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1,901
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_15
| 3
|
Isosceles triangle $ABC$ has $AB = AC = 3\sqrt6$ , and a circle with radius $5\sqrt2$ is tangent to line $AB$ at $B$ and to line $AC$ at $C$ . What is the area of the circle that passes through vertices $A$ $B$ , and $C?$
$\textbf{(A) }24\pi\qquad\textbf{(B) }25\pi\qquad\textbf{(C) }26\pi\qquad\textbf{(D) }27\pi\qquad\textbf{(E) }28\pi$
|
Denote by $O$ the center of the circle that is tangent to line $AB$ at $B$ and to line $AC$ at $C$
Because this circle is tangent to line $AB$ at $B$ , we have $OB \perp AB$ and $OB = 5 \sqrt{2}$
Because this circle is tangent to line $AC$ at $C$ , we have $OC \perp AC$ and $OC = 5 \sqrt{2}$
Because $AB = AC$ $OB = OC$ $AO = AO$ , we get $\triangle ABO \cong \triangle ACO$ . Hence, $\angle BAO = \angle CAO$
Let $AO$ and $BC$ meet at point $D$ .
Because $AB = AC$ $\angle BAO = \angle CAO$ $AD = AD$ , we get $\triangle ABD \cong \triangle ACD$ . Hence, $BD = CD$ and $\angle ADB = \angle ADC = 90^\circ$
Denote $\theta = \angle BAO$ . Hence, $\angle BAC = 2 \theta$
Denote by $R$ the circumradius of $\triangle ABC$ .
In $\triangle ABC$ , following from the law of sines, $2 R = \frac{BC}{\sin \angle BAC}$
Therefore, the area of the circumcircle of $\triangle ABC$ is \begin{align*} \pi R^2 & = \pi \left( \frac{BC}{2 \sin \angle BAC} \right)^2 \\ & = \pi \left( \frac{2 BD}{2 \sin \angle BAC} \right)^2 \\ & = \pi \left( \frac{BD}{\sin 2 \theta} \right)^2 \\ & = \pi \left( \frac{AB \sin \theta }{\sin 2 \theta} \right)^2 \\ & = \pi \left( \frac{AB \sin \theta }{2 \sin \theta \cos \theta} \right)^2 \\ & = \pi \left( \frac{AB }{2 \cos \theta} \right)^2 \\ & = \pi \left( \frac{AO}{2} \right)^2 \\ & = \frac{\pi}{4} \left( AB^2 + OB^2 \right) \\ & = \boxed{26} ~Steven Chen (www.professorchenedu.com)
| 26
|
1,902
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_17
| 1
|
An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$
|
The pillar at $B$ has height $9$ and the pillar at $A$ has height $12.$ Since the solar panel is flat, the inclination from pillar $B$ to pillar $A$ is $3.$ Call the center of the hexagon $G.$ Since $\overrightarrow{CG}\parallel\overrightarrow{BA},$ it follows that the solar panel has height $13$ at $G.$ Since the solar panel is flat, the heights of the solar panel at $B,G,$ and $E$ are collinear. Therefore, the pillar at $E$ has height $9+4+4=\boxed{17}.$
| 17
|
1,903
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_17
| 2
|
An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$
|
Let the height of the pillar at $D$ be $x.$ Notice that the difference between the heights of pillar $C$ and pillar $D$ is equal to the difference between the heights of pillar $A$ and pillar $F.$ So, the height at $F$ is $x+2.$ Now, doing the same thing for pillar $E$ we get the height is $x+3.$ Therefore, we can see the difference between the heights at pillar $C$ and pillar $D$ is half the difference between the heights at $B$ and $E,$ so \begin{align*} x+3-9&=2 \cdot (x-10) \\ x-6&=2 \cdot (x-10) \\ x&=14. \end{align*} The answer is $x+3=\boxed{17}.$
| 17
|
1,904
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_17
| 3
|
An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$
|
