks on a $6 \times 6$ chessboard where both the rows and columns are labeled $1$ to $6$ ; the rooks are placed so that no two rooks are in the same row or the same column. The $value$ of a square is the sum of its row number and column number. The $score$ of an arrangement of rooks is the least value of any occupied square.The average score over all valid configurations is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$
|
We casework to find the number of ways to get each possible score. Note that the lowest possible score is $2$ and the highest possible score is $7$ . Let the bijective function $f(x)=\{1,2,3,4,5,6\} \to \{1,2,3,4,5,6\}$ denote the row number of the rook for the corresponding column number.
Thus, the expected sum is $\dfrac{120 \cdot 2 + 216 \cdot 3 + 222 \cdot 4 + 130 \cdot 5 + 31 \cdot 6 + 1 \cdot 7}{720}= \dfrac{2619}{720}=\dfrac{291}{80}$ , so the desired answer is $291+80=\boxed{371}$
| 371
|
5,848
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_13
| 2
|
Beatrix is going to place six rooks on a $6 \times 6$ chessboard where both the rows and columns are labeled $1$ to $6$ ; the rooks are placed so that no two rooks are in the same row or the same column. The $value$ of a square is the sum of its row number and column number. The $score$ of an arrangement of rooks is the least value of any occupied square.The average score over all valid configurations is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$
|
If the score is $n+1$ , then one of the rooks must appear in the $n$ th antidiagonal, and this is the first antidiagonal in which a rook can appear. To demonstrate this, we draw the following diagram when $n=4$
[asy] for (int i=0;i<7;++i) {draw((0,10*i)--(60,10*i));draw((10*i,0)--(10*i,60));} path x=(1,1)--(9,9),y=(1,9)--(9,1); for (int i=0;i<3;++i) {for (int j=3+i;j<6;++j) {draw(shift(10*i,10*j)*x,linewidth(1));draw(shift(10*i,10*j)*y,linewidth(1));}} for (int i=0;i<4;++i) {filldraw((10*i,20+10*i)--(10*i+10,20+10*i)--(10*i+10,20+10*i+10)--(10*i,20+10*i+10)--cycle,lightgray);} [/asy] We first count the number of arrangements that avoid the squares above the $n$ th diagonal, and then we subtract from these the number of arrangements that avoid all squares above the $(n+1)$ th diagonal. In the first column, there are $7-n$ rows in which to place the rook. In the second column, there is one more possible row, but one of the rows is used up by the rook in the first column, hence there are still $7-n$ places to place the rook. This pattern continues through the $n$ th column, so there are $(7-n)^n$ ways to place the first $n$ rooks while avoiding the crossed out squares. We can similarly compute that there are $(6-n)^n$ ways to place the rooks in the first $n$ columns that avoid both the crossed out and shaded squares. Therefore, there are $(7-n)^n-(6-n)^n$ ways to place the first $n$ rooks such that at least one of them appears in a shaded square.
After this, there are $(6-n)$ rows and $(6-n)$ columns in which to place the remaining rooks, and we can do this in $(6-n)!$ ways. Hence the number of arrangements with a score of $n$ is $((7-n)^n-(6-n)^n)\cdot(6-n)!$ . We also know that $n$ can range from from $1$ to $6$ , so the average score is given by
\[\frac{2\cdot(6^1-5^1)\cdot5!+3\cdot(5^2-4^2)\cdot4!+4\cdot(4^3-3^3)\cdot3!+5\cdot(3^4-2^4)\cdot 2!+6\cdot(2^5-1^5)\cdot 1!+7\cdot(1^6-0^6)\cdot 0!}{6!}=\frac{291}{80}.\] Thus the answer is $\boxed{371}$
| 371
|
5,849
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_13
| 3
|
Beatrix is going to place six rooks on a $6 \times 6$ chessboard where both the rows and columns are labeled $1$ to $6$ ; the rooks are placed so that no two rooks are in the same row or the same column. The $value$ of a square is the sum of its row number and column number. The $score$ of an arrangement of rooks is the least value of any occupied square.The average score over all valid configurations is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$
|
So we first count the number of permutations with score $\ge 2$ . This is obviously $6!=720$ . Then, the number of permutations with score $\ge 3$ can also be computed: in the first column, there are five ways to place a rook- anywhere but the place with score $1$ . In the next column, there are $5$ ways to place a rook- anywhere but the one in the same row as the previous row. We can continue this to obtain that the number of permutations with score $\ge 3$ is $600$ . Doing the same for scores $\ge 4$ $\ge 5$ $\ge 6$ , and $\ge 7$ we obtain that these respective numbers are $384$ $162$ $32$ $1$
Now, note that if $a_k$ is the number of permutations with score $\ge k$ , then $a_k-a_{k-1}=b_{k}$ , where $b_k$ is the number of permutations with score exactly $k$ . Thus, we can compute the number of permutations with scores $2$ $3$ , etc as $120,216,222,130,31,1$ . We then compute \[\frac{120(2)+216(3)+222(4)+130(5)+31(6)+1(7)}{720}=\frac{291}{80}\] leading us to the answer of $291+80=\boxed{371}$ $\blacksquare$
| 371
|
5,850
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_13
| 4
|
Beatrix is going to place six rooks on a $6 \times 6$ chessboard where both the rows and columns are labeled $1$ to $6$ ; the rooks are placed so that no two rooks are in the same row or the same column. The $value$ of a square is the sum of its row number and column number. The $score$ of an arrangement of rooks is the least value of any occupied square.The average score over all valid configurations is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$
|
The problem is asking us to compute $\mathbb{E}[S]$ , where $S$ is the random variable that takes an arrangement of rooks and outputs its score, which is a non-negative integer quantity. For any random variable $S$ with non-negative integer values, we have the tail sum formula \[\mathbb{E}[S] = \sum_{n = 1}^{\infty}\mathbb{P}(S\geq n).\] These probabilities can be computed as in Solution 3, giving us the following table.
Hence \begin{align*} \mathbb{E}[S] &= \frac{720 + 720 + 600 + 384 + 162 + 32 + 1}{720} \\ &= 2 + \frac{1179}{720} = 2 + \frac{131}{80} = \frac{291}{80}, \end{align*} and the final answer is $291 + 80 = \boxed{371}$ as above.
| 371
|
5,851
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14
| 1
|
Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$ . Find $d$
|
The inradius of $\triangle ABC$ is $100\sqrt 3$ and the circumradius is $200 \sqrt 3$ . Now, consider the line perpendicular to plane $ABC$ through the circumcenter of $\triangle ABC$ . Note that $P,Q,O$ must lie on that line to be equidistant from each of the triangle's vertices. Also, note that since $P, Q, O$ are collinear, and $OP=OQ$ , we must have $O$ is the midpoint of $PQ$ . Now, Let $K$ be the circumcenter of $\triangle ABC$ , and $L$ be the foot of the altitude from $A$ to $BC$ . We must have $\tan(\angle KLP+ \angle QLK)= \tan(120^{\circ})$ . Setting $KP=x$ and $KQ=y$ , assuming WLOG $x>y$ , we must have $\tan(120^{\circ})=-\sqrt{3}=\dfrac{\dfrac{x+y}{100 \sqrt{3}}}{\dfrac{30000-xy}{30000}}$ . Therefore, we must have $100(x+y)=xy-30000$ . Also, we must have $\left(\dfrac{x+y}{2}\right)^{2}=\left(\dfrac{x-y}{2}\right)^{2}+120000$ by the Pythagorean theorem, so we have $xy=120000$ , so substituting into the other equation we have $90000=100(x+y)$ , or $x+y=900$ . Since we want $\dfrac{x+y}{2}$ , the desired answer is $\boxed{450}$
| 450
|
5,852
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14
| 2
|
Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$ . Find $d$
|
Draw a good diagram. Draw $CH$ as an altitude of the triangle. Scale everything down by a factor of $100\sqrt{3}$ , so that $AB=2\sqrt{3}$ . Finally, call the center of the triangle U. Draw a cross-section of the triangle via line $CH$ , which of course includes $P, Q$ . From there, we can call $OU=h$ . There are two crucial equations we can thus generate. WLOG set $PU<QU$ , then we call $PU=d-h, QU=d+h$ . First equation: using the Pythagorean Theorem on $\triangle UOB$ $h^2+2^2=d^2$ . Next, using the tangent addition formula on angles $\angle PHU, \angle UHQ$ we see that after simplifying $-d^2+h^2=-4, 2d=3\sqrt{3}$ in the numerator, so $d=\frac{3\sqrt{3}}{2}$ . Multiply back the scalar and you get $\boxed{450}$ . Not that hard, was it?
| 450
|
5,853
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14
| 4
|
Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$ . Find $d$
|
We use the diagram from solution 3. From basic angle chasing, \[180=\angle{QOC}+\angle{CO}P=2\angle{OCP}+2\angle{OCQ}=2\angle{QCP}\] so triangle QCP is a right triangle. This means that triangles $CQI$ and $CPI$ are similar. If we let $\angle{IDQ}=x$ and $\angle{PDI}=y$ , then we know $x+y=120$ and \[\frac{PG}{GC}=\frac{GC}{GQ}\Rightarrow\frac{100\sqrt{3}\tan{y}}{200\sqrt{3}}=\frac{200\sqrt{3}}{100\sqrt{3}\tan{x}}\Rightarrow\tan{x}\tan{y}=4\] We also know that \[PQ=2d=100\sqrt{3}(\tan{x}+\tan{y})\] \[d=50\sqrt{3}(\tan{x}+\tan{y})\] \[\frac{d}{1-\tan{x}\tan{y}}=50\sqrt{3}\cdot\frac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}\] \[\frac{d}{-3}=50\sqrt{3}\tan{(x+y)}\] \[d=-150\sqrt{3}\tan{120}=-150\sqrt{3}(-\sqrt{3})=\boxed{450}\]
| 450
|
5,854
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14
| 5
|
Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$ . Find $d$
|
We use the diagram from solution 3.