We can extend $BA$ and $BC$ to $G$ and $H$ , respectively, such that $AG = CH$ and $E$ lies on $\overline{GH}$ [asy] unitsize(1cm); pair A = (-sqrt(3),1); pair B = (0,2); pair C = (sqrt(3),1); pair D = (sqrt(3),-1); pair E = (0,-2); pair F = (-sqrt(3),-1); draw(A--B--C--D--E--F--cycle); label("$A, 12$", A, NW); label("$B, 9$", B, N); label("$C, 10$", C, NE); label("$D$", D, SE); label("$E$", E, S); label("$F$", F, SW); pair G = (-4*sqrt(3),-2); pair H = (4*sqrt(3),-2); label("$G, 21$", G, S); label("$H, 13$", H, S); draw(A--G, dashed); draw(C--H, dashed); dot(A^^B^^C^^D^^E^^F^^G^^H,linewidth(4.5)); [/asy] Because of hexagon proportions, $\frac{BA}{AG} = \frac{1}{3}$ and $\frac{BC}{CH} = \frac{1}{3}$ . Let $g$ be the height of $G$ . Because $A$ $B$ and $G$ lie on the same line, $\frac{12-9}{g-12} = \frac{1}{3}$ , so $g-12 = 9$ and $g = 21$ . Similarly, the height of $H$ is $13$ $E$ is the midpoint of $GH$ , so we can take the average of these heights to get our answer, $\boxed{17}$
| 17
|
1,905
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_17
| 4
|
An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$
|
Denote by $h_X$ the height of any point $X$
Denote by $M$ the midpoint of $A$ and $C$ .
Hence, \[h_M = \frac{h_A + h_C}{2} = 11.\] Denote by $O$ the center of $ABCDEF$ . Because $ABCDEF$ is a regular hexagon, $O$ is the midpoint of $B$ and $E$ .
Hence, \[h_O = \frac{h_E + h_B}{2} = \frac{h_E + 9}{2}.\] Because $ABCDEF$ is a regular hexagon, $M$ is the midpoint of $B$ and $O$ .
Hence, \[h_M = \frac{h_B + h_O}{2} = \frac{9 + h_O}{2}.\] Solving these equations, we get $h_E = \boxed{17}$
| 17
|
1,906
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_17
| 5
|
An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$
|
In this solution, we define rise as the change of height (in meters) from the solar panel to the ground. It follows that the rise from $B$ to $A$ is $12-9=3,$ and the rise from $B$ to $C$ is $10-9=1.$ Note that $\vec{BE}=2\vec{BA}+2\vec{BC},$ so the rise from $B$ to $E$ is $2\cdot3+2\cdot1=8.$
Together, the height of the pillar at $E$ is $9+8=\boxed{17}$ meters.
| 17
|
1,907
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_17
| 6
|
An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$
|
WLOG, let the side length of the hexagon be $6$
Establish a 3D coordinate system, in which $A=(0,0,0)$ . Let the coordinates of $B$ and $C$ be $(6,0,0)$ $\left(9,-3\sqrt{3},0\right)$ , respectively. Then, the solar panel passes through $P=(0,0,12), Q=(6,0,9), R=\left(9,-3\sqrt{3},10\right)$
The vector $\vec{PQ}=\langle 6,0,-3\rangle$ and $\vec{PR}=\left\langle 9,-3\sqrt{3}, -2\right\rangle$ . Computing $\vec{PQ} \times \vec{PR}$ by the matrix \[\begin{bmatrix} i & j & k \\ 6 & 0 & -3 \\ 9 & -3\sqrt{3} & -2 \end{bmatrix}\] gives the result $-9\sqrt{3}i -15j -18\sqrt{3} k$ . Therefore, a normal vector of the plane of the solar panel is $\left\langle -9\sqrt{3},-15,-18\sqrt{3}\right\rangle$ , and the equation of the plane is $-9\sqrt{3}x-15y-18\sqrt{3}z=k$ . Substituting $(x,y,z)=(0,0,12)$ , we find that $k=-216\sqrt{3}$
Since $E=\left(0,-6\sqrt{3}\right)$ , we substitute $(x,y)=\left(0,-6\sqrt{3}\right)$ into $-9\sqrt{3}x-15y-18\sqrt{3}z=-216\sqrt{3}$ , which gives $z=\boxed{17}$
| 17
|
1,908
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_17
| 7
|
An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$
|
Let the pillars be $AA', BB', \ldots, FF'$ . Since solar panel $A'B'C'D'E'F'$ is a hexagon, the line $B'E'$ hits the midpoint $M$ of $A'C'$ . So, the 3D slope (change in $x$ : change in $y$ : change in $z$ ) of $BE$ is same as $BD$ . If $a$ is side of the hexagonal solar panel, \[B'M' = \frac{1}{2}a, B'E' = a+2\cdot \frac{1}{2}a = 2a\] . So, $B'M:B'E'$ $1:4$ . Since the height of $M$ to ground $ABCDEF$ is $(10+12)/2 = 11$ , the rise (in z) from $B'$ to $M$ is 2 meaning the rise from $B'$ to $E'$ is $8$ . Thus, $EE' = 8+BB' = \boxed{17}$
| 17
|
1,909
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_17
| 8
|
An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$
|
Set the midpoint of $\overline{AC}$ as $M$ [asy] unitsize(1cm); pair A = (-sqrt(3),1); pair B = (0,2); pair C = (sqrt(3),1); pair D = (sqrt(3),-1); pair E = (0,-2); pair F = (-sqrt(3),-1); pair M =(0,1.35); draw(A--B--C--D--E--F--cycle); label("$A, 12$", A, NW); label("$B, 9$", B, N); label("$C, 10$", C, NE); label("$D$", D, SE); label("$E$", E, S); label("$F$", F, SW); dot((0,1)); label("$M$", M); draw((-sqrt(3),1)--(sqrt(3),1),black); [/asy] We know that the height of $M$ is $11$ as it is the midpoint of $\overline{AC}$ , so the height is the average of $A$ and $C$ , which is $\frac{10 + 12}{2}= 11$ . Since $ABCDEF$ is a regular hexagon, $BE = 4\cdot BM$ . Because the increase in height is proportional to the length of the line segments, and the increase in height from $B$ to $M$ is $2$ , the increase in height from $B$ to $E$ is $2\cdot4=8.$ Adding to the height of $B$ , we get $8+9=\boxed{17}$
| 17
|
1,910
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_17
| 9
|
An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$
|
Because the three points given are integers, it is likely that the answer is also an integer. This leaves us with $9$ or $17$ . Because both $A$ and $C$ are greater than $9$ and closer to $E$ than $B$ , we can assume that the height increases as the point gets closer to $E$ . Thus, we know the answer is greater than $9$ . The only choice that satisfies both these criteria is $\boxed{17}$
| 17
|
1,911
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_18
| 1
|
A farmer's rectangular field is partitioned into $2$ by $2$ grid of $4$ rectangular sections as shown in the figure. In each section the farmer will plant one crop: corn, wheat, soybeans, or potatoes. The farmer does not want to grow corn and wheat in any two sections that share a border, and the farmer does not want to grow soybeans and potatoes in any two sections that share a border. Given these restrictions, in how many ways can the farmer choose crops to plant in each of the four sections of the field? [asy] draw((0,0)--(100,0)--(100,50)--(0,50)--cycle); draw((50,0)--(50,50)); draw((0,25)--(100,25)); [/asy] $\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 64 \qquad \textbf{(C)}\ 84 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 144$
|
There are $4$ possibilities for the top-left section. It follows that the top-right and bottom-left sections each have $3$ possibilities, so they have $3^2=9$ combinations. We have two cases:
Together, the answer is $36+48=\boxed{84}.$
| 84
|
1,912
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_18
| 2
|
A farmer's rectangular field is partitioned into $2$ by $2$ grid of $4$ rectangular sections as shown in the figure. In each section the farmer will plant one crop: corn, wheat, soybeans, or potatoes. The farmer does not want to grow corn and wheat in any two sections that share a border, and the farmer does not want to grow soybeans and potatoes in any two sections that share a border. Given these restrictions, in how many ways can the farmer choose crops to plant in each of the four sections of the field? [asy] draw((0,0)--(100,0)--(100,50)--(0,50)--cycle); draw((50,0)--(50,50)); draw((0,25)--(100,25)); [/asy] $\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 64 \qquad \textbf{(C)}\ 84 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 144$
|
We will do casework on the type of crops in the field.
Case 1: all of a kind.
If all four sections have the same type of crop, there are simply $\underline{4}$ ways to choose crops for the sections.
Case 2: $\boldsymbol{3}$ of a kind, $\boldsymbol{1}$ of another kind.