Let $BP = a$ and $BQ = b$ . Then, by Stewart's on $BPQ$ , we find \[2x^3 + 2x^3 = a^2x + b^2x \implies a^2 + b^2 = 4x^2.\]
The altitude from $P$ to $ABC$ is $\sqrt{a^2 - (200\sqrt{3})^2}$ so \[PQ = 2x = \sqrt{a^2 - (200\sqrt{3})^2} + \sqrt{b^2 - (200\sqrt{3})^2}.\]
Furthermore, the altitude from $P$ to $AB$ is $\sqrt{a^2 - 300^2}$ , so, by LoC and the dihedral condition, \[a^2 - 300^2 + b^2 - 300^2 + \sqrt{a^2 - 300^2}\sqrt{b^2-300^2} = 4x^2.\]
Squaring the equation for $PQ$ and substituting $a^2 + b^2 = 4x^2$ yields \[2\sqrt{a^2 - (200\sqrt{3})^2}\sqrt{b^2 - (200\sqrt{3})^2} = 6\cdot 200^2.\]
Substituting $a^2 + b^2 = 4x^2$ into the other equation, \[\sqrt{a^2 - 300^2}\sqrt{b^2-300^2} = 2\cdot 300^2.\]
Squaring both of these gives \[a^2b^2-3\cdot 200^2(a^2 + b^2) + 9\cdot 200^4 = 9\cdot 200^4\] \[a^2b^2 - 300^2(a^2+b^2) + 300^4 = 4\cdot 300^4.\]
Substituting $a^2 + b^2 = 4x^2$ and solving for $x$ gives $\boxed{450}$ , as desired.
| 450
|
5,855
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14
| 6
|
Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$ . Find $d$
|
Let $AB = a, M$ be midpoint $BC, I$ be the center of equilateral $\triangle ABC,$ $IM = b = \frac {a}{2\sqrt{3}}, O$ be the center of sphere $ABCPQ.$ Then \[AI = 2b, AO = BO = PO =QO = d.\] \[QA=QB=QC,PA=PB=PC \implies\] \[POIQ\perp ABC, \angle PMQ = 120^\circ.\] (See upper diagram).
We construct the circle PQMD, use the formulas for intersecting chords and get \[DI = 5b, FI = EO = \frac{3b}{2}\] \[\implies FM = \frac{5b}{2}.\] (See lower diagram).
We apply the Law of Sine to $\triangle PMQ$ and get \[2EM \sin 120^\circ =PQ\] \[\implies r \sqrt{3} = 2d\] \[\implies 3r^2 = 4d^2.\] We apply the Pythagorean Law on $\triangle AOI$ and $\triangle EFM$ and get \[d^2 = 4b^2 + OI^2, r^2 = \frac {25b^2}{4} + EF^2 \implies\] \[r = 3b\implies d = \frac {3a}{2} = \boxed{450}.\] vladimir.shelomovskii@gmail.com, vvsss
| 450
|
5,856
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14
| 7
|
Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$ . Find $d$
|
Let $M$ be the midpoint of $\overline{AB}$ and $X$ the center of $\triangle ABC$ . Then \[P, O, Q, M, X, C\] all lie in the same vertical plane. We can make the following observations:
To make calculations easier, we will denote $100\sqrt{3}=m$ , so that $MX=m$ and $XC=2m$
[asy] unitsize(20); pair P = (0, 12); pair Q = (0, -3); pair O = (P+Q)/2; pair M = (-3, 0); pair X = (0, 0); pair C = (6, 0); draw(P--O--Q); draw(M--X--C); draw(P--M--Q, blue); draw(Q--C--P); draw(circle((0, 4.5), 7.5)); label("$P$", P, N); label("$Q$", Q, S); label("$O$", O, E); dot(O); label("$M$", M, W); label("$X$", X, NE); label("$C$", C, E); label("$m$", (M+X)/2, N); label("$2m$", (X+C)/2, N); [/asy]
Denote $PX=p$ and $QX=q$ , where the tangent addition formula on $\triangle PMQ$ yields \[\frac{\tan\measuredangle PMX+\tan\measuredangle QMX}{1-\tan\measuredangle PMX\tan\measuredangle QMX}=\tan(120^{\circ})=-\sqrt{3}.\] Using $\tan\measuredangle PMX=\frac{p}{m}$ and $\tan\measuredangle QMX=\frac{q}{m}$ , we have \[\frac{\frac{p}{m}+\frac{q}{m}}{1-\frac{p}{m}\cdot\frac{q}{m}}=-\sqrt{3}.\] After multiplying both numerator and denominator by $m^{2}$ we have \[\frac{(p+q)m}{m^{2}-pq}=-\sqrt{3}.\] But note that $pq=(2m)(2m)=4m^{2}$ by power of a point at $X$ , where we deduce by symmetry that $MM^{\prime}=MX=m$ on the diagram below: [asy] unitsize(20); pair P = (0, 12); pair Q = (0, -3); pair O = (P+Q)/2; pair M = (-3, 0); pair Mprime = (-6, 0); pair X = (0, 0); pair C = (6, 0); draw(P--O--Q); draw(Mprime--M--X--C); draw(P--M--Q, blue); draw(Q--C--P); draw(circle((0, 4.5), 7.5)); label("$P$", P, N); label("$Q$", Q, S); label("$O$", O, E); dot(O); label("$M$", M, S); label("$M^{\prime}$", Mprime, W); label("$X$", X, SE); label("$C$", C, E); label("$m$", (Mprime+M)/2, N); label("$m$", (M+X)/2, N); label("$2m$", (X+C)/2, N); [/asy]
Thus \begin{align*} \frac{(p+q)m}{m^{2}-4m^{2}}=-\sqrt{3} \\ \frac{p+q}{-3m}=-\sqrt{3} \\ p+q=\left(-\sqrt{3}\right)\left(-3m\right) \\ p+q=3\sqrt{3}\cdot m.\end{align*} Earlier we assigned the variable $m$ to the length $100\sqrt{3}$ which implies $PQ=\left(3\sqrt{3}\right)\left(100\sqrt{3}\right)=900$ . Thus the distance $d$ is equal to $\frac{PQ}{2}=\boxed{450}$
| 450
|
5,857
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14
| 8
|
Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$ . Find $d$
|
Let $Z$ be the center of $\triangle ABC$ . Let $A’$ be the midpoint of $BC$ . Let $ZA’ = c = 100\sqrt{3}$ and $ZA = 2c = 200\sqrt{3}$ . Let $PZ = a$ and $QZ = b$ . We will be working in the plane that contains the points: $A$ $P$ $A’$ $Q$ $O$ , and $Z$
Since $P$ $O$ , and $Q$ are collinear and $PO = QO = AO$ $\triangle PAQ$ is a right triangle with $\angle PAQ = 90^{\circ}$ . Since $AZ \perp PQ$ $(PZ)(QZ) = (AZ)^2 = ab = (2c)^2 = 120000$
$PA’ = \sqrt{a^2 + c^2}$ $QA’ = \sqrt{b^2 + c^2}$ $PQ = a + b$ , and $\angle PAQ = 120^{\circ}$ . By Law of Cosines \[(a + b)^2 = a^2 + b^2 + 2c^2 + \sqrt{a^2b^2 + a^2c^2 + b^2c^2 + c^4}\] . Substituting $4c^2$ for $ab$ and simplifying, we get \[6c = \sqrt{17c^2 + a^2 + b^2}\] . Squaring and simplifying, we get \[a^2 + b^2 = 19c^2 = 570000\] . Adding $2ab = 8c^2$ to both sides we get $PQ = a + b = 900$ . Since $O$ is the midpoint of $PQ$ $d = PO = \boxed{450}$
| 450
|
5,858
|
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_15
| 1
|
For $1 \leq i \leq 215$ let $a_i = \dfrac{1}{2^{i}}$ and $a_{216} = \dfrac{1}{2^{215}}$ . Let $x_1, x_2, ..., x_{216}$ be positive real numbers such that $\sum_{i=1}^{216} x_i=1$ and $\sum_{1 \leq i < j \leq 216} x_ix_j = \dfrac{107}{215} + \sum_{i=1}^{216} \dfrac{a_i x_i^{2}}{2(1-a_i)}$ . The maximum possible value of $x_2=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
Note that \begin{align*}\sum_{1 \leq i < j \leq 216} x_ix_j &= \frac12\left(\left(\sum_{i=1}^{216} x_i\right)^2-\sum_{i=1}^{216} x_i^2\right)\\&=\frac12\left(1-\sum x_i^2\right).\end{align*} Substituting this into the second equation and collecting $x_i^2$ terms, we find \[\sum_{i=1}^{216}\frac{x_i^2}{1-a_i}=\frac{1}{215}.\] Conveniently, $\sum_{i=1}^{216} 1-a_i=215$ so we find \[\left(\sum_{i=1}^{216} 1-a_i\right)\left(\sum_{i=1}^{216}\frac{x_i^2}{1-a_i}\right)=1=\left(\sum_{i=1}^{216} x_i\right)^2.\] This is the equality case of the Cauchy-Schwarz Inequality, so $x_i=c(1-a_i)$ for some constant $c$ . Summing these equations and using the facts that $\sum_{i=1}^{216} x_i=1$ and $\sum_{i=1}^{216} 1-a_i=215$ , we find $c=\frac{1}{215}$ and thus $x_2=c(1-a_2)=\frac{1}{215}\cdot \left(1-\frac{1}{4}\right)=\frac{3}{860}$ . Hence the desired answer is $860+3=\boxed{863}$
| 863
|
5,859
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_4
| 1
|
Point $B$ lies on line segment $\overline{AC}$ with $AB=16$ and $BC=4$ . Points $D$ and $E$ lie on the same side of line $AC$ forming equilateral triangles $\triangle ABD$ and $\triangle BCE$ . Let $M$ be the midpoint of $\overline{AE}$ , and $N$ be the midpoint of $\overline{CD}$ . The area of $\triangle BMN$ is $x$ . Find $x^2$
|
Let $A$ be the origin, so $B=(16,0)$ and $C=(20,0).$ Using equilateral triangle properties tells us that $D=(8,8\sqrt3)$ and $E=(18,2\sqrt3)$ as well. Therefore, $M=(9,\sqrt3)$ and $N=(14,4\sqrt3).$ Applying the Shoelace Theorem to triangle $BMN$ gives
\[x=\dfrac 1 2 |16\sqrt3+36\sqrt3+0-(0+14\sqrt3+64\sqrt3)| =13\sqrt3,\]
so $x^2=\boxed{507}.$
| 507
|
5,860
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_4
| 2
|
Point $B$ lies on line segment $\overline{AC}$ with $AB=16$ and $BC=4$ . Points $D$ and $E$ lie on the same side of line $AC$ forming equilateral triangles $\triangle ABD$ and $\triangle BCE$ . Let $M$ be the midpoint of $\overline{AE}$ , and $N$ be the midpoint of $\overline{CD}$ . The area of $\triangle BMN$ is $x$ . Find $x^2$
|
Note that $AB=DB=16$ and $BE=BC=4$ . Also, $\angle ABE = \angle DBC = 120^{\circ}$ . Thus, $\triangle ABE \cong \triangle DBC$ by SAS.
From this, it is clear that a $60^{\circ}$ rotation about $B$ will map $\triangle ABE$ to $\triangle DBC$ .
This rotation also maps $M$ to $N$ . Thus, $BM=BN$ and $\angle MBN=60^{\circ}$ . Thus, $\triangle BMN$ is equilateral.
Using the Law of Cosines on $\triangle ABE$ \[AE^2 = 16^2 + 4^2 - 2\cdot 16\cdot 4\cdot\left(-\frac{1}{2}\right)\] \[AE = 4\sqrt{21}\] Thus, $AM=ME=2\sqrt{21}$
Using Stewart's Theorem on $\triangle ABE$ \[AM\cdot ME\cdot AE + AE\cdot BM^2 = BE^2\cdot AM + BA^2\cdot ME\] \[BM = 2\sqrt{13}\]
Calculating the area of $\triangle BMN$ \[[BMN] = \frac{\sqrt{3}}{4} BM^2\] \[[BMN] = 13\sqrt{3}\] Thus, $x=13\sqrt{3}$ , so $x^2 = 507$ . Our final answer is $\boxed{507}$
| 507
|
5,861
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_4
| 3
|
Point $B$ lies on line segment $\overline{AC}$ with $AB=16$ and $BC=4$ . Points $D$ and $E$ lie on the same side of line $AC$ forming equilateral triangles $\triangle ABD$ and $\triangle BCE$ . Let $M$ be the midpoint of $\overline{AE}$ , and $N$ be the midpoint of $\overline{CD}$ . The area of $\triangle BMN$ is $x$ . Find $x^2$
|
$AB = BD, BE = BC, \angle ABE = \angle CBD \implies \triangle ABE \cong \triangle DBC$
Medians are equal, so $MB = MN, \angle ABM = \angle DBN \implies$ $\angle MBN = \angle ABD - \angle ABM + \angle DBN = 60^\circ \implies$
$\triangle MNB$ is equilateral triangle.
The height of $\triangle BCE$ is $2 \sqrt{3},$ distance from $A$ to midpoint $BC$ is $16 + 2 = 18 \implies \frac {AE^2}{4} =\frac{ (16 + 2)^2 +2^2 \cdot 3}{4} = 81 + 3 = 84.$
$BM$ is the median of $\triangle ABE \implies$ $BM^2 = \frac {AB^2}{2} + \frac {BE^2}{2} - \frac {AE^2}{4}=16 \cdot 8 + 4 \cdot 2 - 84 = 52.$
The area of $\triangle BMN$
\[[BMN] = \frac{\sqrt{3}}{4} BM^2 =13 \sqrt{3} \implies \boxed{507}.\]
| 507
|
5,862
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_5
| 1
|
In a drawer Sandy has $5$ pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the $10$ socks in the drawer. On Tuesday Sandy selects $2$ of the remaining $8$ socks at random and on Wednesday two of the remaining $6$ socks at random. The probability that Wednesday is the first day Sandy selects matching socks is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers, Find $m+n$
|
Let the fifth sock be arbitrary; the probability that the sixth sock matches in color is $\dfrac{1}{9}$
Assuming this, then let the first sock be arbitrary; the probability that the second sock does not match is $\dfrac{6}{7}.$
The only "hard" part is the third and fourth sock. But that is simple casework. If the third sock's color matches the color of one of the first two socks (which occurs with probability $\dfrac{2}{6} = \dfrac{1}{3}$ ), then the fourth sock can be arbitrary. Otherwise (with probability $\dfrac{2}{3}$ ), the fourth sock can be chosen with probability $\dfrac{4}{5}$ (5 socks left, 1 sock that can possibly match the third sock's color). The desired probability is thus \[\frac{1}{9} \cdot \frac{6}{7} \cdot \left(\dfrac{1}{3} + \dfrac{2}{3} \cdot \dfrac{4}{5}\right) = \frac{26}{315}.\] The sum is therefore $26+315=\boxed{341}.$
| 341
|
5,863
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_5
| 2
|
In a drawer Sandy has $5$ pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the $10$ socks in the drawer. On Tuesday Sandy selects $2$ of the remaining $8$ socks at random and on Wednesday two of the remaining $6$ socks at random. The probability that Wednesday is the first day Sandy selects matching socks is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers, Find $m+n$
|
The key is to count backwards. First, choose the pair which you pick on Wednesday in $5$ ways. Then there are four pairs of socks for you to pick a pair of on Tuesday, and you don't want to pick a pair. Since there are $4$ pairs, the number of ways to do this is $\dbinom{8}{2}-4$ . Then, there are two pairs and two nonmatching socks for you to pick from on Monday, a total of $6$ socks. Since you don't want to pick a pair, the number of ways to do this is $\dbinom{6}{2}-2$ . Thus the answer is \[\dfrac{\left(5\right)\left(\dbinom{8}{2}-4\right)\left(\dbinom{6}{2}-2\right)}{\dbinom{10}{2}\dbinom{8}{2}\dbinom{6}{2}}=\dfrac{26}{315}.\] $26 + 315 = \boxed{341}$
| 341
|
5,864
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_5
| 3
|
In a drawer Sandy has $5$ pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the $10$ socks in the drawer. On Tuesday Sandy selects $2$ of the remaining $8$ socks at random and on Wednesday two of the remaining $6$ socks at random. The probability that Wednesday is the first day Sandy selects matching socks is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers, Find $m+n$
|
For the first sock, note that to pick two different socks, can complementary count to get the total, $\binom{10}{2}$ minus the number of pairs (5) to get \[\frac{\binom{10}{2} - 5}{\binom{10}{2}}\] The next steps aren't quite as simple, though.
WLOG suppose the socks are (a, a, b, b, c, c, d, d, e, e) and that we chose ab on the first day.
This leaves abccddee as the remaining socks. We can think of our next choice as how many "pairs" we "break".
Note that right now, in abccddee there are 3 pairs, namely {cc, dd, ee}. Our next choice could either leave all of these pairs (if we choose ab), leave 2 of these pairs, (choose say ac to keep dd and ee as pairs), or leave only 1 as a pair, say choosing de to get abccde. \[\bold{\text{Case 1: Break zero pairs}}.\] In this case, we want to keep 3 pairs remaining, so our only choice is to choose AB, which would make the remaining ccddee. The probability that we choose AB is $\frac{1}{\binom{8}{2}}$ and the probability that we get a pair after this is $\frac{3}{\binom{6}{2}}$ since we just choose CC, DD, or EE. This case has a probability $\frac{1}{\binom{8}{2}}\frac{3}{\binom{6}{2}}.$ \[\bold{\text{Case 2: Break one pair}}.\] In this case, we want to choose A and one of {C,C,D,D,E,E} or B and one of {C,C,D,D,E,E}. This yields $\frac{6 + 6}{\binom{8}{2}}$ probability. After this, we end up with 2 pairs, so the total probability of choosing a pair on Wednesday in this case is $\frac{6 + 6}{\binom{8}{2}}\frac{2}{\binom{6}{2}}.$ \[\bold{\text{case 3: Break 2 pairs}}.\] In this case we'd choose two distinct characters from {C,C,D,D,E,E} which would be $\frac{\binom{6}{2} - 3}{\binom{6}{2}}.$ After this, only one pair remains so the total probability for this case is
$\frac{\binom{6}{2} - 3}{\binom{6}{2}} \frac{6}{\binom{8}{2}}\frac{1}{\binom{6}{2}}.$ Adding all the cases and multiplying by $\frac{\binom{10}{2} - 5}{\binom{10}{2}}$ yields \[\frac{\binom{10}{2} - 5}{\binom{10}{2}}\left(\frac{1}{\binom{8}{2}}\frac{3}{\binom{6}{2}}+\frac{6 + 6}{\binom{8}{2}}\frac{2}{\binom{6}{2}}+ \frac{\binom{6}{2} - 3}{\binom{6}{2}}\frac{6}{\binom{8}{2}}\frac{1}{\binom{6}{2}}\right) = \frac{26}{315} \rightarrow 26 + 315 = \boxed{341}\]
| 341
|
5,865
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_7
| 1
|
In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$ . Points $F$ and $G$ lie on $\overline{CE}$ , and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$ , respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$ , and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively, so that $KLMN$ is a square. The area of $KLMN$ is 99. Find the area of $FGHJ$
[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); [/asy]
|
Let us find the proportion of the side length of $KLMN$ and $FJGH$ . Let the side length of $KLMN=y$ and the side length of $FJGH=x$
[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N,P; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); [/asy]
Now, examine $BC$ . We know $BC=BJ+JC$ , and triangles $\Delta BHJ$ and $\Delta JFC$ are similar to $\Delta EDC$ since they are $1-2-\sqrt{5}$ triangles. Thus, we can rewrite $BC$ in terms of the side length of $FJGH$ \[BJ=\frac{1}{\sqrt{5}}HJ=\frac{x}{\sqrt{5}}=\frac{x\sqrt{5}}{5}, JC=\frac{\sqrt{5}}{2}JF=\frac{x\sqrt{5}}{2}\Rightarrow BC=\frac{7x\sqrt{5}}{10}\]
Now examine $AB$ . We can express this length in terms of $x,y$ since $AB=AN+NH+HB$ . By using similar triangles as in the first part, we have \[AB=\frac{1}{\sqrt{5}}y+\frac{\sqrt{5}}{2}y+\frac{2}{\sqrt{5}}x\] \[AB=BC\Rightarrow \frac{7y\sqrt{5}}{10}+\frac{2x\sqrt{5}}{5}=\frac{7x\sqrt{5}}{10}\Rightarrow \frac{7y\sqrt{5}}{10}=\frac{3x\sqrt{5}}{10}\Rightarrow 7y=3x\]
Now, it is trivial to see that $[FJGH]=\left(\frac{x}{y}\right)^2[KLMN]=\left(\frac{7}{3}\right)^2\cdot 99=\boxed{539}.$
| 539
|
5,866
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_7
| 2
|
In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$ . Points $F$ and $G$ lie on $\overline{CE}$ , and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$ , respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$ , and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively, so that $KLMN$ is a square. The area of $KLMN$ is 99. Find the area of $FGHJ$
[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); [/asy]
|
[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N,P; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); P=foot(E,M,L); draw(P--E); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); label("$P$",P,dir(235)); [/asy]
We begin by denoting the length $ED$ $a$ , giving us $DC = 2a$ and $EC = a\sqrt5$ . Since angles $\angle DCE$ and $\angle FCJ$ are complementary, we have that $\triangle CDE \sim \triangle JFC$ (and similarly the rest of the triangles are $1-2-\sqrt5$ triangles). We let the sidelength of $FGHJ$ be $b$ , giving us:
\[JC = \sqrt5 \cdot FC = \sqrt5 \cdot FJ/2 = \frac{b\sqrt 5}{2}\] and \[BJ = \frac{1}{\sqrt5} \cdot HJ = \frac{b}{\sqrt5}\]
Since $BC = CJ + BJ$
\[2a = \frac{b\sqrt 5}{2} + \frac{b}{\sqrt5}\]
Solving for $b$ in terms of $a$ yields \[b = \frac{4a\sqrt5}{7}\]
We now use the given that $[KLMN] = 99$ , implying that $KL = LM = MN = NK = 3\sqrt{11}$ . We also draw the perpendicular from $E$ to $ML$ and label the point of intersection $P$ as in the diagram at the top
This gives that \[AM = 2 \cdot AN = 2 \cdot \frac{3\sqrt{11}}{\sqrt5}\] and \[ME = \sqrt5 \cdot MP = \sqrt5 \cdot \frac{EP}{2} = \sqrt5 \cdot \frac{LG}{2} = \sqrt5 \cdot \frac{HG - HK - KL}{2} = \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2}\]
Since $AE$ $AM + ME$ , we get
\[2 \cdot \frac{3\sqrt{11}}{\sqrt5} + \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2} = a\]
\[\Rightarrow 12\sqrt{11} + 5(\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}) = 2\sqrt5a\]
\[\Rightarrow \frac{-21}{2}\sqrt{11} + \frac{20a\sqrt5}{7} = 2\sqrt5a\]
\[\Rightarrow -21\sqrt{11} = 2\sqrt5a\frac{14 - 20}{7}\]
\[\Rightarrow \frac{49\sqrt{11}}{4} = \sqrt5a\]
\[\Rightarrow 7\sqrt{11} = \frac{4a\sqrt{5}}{7}\]
So our final answer is $(7\sqrt{11})^2 = \boxed{539}$
| 539
|
5,867
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_8
| 1
|
For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$
|
You know whatever $n$ is, it has to have 3 digits, because if it had only two, the maximum of $s(n)$ is 18.