Since the one of another kind must be adjacent to two of the other crops, when choosing the type of crops in this case, we cannot choose soybeans and potatoes, or corn and wheat. Therefore, there are $4 \cdot 3 - 2 \cdot 2 = 8$ choices for the two crops we choose for the section (notice we did not choose by $2$ , since the crop we pick first will be the unique one), and $4$ ways to choose which section the unique crop is planted on. This gives us a total of $4 \cdot 8 = \underline{32}$ ways to choose crops for the sections.
Case 3: $\boldsymbol{2}$ of a kind, $\boldsymbol{2}$ of another kind.
We cannot choose corn and wheat, or soybeans and potatoes, once again, because if we do, the two would have to be adjacent in some way, which the problem disallows. So, there are ${4 \choose 2} - 2 = 4$ ways to choose our two crops (notice that we did choose by $2$ , since there are two of both crops). There are ${4 \choose 2} = 6$ ways to choose where one of the crops go, so there are $4 \cdot 6 = \underline{24}$ ways to choose crops for the sections.
Case 4: $\boldsymbol{2}$ of a kind, $\boldsymbol{1}$ of another kind, $\boldsymbol{1}$ of another kind.
In cases 2 and 3, we excluded the possibility of choosing bad pairs for our crops (i.e. soybeans and potatoes, or corn and wheat). In this case, it is inevitable that we choose a bad pair, because we are choosing $3$ crops this time. The two sections of the same kind must contain the crop that is not part of the bad pair in the trio: for example, if we choose corn, soybeans and potatoes as our three crop types, nor soybeans and potatoes can be the type which occupies two sections in this case; corn must be the one to do so. There are $4$ ways to choose the crop that is not part of the bad pair, and then $1$ way to choose the bad pair, giving us $4 \cdot 1 = 4$ ways to choose the crops. To separate the bad pair of crops, the two of a kind must be diagonally placed. There are $2$ ways to choose where the two of a kind go, and $2$ ways to choose which of the bad pair goes where, giving us $2 \cdot 2 = 4$ ways to choose the positions for the crops. In total, there are $4 \cdot 4 = \underline{16}$ ways to choose crops for the sections.
Case 5: every single crop.
Bad pairs must be on the same diagonal, so there are $2$ ways to choose which pair gets which diagonal, and $2 \cdot 2 = 4$ ways to choose which of each pair goes where on the diagonal, giving us $2 \cdot 4 = \underline{8}$ ways to choose crops for the sections.
Adding up all our values, we get our final answer of $4+32+24+16+8 = \boxed{84}$
| 84
|
1,913
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_18
| 3
|
A farmer's rectangular field is partitioned into $2$ by $2$ grid of $4$ rectangular sections as shown in the figure. In each section the farmer will plant one crop: corn, wheat, soybeans, or potatoes. The farmer does not want to grow corn and wheat in any two sections that share a border, and the farmer does not want to grow soybeans and potatoes in any two sections that share a border. Given these restrictions, in how many ways can the farmer choose crops to plant in each of the four sections of the field? [asy] draw((0,0)--(100,0)--(100,50)--(0,50)--cycle); draw((50,0)--(50,50)); draw((0,25)--(100,25)); [/asy] $\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 64 \qquad \textbf{(C)}\ 84 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 144$
|
To lighten notation, we use C, W, S, P to denote corn, wheat, soybeans, and potatoes, respectively.
We use I, II, III, IV to denote four quadrants, respectively.
We determine an arrangement in the following steps.
Step 1: Determine the crop planted in I.
The number of ways is $4$
Step 2: Determine the crops planed in II, III, IV.
To find the number of arrangements in this step, without loss of generality, we assume that we plant C in I.
We do the following casework analysis.
Case 1: Both II and IV are planted with C.
In this case, the number of ways to plant in III is $3$
Case 2: In II and IV, only one quadrant is planted with C, and another quadrant is planted with either S or P.
In this case, we determine an arrangement in the following steps.
Step 2.1: Determine whether C is planted in II or IV.
The number of ways is $2$
Step 2.2: In II or IV not planted with C, determine whether it is planted with S or P.
The number of ways is $2$
Step 2.3: Determine the crop planted in III.