Now let $n=100a_2+10a_1+a_0$
So first we know, $a_2+a_1+a_0=20$ . Okay now we have to split into cases based on which digit gets carried. This meaning, when you add a 3 digit number to 864, we have to know when to carry the digits. Note that if you don't understand any of the steps I take, just try adding any 3-digit number to 864 regularly (using the old-fashioned "put one number over the other" method, not mental calculation), and observe what you do at each step.
(1) $\textcolor{red}{*}$ None of the digits get carried over to the next space:
So this means $a_2<2, a_1<4$ and $a_0<6$ . So
$s(864+n)=(8+a_2)+(6+a_1)+(4+a_0)=38$ So it doesn't work. Now:
(2) $a_2+8$ is the only one that carries over
So this means $a_2>1, a_1<4$ and $a_0<6$ . So
$s(864+n)=1+(8+a_2-10)+(6+a_1)+(a_0+4)=29$
(3) $\textcolor{red}{*}$ $a_0+4$ is the only one that carries over.
So
$s(864+n)=(8+a_2)+(6+a_1+1)+(4+a_0-10)=29$
(4)The first and second digit carry over (but not the third)
$s(864+n)=1+(8+a_2-10+1)+(6+a_1-10)+(4+a_0)=20$
Aha! This case works but we still have to make sure it's possible for $a_2+a_1+a_0=20$ (We assumed this is true, so we have to find a number that works.) Since only the second and first digit carry over, $a_2>0, a_1>3$ and $a_0<6$ . The smallest value we can get with this is 695. Let's see if we can find a smaller one:
(5)The first and third digit carry over (but not the second)
$s(864+n)=1+(8+a_2-10)+(7+a_1)+(4+a_0-10)=20$
The largest value for the middle digit is 2, so the other digits have to be both 9's. So the smallest possible value is 929
(6) All the digits carry over
$s(864+n)=1+(9+a_2-10)+(7+a_1-10)+(4+a_0-10)=\text{Way less than 20}$
So the answer is $\boxed{695}$ which after a quick test, does indeed work.
| 695
|
5,868
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_8
| 2
|
For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$
|
First, it is easy to verify that $695$ works and that no other numbers beginning with the digit 6 work (i.e. $686, 677, 668, 659$ do not work).
Suppose by contradiction that there is a smaller valid $n$ , where the leading digit of the three-digit number $n$ is 5 or less. (Two-digit $n$ obviously do not work because 9 + 9 < 20.) Clearly $n > 200$ because the smallest three-digit number whose digits sum to 20 is $299$ . Also, because the second digit is at most 9, the units digit is at least 6, which means that the addition $N = n + 864$ regroups in the ones place. Then the units digit of $N$ is clearly less than 4. But as $1000 < 200 + 864 < N < 600 + 864 = 1464$ , the sum of the thousands digit and the hundredth digit is at most 5. Because the second digit is at most 9, the sum of the digits of $N$ is at most $5 + 9 + 4 < 20$ , contradiction. Hence $\boxed{695}$ is the answer.
| 695
|
5,869
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_8
| 3
|
For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$
|
First of all, notice that the smallest $n$ with $s(n) = 20$ is $299$ . Also, if $s(n + 864) = 20$ $s(n - 136) = 19$ (because subtracting $1000$ from the number removes the $1$ in the thousands place). After checking $s(n - 136)$ for various $n$ with $s(n) = 20$ , we see that we need to have a carry when subtracting $136$ . To have this, we must either have a $2$ in the tens place or a $5$ in the units place. The minimum $n$ for the former is $929$ , and for the latter it is $695$ . We check and see that $s(695-136) = s(559) = 19$ , so our answer is $\boxed{695}$
| 695
|
5,870
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_8
| 4
|
For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$
|
Observation (Lemma) : If r is the number of regroups in the addition of n+k, $S(n+k) = S(n)+S(k)-9r$
Proof : When you add two numbers, and you do a carry, you are taking away 10 from 1 column, and adding 1 to another column, giving a net loss of 9 to the total.
Thus, we can see that we need to regroup exactly twice when we add 864. And, the lowest possible n is 299, so let's start from there.
299 gets three regroups, so we are going to need to take away from digits, and dump the excess in the hundred's place, since the hundreds are going to regroup anyways.
So, if we take away from the tens digit, we need to take away until we get 2 in the tens digit(since the ones will regroup). So, we get the number 929, which works (929+864 = 1793), but is not the smallest.
If we take away from the ones digit, we only have to take away 4, turning the unit's place to 5. 5+4 is 9, so it won't regroup. Dump the ones into the hundred's place, and we get the number $\boxed{695}$
| 695
|
5,871
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_8
| 5
|
For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$
|
Although this solution doesn't directly solve the problem, it greatly hastens the bashing process.
Call the three digits a, b, and c.
When you add each of 8, 6, and 4 to a, b, and c the resultant will either get smaller or larger, depending on the original number.
e.g. If c is 7, then adding 4 will reduce the 7 to a 1, whilst leaving a one for b.
If c is 3, then adding 4 will simply add four to the total, and make the 3 a 7.
Each of 8, 6, and 4 all can reduce the original number by a certain amount and can increase the original number by a certain amount.
8 can reduce by 2 for all numbers greater than 1, 6 can reduce numbers by 4, and 4 can reduce numbers by 6.
Possibilities:
Also, realize that if the number is reduced, then a one will be carried to the following decimal place on the left, consequently reducing that amount they reduce. It's like a puzzle! Within no time you should find that if you add 4 to c, subtract 4 from b, subtract 1 from a, and leave a 1 in the thousands place, the total is equated to zero. This is optimal because most of the addition is kept to the left, where the effect to real value is less. (e.g. 299 is smaller than 992)
Now you have +1, -1, -4, +4 in the decimals, and a VERY fast trial and error gives $\boxed{695}.$
| 695
|
5,872
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_8
| 6
|
For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$
|
First, note that to compute $s(m+n)$ (for any positive integers $m$ and $n$ ), one can simply find the sum of $s(m)$ and $s(n)$ minus 9 times the number of times one regroups when adding $m$ to $n$ . One can see why this is by noticing that if one were to "forget" to regroup, and leave, say, a 10 in the ones' place, the sum of the digits would be 9 higher than if one did regroup. Anyway, one can see that the smallest 3-digit number (on AIME, all the answers are integers from 0 to 999) whose digits sum to 20 is 299. If we add this number to 864, we have to regroup 3 times, so $s(299+864)$ will be $9=9\cdot3-(8+6+4)$ smaller than $s(299)$ . We want this difference to be 0, so we need to find a way to only regroup two times.
We now notice that regrouping the hundreds is inevitable, so we must either prevent regrouping the ones or the tens. Preventing regrouping the tens would require moving many of the tens to the hundreds' place (the ones' place is already full), which is bad when we are trying to minimize the number, but preventing regrouping the ones requires moving fewer ones to the hundreds' place.
We find that to preventing regrouping the ones, the ones' place of our number must be at most 5 (a larger number would sum to ten when added to 4). Because we want to minimize the number of ones we move to the hundreds' place, we leave exactly 5 by moving four ones to the hundreds' place: $299\to\boxed{695}$
| 695
|
5,873
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_8
| 7
|
For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$
|
Bashing out modulo $9$ and getting lucky we get that if the $8$ and $6$ carry over when adding $n$ and $864$ , that $100a+10b+c \equiv 1+100(a-1)+10(b-4)+c+4 \pmod{9}$ such that $n=100a+10b+c$ and after maximizing $b$ and $c$ such that $c<6$ to not make the $4$ carry over to minimize $a$ we get that $\boxed{695}$ is our answer after confirming there are no lesser solutions.
| 695
|
5,874
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_9
| 1
|
Let $S$ be the set of all ordered triple of integers $(a_1,a_2,a_3)$ with $1 \le a_1,a_2,a_3 \le 10$ . Each ordered triple in $S$ generates a sequence according to the rule $a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |$ for all $n\ge 4$ . Find the number of such sequences for which $a_n=0$ for some $n$
|
Let $a_1=x, a_2=y, a_3=z$ . First note that if any absolute value equals 0, then $a_n=0$ .
Also note that if at any position, $a_n=a_{n-1}$ , then $a_{n+2}=0$ .
Then, if any absolute value equals 1, then $a_n=0$ .
Therefore, if either $|y-x|$ or $|z-y|$ is less than or equal to 1, then that ordered triple meets the criteria.
Assume that to be the only way the criteria is met.
To prove, let $|y-x|>1$ , and $|z-y|>1$ . Then, $a_4 \ge 2z$ $a_5 \ge 4z$ , and $a_6 \ge 4z$ .
However, since the minimum values of $a_5$ and $a_6$ are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be $z=1$ $|y-x|=2$ . Again assume that any other scenario will not meet criteria.
To prove, divide the other scenarios into two cases: $z>1$ $|y-x|>1$ , and $|z-y|>1$ ; and $z=1$ $|y-x|>2$ , and $|z-y|>1$ .
For the first one, $a_4 \ge 2z$ $a_5 \ge 4z$ $a_6 \ge 8z$ , and $a_7 \ge 16z$ , by which point we see that this function diverges.
For the second one, $a_4 \ge 3$ $a_5 \ge 6$ $a_6 \ge 18$ , and $a_7 \ge 54$ , by which point we see that this function diverges.
Therefore, the only scenarios where $a_n=0$ is when any of the following are met: $|y-x|<2$ (280 options) $|z-y|<2$ (280 options, 80 of which coincide with option 1) $z=1$ $|y-x|=2$ . (16 options, 2 of which coincide with either option 1 or option 2)
Adding the total number of such ordered triples yields $280+280-80+16-2=\boxed{494}$
| 494
|
5,875
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_11
| 1
|
Triangle $ABC$ has positive integer side lengths with $AB=AC$ . Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$ . Suppose $BI=8$ . Find the smallest possible perimeter of $\triangle ABC$
|
Let $D$ be the midpoint of $\overline{BC}$ . Then by SAS Congruence, $\triangle ABD \cong \triangle ACD$ , so $\angle ADB = \angle ADC = 90^o$
Now let $BD=y$ $AB=x$ , and $\angle IBD = \dfrac{\angle ABD}{2} = \theta$
Then $\mathrm{cos}{(\theta)} = \dfrac{y}{8}$
and $\mathrm{cos}{(2\theta)} = \dfrac{y}{x} = 2\mathrm{cos^2}{(\theta)} - 1 = \dfrac{y^2-32}{32}$
Cross-multiplying yields $32y = x(y^2-32)$
Since $x,y>0$ $y^2-32$ must be positive, so $y > 5.5$
Additionally, since $\triangle IBD$ has hypotenuse $\overline{IB}$ of length $8$ $BD=y < 8$
Therefore, given that $BC=2y$ is an integer, the only possible values for $y$ are $6$ $6.5$ $7$ , and $7.5$
However, only one of these values, $y=6$ , yields an integral value for $AB=x$ , so we conclude that $y=6$ and $x=\dfrac{32(6)}{(6)^2-32}=48$
Thus the perimeter of $\triangle ABC$ must be $2(x+y) = \boxed{108}$
| 108
|
5,876
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_11
| 3
|
Triangle $ABC$ has positive integer side lengths with $AB=AC$ . Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$ . Suppose $BI=8$ . Find the smallest possible perimeter of $\triangle ABC$
|
Let $M$ be midpoint $BC, BM = x, AB = y, \angle IBM = \alpha.$
$BI$ is the bisector of $\angle ABM$ in $\triangle ABM.$ $BI = \frac {2 xy \cos \alpha}{x+y} = 8, \cos \alpha = \frac {x}{8} \implies \frac {x^2 y}{x+y} = 32.$ \[y = \frac {32 x} {x^2 - 32}.\] $BC = 2x$ is integer, $5.5^2 < 32 \implies x \ge 6.$ $BM < BI \implies x =\{ 6, 6.5, 7, 7.5 \}.$
If $x > 6$ then $y$ is not integer. \[x = 6 \implies y = 48 \implies 2(x+y) = \boxed{108}.\] vladimir.shelomovskii@gmail.com, vvsss
| 108
|
5,877
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_12
| 1
|
Consider all 1000-element subsets of the set $\{1, 2, 3, ... , 2015\}$ . From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$
|
Let $M$ be the desired mean. Then because $\dbinom{2015}{1000}$ subsets have 1000 elements and $\dbinom{2015 - i}{999}$ have $i$ as their least element, \begin{align*} \binom{2015}{1000} M &= 1 \cdot \binom{2014}{999} + 2 \cdot \binom{2013}{999} + \dots + 1016 \cdot \binom{999}{999} \\ &= \binom{2014}{999} + \binom{2013}{999} + \dots + \binom{999}{999} \\ & + \binom{2013}{999} + \binom{2012}{999} + \dots + \binom{999}{999} \\ & \dots \\ & + \binom{999}{999} \\ &= \binom{2015}{1000} + \binom{2014}{1000} + \dots + \binom{1000}{1000} \\ &= \binom{2016}{1001}. \end{align*} Using the definition of binomial coefficient and the identity $n! = n \cdot (n-1)!$ , we deduce that \[M = \frac{2016}{1001} = \frac{288}{143}.\] The answer is $\boxed{431}.$
| 431
|
5,878
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_12
| 2
|
Consider all 1000-element subsets of the set $\{1, 2, 3, ... , 2015\}$ . From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$
|
Each 1000-element subset $\left\{ a_1, a_2,a_3,...,a_{1000}\right\}$ of $\left\{1,2,3,...,2015\right\}$ with $a_1<a_2<a_3<...<a_{1000}$ contributes $a_1$ to the sum of the least element of each subset. Now, consider the set $\left\{a_1+1,a_2+1,a_3+1,...,a_{1000}+1\right\}$ . There are $a_1$ ways to choose a positive integer $k$ such that $k<a_1+1<a_2+1,a_3+1<...<a_{1000}+1$ $k$ can be anything from $1$ to $a_1$ inclusive). Thus, the number of ways to choose the set $\left\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\right\}$ is equal to the sum. But choosing a set $\left\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\right\}$ is the same as choosing a 1001-element subset from $\left\{1,2,3,...,2016\right\}$
Thus, the average is $\frac{\binom{2016}{1001}}{\binom{2015}{1000}}=\frac{2016}{1001}=\frac{288}{143}$ . Our answer is $p+q=288+143=\boxed{431}$
| 431
|
5,879
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_12
| 3
|
Consider all 1000-element subsets of the set $\{1, 2, 3, ... , 2015\}$ . From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$
|
Let $p$ be the size of the large set and $q$ be the size of the subset (i.e. in this problem, $p = 2015$ and $q = 1000$ ). We can easily find the answers for smaller values of $p$ and $q$
For $p = 2$ and $q = 2$ , the answer is $1$
For $p = 3$ and $q = 2$ , the answer is $\frac43$
For $p = 4$ and $q = 2$ , the answer is $\frac53$
For $p = 3$ and $q = 3$ , the answer is $1$
For $p = 4$ and $q = 3$ , the answer is $\frac54$
For $p = 5$ and $q = 3$ , the answer is $\frac32$
At this point, we can see a pattern: our desired answer is always $\frac{p+1}{q+1}$ . Plugging in $p = 2015$ and $q = 1000$ , the answer is $\frac{2016}{1001}=\frac{288}{143}$ , so $288 + 143 = \boxed{431}$
| 431
|
5,880
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_13
| 1
|
With all angles measured in degrees, the product $\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$ , where $m$ and $n$ are integers greater than 1. Find $m+n$
|
Let $x = \cos 1^\circ + i \sin 1^\circ$ . Then from the identity \[\sin 1 = \frac{x - \frac{1}{x}}{2i} = \frac{x^2 - 1}{2 i x},\] we deduce that (taking absolute values and noticing $|x| = 1$ \[|2\sin 1| = |x^2 - 1|.\] But because $\csc$ is the reciprocal of $\sin$ and because $\sin z = \sin (180^\circ - z)$ , if we let our product be $M$ then \[\frac{1}{M} = \sin 1^\circ \sin 3^\circ \sin 5^\circ \dots \sin 177^\circ \sin 179^\circ\] \[= \frac{1}{2^{90}} |x^2 - 1| |x^6 - 1| |x^{10} - 1| \dots |x^{354} - 1| |x^{358} - 1|\] because $\sin$ is positive in the first and second quadrants. Now, notice that $x^2, x^6, x^{10}, \dots, x^{358}$ are the roots of $z^{90} + 1 = 0.$ Hence, we can write $(z - x^2)(z - x^6)\dots (z - x^{358}) = z^{90} + 1$ , and so \[\frac{1}{M} = \dfrac{1}{2^{90}}|1 - x^2| |1 - x^6| \dots |1 - x^{358}| = \dfrac{1}{2^{90}} |1^{90} + 1| = \dfrac{1}{2^{89}}.\] It is easy to see that $M = 2^{89}$ and that our answer is $2 + 89 = \boxed{91}$
| 91
|
5,881
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_14
| 1
|
For each integer $n \ge 2$ , let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1\le x \le n$ and $0\le y \le x \left\lfloor \sqrt x \right\rfloor$ , where $\left\lfloor \sqrt x \right\rfloor$ is the greatest integer not exceeding $\sqrt x$ . Find the number of values of $n$ with $2\le n \le 1000$ for which $A(n)$ is an integer.
|
Let $n\ge 2$ and define $a(n) = \left\lfloor \sqrt n \right\rfloor$ . For $2\le n \le 1000$ , we have $1\le a(n)\le 31$
For $a^2 \le x < (a+1)^2$ we have $y=ax$ . Thus $A(n+1)-A(n)=a(n+\tfrac 12) = \Delta_n$ (say), and $\Delta_n$ is an integer if $a$ is even; otherwise $\Delta_n$ is an integer plus $\tfrac 12$
If $a=1$ $n\in \{1,2,3\}$ and $\Delta_n$ is of the form $k(n)+\tfrac 12$ so $A(n)$ is an integer when $n$ is even.
If $a=2$ $n\in\{4,\ldots , 8\}$ and $\Delta_n$ is an integer for all $n$ . Since $A(3)$ is not an integer, so $A(n)$ is not an integer for any $n$
If $a=3$ $n\in\{9,\ldots , 15\}$ and $\Delta_n$ is of the form $k(n)+\tfrac 12$ . Since $A(8)$ is of the form $k+\tfrac 12$ so $A(n)$ is an integer only when $n$ is odd.