The number of ways is $2$
Following from the rule of product, the number of ways in this case is $2 \cdot 2 \cdot 2 = 8$
Case 3: II and IV are both planted with S or W.
In this case, we determine an arrangement in the following steps.
Step 2.1: Determine whether S or W is planted in II and IV.
The number of ways is $2$
Step 2.2: Determine the crop planted in III.
The number of ways is $3$
Following from the rule of product, the number of ways in this case is $2 \cdot 3 = 6$
Case 4: In II and IV, exactly one quadrant is planted with S and another one is planted with W.
Step 2.1: Determine which quadrant in II and IV is planted with S.
The number of ways is $2$
Step 2.2: Determine the crop planted in III.
The number of ways is $2$
Following from the rule of product, the number of ways in this case is $2 \cdot 2 = 4$
Putting all cases together, the total number of arrangements in Step 2 is $3 + 8 + 6 + 4 = 21$
Following from the rule of product, the total number of arrangements is $4 \cdot 21 = 84$
Therefore, the answer is $\boxed{84}$
| 84
|
1,914
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_18
| 4
|
A farmer's rectangular field is partitioned into $2$ by $2$ grid of $4$ rectangular sections as shown in the figure. In each section the farmer will plant one crop: corn, wheat, soybeans, or potatoes. The farmer does not want to grow corn and wheat in any two sections that share a border, and the farmer does not want to grow soybeans and potatoes in any two sections that share a border. Given these restrictions, in how many ways can the farmer choose crops to plant in each of the four sections of the field? [asy] draw((0,0)--(100,0)--(100,50)--(0,50)--cycle); draw((50,0)--(50,50)); draw((0,25)--(100,25)); [/asy] $\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 64 \qquad \textbf{(C)}\ 84 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 144$
|
The number of cases with at least one pair of corn and wheat adjacent is $4 \cdot 2 \cdot 4^2 - 4 \cdot 2 \cdot 4 + 0 - 2 = 94$ possible fields (You can easily see this for yourself using PIE.), and WLOG, the same goes for soybean and potatoes. Now, applying PIE on both sets (number of cases with at least one pair of corn and wheat and the number of cases with at least one pair of soybeans and potatoes) yields $2\cdot94 - 16 = 172$ . We want the number of cases without adjacent pairs of soybeans and potatoes or wheat and corn, so we subtract $256 - 172$ , yielding an answer of $\boxed{84}.$
| 84
|
1,915
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_18
| 5
|
A farmer's rectangular field is partitioned into $2$ by $2$ grid of $4$ rectangular sections as shown in the figure. In each section the farmer will plant one crop: corn, wheat, soybeans, or potatoes. The farmer does not want to grow corn and wheat in any two sections that share a border, and the farmer does not want to grow soybeans and potatoes in any two sections that share a border. Given these restrictions, in how many ways can the farmer choose crops to plant in each of the four sections of the field? [asy] draw((0,0)--(100,0)--(100,50)--(0,50)--cycle); draw((50,0)--(50,50)); draw((0,25)--(100,25)); [/asy] $\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 64 \qquad \textbf{(C)}\ 84 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 144$
|
We can create a tree representing an arbitrary box we start with and the possibilities for other boxes around the grid. We can just designate the crop we start with as $1$ , and the other crops as $2$ $3$ , and $4$ , where $1$ cannot be next to $2$ and $3$ cannot be next to $4$ in the grid. [asy] size(400); draw((0, 0)--(-10, -2)--(-13, -4)--(-14, -6)); draw((-13, -4)--(-13,-6)); draw((-13, -4)--(-12,-6)); draw((-10, -2)--(-10, -4)--(-10.5, -6)); draw((-10, -4)--(-9.5, -6)); draw((-10, -2)--(-7, -4)--(-7.5, -6)); draw((-7, -4)--(-6.5, -6)); label("$1$", (0, 0), N); label("$1$", (-10, -2), UnFill(0.5mm)); label("$1$", (-13, -4), UnFill(0.5mm)); label("$1$", (-14, -6), UnFill(0.5mm)); label("$3$", (-13, -6), UnFill(0.5mm)); label("$4$", (-12, -6), UnFill(0.5mm)); label("$3$", (-10, -4), UnFill(0.5mm)); label("$1$", (-10.5, -6), UnFill(0.5mm)); label("$3$", (-9.5, -6), UnFill(0.5mm)); label("$4$", (-7, -4), UnFill(0.5mm)); label("$1$", (-7.5, -6), UnFill(0.5mm)); label("$4$", (-6.5, -6), UnFill(0.5mm)); draw((0, 0)--(0, -2)--(-3, -4)--(-4, -6)); draw((-3, -4)--(-3, -6)); draw((-3, -4)--(-2, -6)); draw((0, -2)--(0, -4)--(-0.5, -6)); draw((0, -4)--(0.5, -6)); draw((0, -2)--(3, -4)--(2.5, -6)); draw((3, -4)--(3.5, -6)); label("$3$", (0, -2), UnFill(0.5mm)); label("$1$", (-3, -4), UnFill(0.5mm)); label("$2$", (0, -4), UnFill(0.5mm)); label("$3$", (3, -4), UnFill(0.5mm)); label("$4$", (-2, -6), UnFill(0.5mm)); label("$3$", (-3, -6), UnFill(0.5mm)); label("$1$", (-4, -6), UnFill(0.5mm)); label("$3$", (-0.5, -6), UnFill(0.5mm)); label("$4$", (0.5, -6), UnFill(0.5mm)); label("$1$", (2.5, -6), UnFill(0.5mm)); label("$3$", (3.5, -6), UnFill(0.5mm)); draw((0, 0)--(10, -2)--(13, -4)--(13.5, -6)); draw((13, -4)--(12.5,-6)); draw((10, -2)--(10, -4)--(10.5, -6)); draw((10, -4)--(9.5, -6)); draw((10, -2)--(7, -4)--(8, -6)); draw((7, -4)--(7, -6)); draw((7, -4)--(6, -6)); label("$4$", (10, -2), UnFill(0.5mm)); label("$4$", (13, -4), UnFill(0.5mm)); label("$4$", (13.5, -6), UnFill(0.5mm)); label("$1$", (12.5, -6), UnFill(0.5mm)); label("$2$", (10, -4), UnFill(0.5mm)); label("$4$", (10.5, -6), UnFill(0.5mm)); label("$3$", (9.5, -6), UnFill(0.5mm)); label("$1$", (7, -4), UnFill(0.5mm)); label("$4$", (8, -6), UnFill(0.5mm)); label("$3$", (7, -6), UnFill(0.5mm)); label("$1$", (6, -6), UnFill(0.5mm)); [/asy] As we can see, there are $21$ possibilities. We multiply this by the $4$ (the number of possibilities there are for the crop we started with), so our answer is $\boxed{84}$
| 84
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1,916
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_18
| 6
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A farmer's rectangular field is partitioned into $2$ by $2$ grid of $4$ rectangular sections as shown in the figure. In each section the farmer will plant one crop: corn, wheat, soybeans, or potatoes. The farmer does not want to grow corn and wheat in any two sections that share a border, and the farmer does not want to grow soybeans and potatoes in any two sections that share a border. Given these restrictions, in how many ways can the farmer choose crops to plant in each of the four sections of the field? [asy] draw((0,0)--(100,0)--(100,50)--(0,50)--cycle); draw((50,0)--(50,50)); draw((0,25)--(100,25)); [/asy] $\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 64 \qquad \textbf{(C)}\ 84 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 144$
|
The top right box has $4$ choices and the top left box has $3$ choices. Thus, it is reasonable to assume that the answer is a multiple of $12$ . We know that the answer will not be too small or too large, so the answer is $\boxed{84}$
| 84
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1,917
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_19
| 1
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A disk of radius $1$ rolls all the way around the inside of a square of side length $s>4$ and sweeps out a region of area $A$ . A second disk of radius $1$ rolls all the way around the outside of the same square and sweeps out a region of area $2A$ . The value of $s$ can be written as $a+\frac{b\pi}{c}$ , where $a,b$ , and $c$ are positive integers and $b$ and $c$ are relatively prime. What is $a+b+c$
$\textbf{(A)} ~10\qquad\textbf{(B)} ~11\qquad\textbf{(C)} ~12\qquad\textbf{(D)} ~13\qquad\textbf{(E)} ~14$
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The side length of the inner square traced out by the disk with radius $1$ is $s-4.$ However, there is a piece at each corner (bounded by two line segments and one $90^\circ$ arc) where the disk never sweeps out. The combined area of these four pieces is $(1+1)^2-\pi\cdot1^2=4-\pi.$ As a result, we have \[A=s^2-(s-4)^2-(4-\pi)=8s-20+\pi.\] Now, we consider the second disk. The part it sweeps is comprised of four quarter circles with radius $2$ and four rectangles with side lengths of $2$ and $s.$ When we add it all together, we have $2A=8s+4\pi,$ or \[A=4s+2\pi.\] We equate the expressions for $A,$ and then solve for $s:$ \[8s-20+\pi=4s+2\pi.\] We get $s=5+\frac{\pi}{4},$ so the answer is $5+1+4=\boxed{10}.$
| 10
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1,918
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https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_20
| 2
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For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$
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Similar to Solution 1, use the discriminant to get $b^2\leq 4c$ and $c^2\leq 4b$ . These can be rearranged to $c\geq \frac{1}{4}b^2$ and $b\geq \frac{1}{4}c^2$ . Now, we can roughly graph these two inequalities, letting one of them be the $x$ axis and the other be $y$ .