If $a=4$ $n\in\{16,\ldots , 24\}$ and $\Delta_n$ is an integer for all $n$ . Since $A(15)$ is an integer so $A(n)$ is an integer for all $n$
Now we are back to where we started; i.e., the case $a=5$ will be the same as $a=1$ and so on. Thus, \begin{align} a(n)\equiv 1\pmod 4 \qquad &\Longrightarrow \qquad A(n) \in \mathbb{Z} \textrm{ for even } n, \\ a(n)\equiv 2\pmod 4 \qquad &\Longrightarrow \qquad A(n) \not\in \mathbb{Z} \textrm{ for any } n, \\ a(n)\equiv 3\pmod 4 \qquad &\Longrightarrow \qquad A(n) \in \mathbb{Z} \textrm{ for odd } n, \\ a(n)\equiv 0\pmod 4 \qquad &\Longrightarrow \qquad A(n) \in \mathbb{Z} \textrm{ for all } n. \end{align}
For each $a$ there are $2a+1$ corresponding values of $n$ : i.e., $n\in \{a^2, \ldots , (a+1)^2-1\}$
Thus, the number of values of $n$ corresponding to $(4)$ (i.e., $a(n)\equiv 0\pmod 4$ ) is given by \[\sum_{\substack{a=4k \\ a\le 31}}(2a+1) = \sum_{k=1}^7 (8k+1)=231.\]
The cases $(1)$ and $(3)$ combine to account for half the values of $n$ corresponding to odd values of $a(n)$ ; i.e., \[\frac 12 \cdot \sum_{\substack{a=2k+1 \\ a\le 31}} (2a+1) = \sum_{k=0}^{15} (2k+\tfrac 32) = 264\] However, this also includes the odd integers in $\{1001, \ldots , 1023\}$ . Subtracting $12$ to account for these, we get the number of values of $n$ corresponding to cases $(1)$ and $(3)$ to be $264-12=252$
Adding the contributions from all cases we get our answer to be $231+252= \boxed{483}$
| 483
|
5,882
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_14
| 2
|
For each integer $n \ge 2$ , let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1\le x \le n$ and $0\le y \le x \left\lfloor \sqrt x \right\rfloor$ , where $\left\lfloor \sqrt x \right\rfloor$ is the greatest integer not exceeding $\sqrt x$ . Find the number of values of $n$ with $2\le n \le 1000$ for which $A(n)$ is an integer.
|
By considering the graph of this function, it is shown that the graph is composed of trapezoids ranging from $a^2$ to $(a+1)^2$ with the top made of diagonal line $y=ax$ . The width of each trapezoid is $3, 5, 7$ , etc. Whenever $a$ is odd, the value of $A(n)$ increases by an integer value, plus $\frac{1}{2}$ . Whenever $a$ is even, the value of $A(n)$ increases by an integer value. Since each trapezoid always has an odd width, every value of $n$ is not an integer when $a \pmod{4} \equiv 2$ , and is an integer when $a \pmod{4} \equiv 0$ . Every other value is an integer when $a$ is odd. Therefore, it is simply a matter of determining the number of values of $n$ where $a \pmod{4} \equiv 0$ $(5^2-4^2)+(9^2-8^2)+...+(29^2-28^2)$ ), and adding the number of values of $n$ where $a$ is odd ( $\frac{(2^2-1^2)+(4^2-3^2)+...+(30^2-29^2)+(1000-31^2)}{2}$ ). Adding the two values gives $231+252=\boxed{483}$
| 483
|
5,883
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_14
| 3
|
For each integer $n \ge 2$ , let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1\le x \le n$ and $0\le y \le x \left\lfloor \sqrt x \right\rfloor$ , where $\left\lfloor \sqrt x \right\rfloor$ is the greatest integer not exceeding $\sqrt x$ . Find the number of values of $n$ with $2\le n \le 1000$ for which $A(n)$ is an integer.
|
First, draw a graph of the function. Note that it is just a bunch of line segments. Since we only need to know whether or not $A(n)$ is an integer, we can take the area of each piece from some $x$ to $x+1$ (mod 1), aka the piece from $2$ to $3$ has area $\frac{1}{2} (\mod 1)$ . There are some patterns. Every time we increase $n$ starting with $2$ , we either add $0 (\mod 1)$ or $\frac{1}{2} (\mod 1)$ . We look at $\lfloor \sqrt{x} \rfloor$ for inspiration. Every time this floor (which is really the slope) is odd, there is always an addition of $\frac{1}{2} (\mod 1)$ , and whenever that slope is even, that addition is zero.
Take a few cases. For slope $=1$ , we see that only one value satisfies. Because the last value, $n=4$ , fails, and the numbers $n$ which have a slope of an even number don't change this modulus, all these do not satisfy the criterion. The pattern then comes back to the odds, and this time $\lfloor \frac{7}{2} \rfloor + 1 = 4$ values work. Since the work/fail pattern alternates, all the $n$ s with even slope, $[17, 25]$ , satisfy the criterion. This pattern is cyclic over period 4 of slopes.
Even summation of working cases: $9+17+25+...+57 = 231$ .
Odd summation: $1+4+5+8+9+12...+29$ and plus the $20$ cases from $n=[962, 1000]$ $252$ . Answer is $\boxed{483}$
| 483
|
5,884
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_1
| 1
|
Let $N$ be the least positive integer that is both $22$ percent less than one integer and $16$ percent greater than another integer. Find the remainder when $N$ is divided by $1000$
|
If $N$ is $22$ percent less than one integer $k$ , then $N=\frac{78}{100}k=\frac{39}{50}k$ . In addition, $N$ is $16$ percent greater than another integer $m$ , so $N=\frac{116}{100}m=\frac{29}{25}m$ . Therefore, $k$ is divisible by $50$ and $m$ is divisible by $25$ . Setting these two equal, we have $\frac{39}{50}k=\frac{29}{25}m$ . Multiplying by $50$ on both sides, we get $39k=58m$
The smallest integers $k$ and $m$ that satisfy this are $k=1450$ and $m=975$ , so $N=1131$ . The answer is $\boxed{131}$
| 131
|
5,885
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_1
| 2
|
Let $N$ be the least positive integer that is both $22$ percent less than one integer and $16$ percent greater than another integer. Find the remainder when $N$ is divided by $1000$
|
Continuing from Solution 1, we have $N=\frac{39}{50}k$ and $N=\frac{29}{25}m$ . It follows that $k=\frac{50}{39}N$ and $m=\frac{25}{29}N$ . Both $m$ and $k$ have to be integers, so, in order for that to be true, $N$ has to cancel the denominators of both $\frac{50}{39}$ and $\frac{25}{29}$ . In other words, $N$ is a multiple of both $29$ and $39$ . That makes $N=\operatorname{lcm}(29,39)=29\cdot39=1131$ . The answer is $\boxed{131}$
| 131
|
5,886
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_3
| 1
|
Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$ . Find $m$
|
The three-digit integers divisible by $17$ , and their digit sum: \[\begin{array}{c|c} m & s(m)\\ \hline 102 & 3 \\ 119 & 11\\ 136 & 10\\ 153 & 9\\ 170 & 8\\ 187 & 16\\ 204 & 6\\ 221 & 5\\ 238 & 13\\ 255 & 12\\ 272 & 11\\ 289 & 19\\ 306 & 9\\ 323 & 8\\ 340 & 7\\ 357 & 15\\ 374 & 14\\ 391 & 13\\ 408 & 12\\ 425 & 11\\ 442 & 10\\ 459 & 18\\ 476 & 17 \end{array}\]
Thus the answer is $\boxed{476}$
| 476
|
5,887
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_3
| 2
|
Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$ . Find $m$
|
We can do the same thing as solution 1, except note the following fact: $102$ is a multiple of $17$ and its digits sum to $3$
Therefore, we can add it onto an existing multiple of $17$ that we know of to have $s(m) = 14$ , shown in the right-hand column, provided that its units digit is less than $8$ and its hundreds digit is less than $9$ . Unfortunately, $68$ does not fit the criteria, but $374$ does, meaning that, instead of continually adding multiples of $17$ , we can stop here and simply add $102$ to reach our final answer of $\boxed{476}$
| 476
|
5,888
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_3
| 4
|
Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$ . Find $m$
|
Since the sum of the digits in the base-10 representation of $m$ is $17$ , we must have $m\equiv 17 \pmod{9}$ or $m\equiv -1\pmod{9}$ .
We also know that since $m$ is divisible by 17, $m\equiv 0 \pmod{17}$
To solve this system of linear congruences, we can use the Chinese Remainder Theorem. If we set $m\equiv (-1)(17)(8)\pmod {153}$ , we find that $m\equiv 0\pmod{17}$ and $m\equiv -1\pmod{9}$ , because $17\cdot 8\equiv 136 \equiv 1\pmod{9}$ . The trick to getting here was to find the number $x$ such that $17x\equiv 1\pmod{9}$ , so that when we take things $\pmod{9}$ , the $17$ goes away. We can do this using the Extended Euclidean Algorithm or by guess and check to find that $x\equiv 8\pmod{9}$
Finally, since $m\equiv 17\pmod{153}$ , we repeatedly add multiples of $153$ until we get a number in which its digits sum to 17, which first happens when $m=\boxed{476}$
| 476
|
5,889
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_3
| 5
|
Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$ . Find $m$
|
We proceed by casework on the number of digits. Clearly the answer must have at least two digits, seeing as the maximum digit sum for a one-digit number is 9. The answer must also have less than 4 digits, because this is the AIME.
Case 1: The answer is a 2-digit number.
Represent the number as $10a + b$ , where $0 < a \leq 9$ and $0 \leq b \leq 9$ . The conditions of the problem translated into algebra are: \[17|10a+b\] \[a+b=17\] By the Euclidean Algorithm, this is equivalent to: \[17|9a\] 9 is not a factor of 17, so $17|a$ . So $a$ must be a multiple of 17, but this is impossible because of the conditions we placed on $a$ and $b$ .
(Alternatively, note that the only possible options are 89 and 98, and neither works.)
Case 2: The answer is a 3-digit number.
Represent the number as $100a+10b+c$ , where $0 < a \leq 9$ and $0 \leq b,c \leq 9$ . Translating the conditions again: \[17|100a+10b+c\] \[a+b+c=17\] \[17|99a+9b\] \[17|9(11a+b)\] \[17|11a+b\] Testing multiples of 17 yields $(4, 7, 6)$ as the minimal solution for $(a, b, c)$ and thus the answer is $\boxed{476}$
| 476
|
5,890
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_4
| 2
|
In an isosceles trapezoid, the parallel bases have lengths $\log 3$ and $\log 192$ , and the altitude to these bases has length $\log 16$ . The perimeter of the trapezoid can be written in the form $\log 2^p 3^q$ , where $p$ and $q$ are positive integers. Find $p + q$
|
Set the base of the log as 2. Then call the trapezoid $ABCD$ with $CD$ as the longer base. Then have the two feet of the altitudes be $E$ and $F$ , with $E$ and $F$ in position from left to right respectively. Then, $CF$ and $ED$ are $\log 192 - \log 3 = \log 64$ (from the log subtraction identity. Then $CF=EF=3$ (isosceles trapezoid and $\log 64$ being 6. Then the 2 legs of the trapezoid is $\sqrt{3^2+4^2}=5=\log 32$
And we have the answer:
$\log 192 + \log 32 + \log 32 + \log 3 = \log(192 \cdot 32 \cdot 32 \cdot 3) = \log(2^6 \cdot 3 \cdot 2^5 \cdot 2^5 \cdot 3) = \log(2^{16} \cdot 3^2) \Rightarrow 16+2 = \boxed{18}$
| 18
|
5,891
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_4
| 3
|
In an isosceles trapezoid, the parallel bases have lengths $\log 3$ and $\log 192$ , and the altitude to these bases has length $\log 16$ . The perimeter of the trapezoid can be written in the form $\log 2^p 3^q$ , where $p$ and $q$ are positive integers. Find $p + q$
|
Let $ABCD$ be the trapezoid, where $\overline{AB} || \overline{CD}$ and $AB = \log 3$ and $CD = \log 192$ . Draw altitudes from $A$ and $B$ to $\overline{CD}$ with feet at $E$ and $F$ , respectively. $AB = \log 3$ , so $EF = \log 3$ . Now, we attempt to find $DE + FC$ , or what's left of $CD$ after we take out $EF$ . We make use of the two logarithmic rules:
\[\log(xy) = \log x + \log y\]
\[\log(x^a) = a\log(x)\]
\[CD = \log 192 = \log (3 \cdot 2^6) = \log 3 + \log(2^6) = \log 3 + 6\log 2\]
Thus, since $CD = DE + EF + FC = \log 3 + 6\log 2$ $CD - EF = \log 3 + 6\log 2 - \log 3 = 6\log 2 = DE + FC$
Now, why was finding $DE + FC$ important? Absolutely no reason! Just kidding, lol 🤣 Now, we essentially "glue" triangles $\triangle DAE$ and $\triangle BFC$ together to get $\triangle XC'D'$ , where $X$ is the point where $A$ and $B$ became one. Note we can do this because $\triangle DAE$ and $\triangle BFC$ are both right triangles with a common leg length (the altitude of trapezoid $ABCD$ ).
Triangle $XC'D'$ has a base of $C'D'$ , which is just equal to $DE + FC = 6\log 2$ . It is equal to $DE + FC$ because when we brought triangles $\triangle DAE$ and $\triangle BFC$ together, the length of $CD$ was not changed except for taking out $EF$
$XC' = XD'$ since $AD = BC$ because the problem tells us we have an isosceles trapezoid. Drop and altitude from $X$ to $C'D'$ The altitude has length $\log 16 = 4\log 2$ . The altitude also bisects $C'D'$ since $\triangle XC'D'$ is isosceles. Let the foot of the altitude be $M$ . Then $MD' = 3\log 2$ (Remember that C'D' was $6\log 2$ , and then it got bisected by the altitude). Thus, the hypotenuse, $XD'$ must be $5\log 2$ from the Pythagorean Theorem or by noticing that you have a 3-4-5 right triangle with a similarity ratio of $\log 2$ . Since $XD' = XC' = BC = AD$ $BC = AD = 5\log 2 = \log 2^5$
Now, we have $CD = \log (3 \cdot 2^6)$ $AB = \log 3$ , and $BC = AD = \log 2^5$ . Thus, their sum is
\[\log (3 \cdot 2^6) + \log 3 + \log 2^5 + \log 2^5 = \log (2^16 \cdot 3^2)\]
Thus, $p + q = 16 + 2 = \boxed{18}$ . ~Extremelysupercooldude
| 18
|
5,892
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_6
| 1
|
Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form $P(x) = 2x^3-2ax^2+(a^2-81)x-c$ for some positive integers $a$ and $c$ . Can you tell me the values of $a$ and $c$ ?"
After some calculations, Jon says, "There is more than one such polynomial."
Steve says, "You're right. Here is the value of $a$ ." He writes down a positive integer and asks, "Can you tell me the value of $c$ ?"
Jon says, "There are still two possible values of $c$ ."
Find the sum of the two possible values of $c$
|
We call the three roots (some may be equal to one another) $x_1$ $x_2$ , and $x_3$ . Using Vieta's formulas, we get $x_1+x_2+x_3 = a$ $x_1 \cdot x_2+x_1 \cdot x_3+x_2 \cdot x_3 = \frac{a^2-81}{2}$ , and $x_1 \cdot x_2 \cdot x_3 = \frac{c}{2}$
Squaring our first equation we get $x_1^2+x_2^2+x_3^2+2 \cdot x_1 \cdot x_2+2 \cdot x_1 \cdot x_3+2 \cdot x_2 \cdot x_3 = a^2$
We can then subtract twice our second equation to get $x_1^2+x_2^2+x_3^2 = a^2-2 \cdot \frac{a^2-81}{2}$
Simplifying the right side:
\begin{align*} a^2-2 \cdot \frac{a^2-81}{2} &= a^2-a^2+81\\ &= 81.\\ \end{align*}
So, we know $x_1^2+x_2^2+x_3^2 = 81$
We can then list out all the triples of positive integers whose squares sum to $81$
We get $(1, 4, 8)$ $(3, 6, 6)$ , and $(4, 4, 7)$
These triples give $a$ values of $13$ $15$ , and $15$ , respectively, and $c$ values of $64$ $216$ , and $224$ , respectively.
We know that Jon still found two possible values of $c$ when Steve told him the $a$ value, so the $a$ value must be $15$ . Thus, the two $c$ values are $216$ and $224$ , which sum to $\boxed{440}$
| 440
|
5,893
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_6
| 2
|
Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form $P(x) = 2x^3-2ax^2+(a^2-81)x-c$ for some positive integers $a$ and $c$ . Can you tell me the values of $a$ and $c$ ?"
After some calculations, Jon says, "There is more than one such polynomial."
Steve says, "You're right. Here is the value of $a$ ." He writes down a positive integer and asks, "Can you tell me the value of $c$ ?"
Jon says, "There are still two possible values of $c$ ."
Find the sum of the two possible values of $c$
|
First things first. Vietas gives us the following:
\begin{align} x_1+x_2+x_3 = a\\ x_1 \cdot x_2+x_1 \cdot x_3+x_2 \cdot x_3 = \frac{a^2-81}{2}\\ x_1 \cdot x_2 \cdot x_3 = \frac{c}{2} \end{align}
From $(2)$ $a$ must have odd parity, meaning $a^2-81$ must be a multiple of $4$ , which implies that both sides of $(2)$ are even. Then, from $(1)$ , we see that an odd number of $x_1$ $x_2$ , and $x_3$ must be odd, because we have already deduced that $a$ is odd. In order for both sides of $(2)$ to be even, there must only be one odd number and two even numbers.
Now, the theoretical maximum value of the left side of $(2)$ is $3 \cdot \biggl(\frac{a}{3}\biggr)^2=\frac{a^2}{3}$ . That means that the maximum bound of $a$ is where \[\frac{a^2}{3} > \frac{a^2-81}{2},\] which simplifies to $\sqrt{243} > a$ , meaning \[16 > a.\] So now we have that $9<a$ from $(2)$ $a<16$ , and $a$ is odd from $(2)$ . This means that $a$ could equal $11$ $13$ , or $15$ . At this point, we have simplified the problem to the point where we can casework+ brute force. As said above, we arrive at our solutions of $(1, 4, 8)$ $(3, 6, 6)$ , and $(4, 4, 7)$ , of which the last two return equal $a$ values. Then, $2(3 \cdot 6 \cdot 6+4 \cdot 4 \cdot 7)=\boxed{440}$ AWD.
| 440
|
5,894
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_6
| 3
|
Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form $P(x) = 2x^3-2ax^2+(a^2-81)x-c$ for some positive integers $a$ and $c$ . Can you tell me the values of $a$ and $c$ ?"
After some calculations, Jon says, "There is more than one such polynomial."
Steve says, "You're right. Here is the value of $a$ ." He writes down a positive integer and asks, "Can you tell me the value of $c$ ?"
Jon says, "There are still two possible values of $c$ ."