The graph of solutions should be above the parabola and under its inverse, meaning we want points on the graph or in the first area enclosed by the two graphs: [asy] unitsize(2); Label f; f.p=fontsize(6); xaxis("$x$",0,5,Ticks(f, 1.0)); yaxis("$y$",0,5,Ticks(f, 1.0)); real f(real x) { return 0.25x^2; } real g(real x) { return 2*sqrt(x); } dot((1,1)); dot((2,1)); dot((1,2)); dot((2,2)); dot((3,3)); dot((4,4)); draw(graph(f,0,sqrt(20))); draw(graph(g,0,5)); [/asy] We are looking for lattice points (since $b$ and $c$ are positive integers), of which we can count $\boxed{6}$
| 6
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1,919
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_20
| 3
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For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$
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We need to solve the following system of inequalities: \[ \left\{ \begin{array}{ll} b^2 - 4 c \leq 0 \\ c^2 - 4 b \leq 0 \end{array} \right.. \] Feasible solutions are in the region formed between two parabolas $b^2 - 4 c = 0$ and $c^2 - 4 b = 0$
Define $f \left( b \right) = \frac{b^2}{4}$ and $g \left( b \right) = 2 \sqrt{b}$ .
Therefore, all feasible solutions are in the region formed between the graphs of these two functions.
For $b = 1$ , we have $f(b) = \frac{1}{4}$ and $g(b) = 2$ .
Hence, the feasible $c$ are $1, 2$
For $b = 2$ , we have $f(b) = 1$ and $g(b) = 2 \sqrt{2}$ .
Hence, the feasible $c$ are $1, 2$
For $b = 3$ , we have $f(b) = \frac{9}{4}$ and $g(b) = 2 \sqrt{3}$ .
Hence, the feasible $c$ is $3$
For $b = 4$ , we have $f(b) = 4$ and $g(b) = 4$ .
Hence, the feasible $c$ is $4$
For $b > 4$ , we have $f(b) > g(b)$ . Hence, there is no feasible $c$
Putting all cases together, the correct answer is $\boxed{6}$
| 6
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1,920
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https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_20
| 4
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For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$
|
A quadratic equation $Ax^2+Bx+C=0$ has one real solution if and only if $\sqrt{B^2-4AC}=0.$ Similarly, it has imaginary solutions if and only if $\sqrt{B^2-4AC}<0.$ We proceed as following:
We want both $x^2+bx+c$ to be $1$ value or imaginary and $x^2+cx+b$ to be $1$ value or imaginary. $x^2+4x+4$ is one such case since $\sqrt {b^2-4ac}$ is $0.$ Also, $x^2+3x+3, x^2+2x+2, x^2+x+1$ are always imaginary for both $b$ and $c.$ We also have $x^2+x+2$ along with $x^2+2x+1$ since the latter has one solution, while the first one is imaginary. Therefore, we have $\boxed{6}$ total ordered pairs of integers.