Find the sum of the two possible values of $c$
|
Since each of the roots is positive, the local maximum of the function must occur at a positive value of $x$ . Taking $\frac{d}{dx}$ of the polynomial yields $6x^2-4ax+a^2-81$ , which is equal to $0$ at the local maximum. Since this is a quadratic in $a$ , we can find an expression for $a$ in terms of $x$ . The quadratic formula gives $a=\frac{4x\pm\sqrt{324-8x^2}}{2}$ , which simplifies to $a=2x\pm\sqrt{81-2x^2}$ . We know that $a$ is a positive integer, and testing small positive integer values of $x$ yields $a=15$ or $a=1$ when $x=4$ , and $a=15$ or $a=9$ when $x=6$ . Because the value of $a$ alone does not determine the polynomial, $a$ $a$ must equal $15$
Now our polynomial equals $2x^3-30x^2+144x-c$ . Because one root is less than (or equal to) the $x$ value at the local maximum (picture the graph of a cubic equation), it suffices to synthetically divide by small integer values of $x$ to see if the resulting quadratic also has positive integer roots. Dividing by $x=3$ leaves a quotient of $2x^2-24x+72=2(x-6)^2$ , and dividing by $x=4$ leaves a quotient of $2x^2-22x+56=2(x-4)(x-7)$ . Thus, $c=2\cdot 3\cdot 6\cdot 6=216$ , or $c=2\cdot 4\cdot 4\cdot 7=224$ . Our answer is $216+224=\boxed{440}$ ~bad_at_mathcounts
| 440
|
5,895
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_7
| 1
|
Triangle $ABC$ has side lengths $AB = 12$ $BC = 25$ , and $CA = 17$ . Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$ , vertex $Q$ on $\overline{AC}$ , and vertices $R$ and $S$ on $\overline{BC}$ . In terms of the side length $PQ = \omega$ , the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS) = \alpha \omega - \beta \omega^2.\]
Then the coefficient $\beta = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
If $\omega = 25$ , the area of rectangle $PQRS$ is $0$ , so
\[\alpha\omega - \beta\omega^2 = 25\alpha - 625\beta = 0\]
and $\alpha = 25\beta$ . If $\omega = \frac{25}{2}$ , we can reflect $APQ$ over $PQ$ $PBS$ over $PS$ , and $QCR$ over $QR$ to completely cover rectangle $PQRS$ , so the area of $PQRS$ is half the area of the triangle. Using Heron's formula, since $s = \frac{12 + 17 + 25}{2} = 27$
\[[ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90\]
so
\[45 = \alpha\omega - \beta\omega^2 = \frac{625}{2} \beta - \beta\frac{625}{4} = \beta\frac{625}{4}\]
and
\[\beta = \frac{180}{625} = \frac{36}{125}\]
so the answer is $m + n = 36 + 125 = \boxed{161}$
| 161
|
5,896
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_7
| 2
|
Triangle $ABC$ has side lengths $AB = 12$ $BC = 25$ , and $CA = 17$ . Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$ , vertex $Q$ on $\overline{AC}$ , and vertices $R$ and $S$ on $\overline{BC}$ . In terms of the side length $PQ = \omega$ , the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS) = \alpha \omega - \beta \omega^2.\]
Then the coefficient $\beta = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
[asy] unitsize(20); pair A,B,C,E,F,P,Q,R,S; A=(48/5,36/5); B=(0,0); C=(25,0); E=(48/5,0); F=(48/5,18/5); P=(24/5,18/5); Q=(173/10,18/5); S=(24/5,0); R=(173/10,0); draw(A--B--C--cycle); draw(P--Q); draw(Q--R); draw(R--S); draw(S--P); draw(A--E,dashed); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$E$",E,SE); label("$F$",F,NE); label("$P$",P,NW); label("$Q$",Q,NE); label("$R$",R,SE); label("$S$",S,SW); draw(rightanglemark(B,E,A,12)); dot(E); dot(F); [/asy]
Similar triangles can also solve the problem.
First, solve for the area of the triangle. $[ABC] = 90$ . This can be done by Heron's Formula or placing an $8-15-17$ right triangle on $AC$ and solving. (The $8$ side would be collinear with line $AB$
After finding the area, solve for the altitude to $BC$ . Let $E$ be the intersection of the altitude from $A$ and side $BC$ . Then $AE = \frac{36}{5}$ .
Solving for $BE$ using the Pythagorean Formula, we get $BE = \frac{48}{5}$ . We then know that $CE = \frac{77}{5}$
Now consider the rectangle $PQRS$ . Since $SR$ is collinear with $BC$ and parallel to $PQ$ $PQ$ is parallel to $BC$ meaning $\Delta APQ$ is similar to $\Delta ABC$
Let $F$ be the intersection between $AE$ and $PQ$ . By the similar triangles, we know that $\frac{PF}{FQ}=\frac{BE}{EC} = \frac{48}{77}$ . Since $PF+FQ=PQ=\omega$ . We can solve for $PF$ and $FQ$ in terms of $\omega$ . We get that $PF=\frac{48}{125} \omega$ and $FQ=\frac{77}{125} \omega$
Let's work with $PF$ . We know that $PQ$ is parallel to $BC$ so $\Delta APF$ is similar to $\Delta ABE$ . We can set up the proportion:
$\frac{AF}{PF}=\frac{AE}{BE}=\frac{3}{4}$ . Solving for $AF$ $AF = \frac{3}{4} PF = \frac{3}{4} \cdot \frac{48}{125} \omega = \frac{36}{125} \omega$
We can solve for $PS$ then since we know that $PS=FE$ and $FE= AE - AF = \frac{36}{5} - \frac{36}{125} \omega$
Therefore, $[PQRS] = PQ \cdot PS = \omega (\frac{36}{5} - \frac{36}{125} \omega) = \frac{36}{5}\omega - \frac{36}{125} \omega^2$
This means that $\beta = \frac{36}{125} \Rightarrow (m,n) = (36,125) \Rightarrow m+n = \boxed{161}$
| 161
|
5,897
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_7
| 3
|
Triangle $ABC$ has side lengths $AB = 12$ $BC = 25$ , and $CA = 17$ . Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$ , vertex $Q$ on $\overline{AC}$ , and vertices $R$ and $S$ on $\overline{BC}$ . In terms of the side length $PQ = \omega$ , the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS) = \alpha \omega - \beta \omega^2.\]
Then the coefficient $\beta = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
Using the diagram from Solution 2 above, label $AF$ to be $h$ . Through Heron's formula, the area of $\triangle ABC$ turns out to be $90$ , so using $AE$ as the height and $BC$ as the base yields $AE=\frac{36}{5}$ . Now, through the use of similarity between $\triangle APQ$ and $\triangle ABC$ , you find $\frac{w}{25}=\frac{h}{36/5}$ . Thus, $h=\frac{36w}{125}$ . To find the height of the rectangle, subtract $h$ from $\frac{36}{5}$ to get $\left(\frac{36}{5}-\frac{36w}{125}\right)$ , and multiply this by the other given side $w$ to get $\frac{36w}{5}-\frac{36w^2}{125}$ for the area of the rectangle. Finally, $36+125=\boxed{161}$
| 161
|
5,898
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_7
| 4
|
Triangle $ABC$ has side lengths $AB = 12$ $BC = 25$ , and $CA = 17$ . Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$ , vertex $Q$ on $\overline{AC}$ , and vertices $R$ and $S$ on $\overline{BC}$ . In terms of the side length $PQ = \omega$ , the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS) = \alpha \omega - \beta \omega^2.\]
Then the coefficient $\beta = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
Using the diagram as shown in Solution 2, let $AE=h$ and $AP=L$ Now, by Heron's formula, we find that the $[ABC]=90$ . Hence, $h=\frac{36}{5}$
Now, we see that $\sin{B}=\frac{PS}{12-L}\implies PS=\sin{B}(12-L)$ We easily find that $\sin{B}=\frac{3}{5}$
Hence, $PS=\frac{3}{5}(12-L)$
Now, we see that $[PQRS]=\frac{3}{5}(12-L)(w)$
Now, it is obvious that we want to find $L$ in terms of $W$
Looking at the diagram, we see that because $PQRS$ is a rectangle, $\triangle{APQ}\sim{\triangle{ABC}}$
Hence.. we can now set up similar triangles.
We have that $\frac{AP}{AB}=\frac{PQ}{BC}\implies \frac{L}{12}=\frac{W}{25}\implies 25L=12W\implies L=\frac{12W}{25}$
Plugging back in..
$[PQRS]=\frac{3w}{5}(12-(\frac{12W}{25}))\implies \frac{3w}{5}(\frac{300-12W}{25})\implies \frac{900W-36W^2}{125}$
Simplifying, we get $\frac{36W}{5}-\frac{36W^2}{125}$
Hence, $125+36=\boxed{161}$
| 161
|
5,899
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_7
| 5
|
Triangle $ABC$ has side lengths $AB = 12$ $BC = 25$ , and $CA = 17$ . Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$ , vertex $Q$ on $\overline{AC}$ , and vertices $R$ and $S$ on $\overline{BC}$ . In terms of the side length $PQ = \omega$ , the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS) = \alpha \omega - \beta \omega^2.\]
Then the coefficient $\beta = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
Proceed as in solution 1. When $\omega$ is equal to zero, $\alpha - \beta\omega=\alpha$ is equal to the altitude. This means that $25\beta$ is equal to $\frac{36}{5}$ , so $\beta = \frac{36}{125}$ , yielding $\boxed{161}$
| 161
|
5,900
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_9
| 1
|
A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$
[asy] import three; import solids; size(5cm); currentprojection=orthographic(1,-1/6,1/6); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,360)),white,nolight); triple A =(8*sqrt(6)/3,0,8*sqrt(3)/3), B = (-4*sqrt(6)/3,4*sqrt(2),8*sqrt(3)/3), C = (-4*sqrt(6)/3,-4*sqrt(2),8*sqrt(3)/3), X = (0,0,-2*sqrt(2)); draw(X--X+A--X+A+B--X+A+B+C); draw(X--X+B--X+A+B); draw(X--X+C--X+A+C--X+A+B+C); draw(X+A--X+A+C); draw(X+C--X+C+B--X+A+B+C,linetype("2 4")); draw(X+B--X+C+B,linetype("2 4")); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,240)),white,nolight); draw((-2,-2*sqrt(3),0)..(4,0,0)..(-2,2*sqrt(3),0)); draw((-4*cos(atan(5)),-4*sin(atan(5)),0)--(-4*cos(atan(5)),-4*sin(atan(5)),-10)..(4,0,-10)..(4*cos(atan(5)),4*sin(atan(5)),-10)--(4*cos(atan(5)),4*sin(atan(5)),0)); draw((-2,-2*sqrt(3),0)..(-4,0,0)..(-2,2*sqrt(3),0),linetype("2 4")); [/asy]
|
Our aim is to find the volume of the part of the cube submerged in the cylinder.
In the problem, since three edges emanate from each vertex, the boundary of the cylinder touches the cube at three points. Because the space diagonal of the cube is vertical, by the symmetry of the cube, the three points form an equilateral triangle. Because the radius of the circle is $4$ , by the Law of Cosines, the side length s of the equilateral triangle is
\[s^2 = 2\cdot(4^2) - 2\cdot(4^2)\cos(120^{\circ}) = 3(4^2)\]
so $s = 4\sqrt{3}$ .* Again by the symmetry of the cube, the volume we want to find is the volume of a tetrahedron with right angles on all faces at the submerged vertex, so since the lengths of the legs of the tetrahedron are $\frac{4\sqrt{3}}{\sqrt{2}} = 2\sqrt{6}$ (the three triangular faces touching the submerged vertex are all $45-45-90$ triangles) so
\[v = \frac{1}{3}(2\sqrt{6})\left(\frac{1}{2} \cdot (2\sqrt{6})^2\right) = \frac{1}{6} \cdot 48\sqrt{6} = 8\sqrt{6}\]
so
\[v^2 = 64 \cdot 6 = \boxed{384}.\]
| 384
|
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