| 6
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1,921
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https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_20
| 5
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For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$
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We see that $b^2 \leq 4c$ and $c^2 \leq 4b.$ WLOG, assume that $b \geq c.$ Then we have that $b^2 \leq 4c \leq 4b$ , so $b^2 \leq 4b$ and therefore $b \leq 4$ , also meaning that $c \leq 4.$ This means that we only need to try 16 cases. Now we can get rid of the assumption that $b \geq c$ , because we want ordered pairs. For $b = 1$ and $b = 2$ $c = 1$ and $c = 2$ work. When $b = 3$ $c$ can only be $3$ , and when $b = 4$ , only $c = 4$ works, for a total of $\boxed{6}$ ordered pairs of integers.
| 6
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1,922
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_20
| 6
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For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$
|
We need both $b^2\leq 4c$ and $c^2\leq 4b$
If $b=c$ then the above become $b^2\leq 4b\iff b\leq 4$ , so we have four solutions $(k,k)$ , where $k=1$ $2$ $3$ $4$
If $b<c$ then we only need $c^2\leq 4b$ since it implies $b^2< 4c$ . Now $c^2\leq 4b\leq 4(c-1) \implies (c-2)^2\leq 0 \implies c=2$ , so $b=1$ . We plug $b=1$ $c=2$ back into $c^2\leq 4b$ and it works. So there is another solution $(1,2)$
By symmetry, if $b>c$ then $(b,c)=(2,1)$
Therefore the total number of solutions is $\boxed{6}$
| 6
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1,923
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_20
| 7
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For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$
|
Since $b^{2} - 4c \le 0$ and $c^{2} - 4b \le 0$ , adding the two together yields $b^{2} + c^{2} \le 4(c+b)$ . Obviously, this is not true if either $b$ or $c$ get too large, and they are equal when $b = c = 4$ , so the greatest pair is $(4,4)$ and both numbers must be lesser for further pairs. For there to be two distinct real solutions, we can test all these pairs where $(b,c)$ are less than 4 (except for the already valid solution) on the original quadratics, and we find the working pairs are $(1,1)$ $(2,1)$ $(2,1)$ $(2,2)$ $(3,3)$ $(4,4)$ meaning there are $\boxed{6}$ pairs.
| 6
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1,924
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_21
| 1
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Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 16$
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For simplicity purposes, we assume that the balls and the bins are both distinguishable.
Recall that there are $5^{20}$ ways to distribute $20$ balls into $5$ bins. We have \[p=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{5^{20}} \text{ and } q=\frac{\binom{20}{4,4,4,4,4}}{5^{20}}.\] Therefore, the answer is \[\frac pq=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{\binom{20}{4,4,4,4,4}}=\frac{5\cdot4\cdot\frac{20!}{3!\cdot5!\cdot4!\cdot4!\cdot4!}}{\frac{20!}{4!\cdot4!\cdot4!\cdot4!\cdot4!}}=\frac{5\cdot4\cdot(4!\cdot4!\cdot4!\cdot4!\cdot4!)}{3!\cdot5!\cdot4!\cdot4!\cdot4!}=\frac{5\cdot4\cdot4}{5}=\boxed{16}.\] ~MRENTHUSIASM ~Jesshuang
| 16
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1,925
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_21
| 2
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Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 16$
|
For simplicity purposes, we assume that the balls and the bins are both distinguishable.
Let $q=\frac{x}{a},$ where $a$ is the total number of combinations and $x$ is the number of cases where every bin ends up with $4$ balls.
We can take $1$ ball from one bin and place it in another bin so that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Note that one configuration of $4{-}4{-}4{-}4{-}4$ corresponds to $5\cdot4\cdot4=80$ configurations of $3{-}5{-}4{-}4{-}4.$ On the other hand, one configuration of $3{-}5{-}4{-}4{-}4$ corresponds to $5$ configurations of $4{-}4{-}4{-}4{-}4.$
Therefore, we have \[p = \frac{80}{5}\cdot\frac{x}{a} = 16\cdot\frac{x}{a},\] from which $\frac{p}{q} = \boxed{16}.$
| 16
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1,926
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https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_21
| 3
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Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 16$
|
Since both of the cases will have $3$ bins with $4$ balls in them, we can leave those out. There are $2 \cdot \binom {5}{2} = 20$ ways to choose where to place the $3$ and the $5$ . After that, there are $\binom {8}{3} = 56$ ways to put the $3$ and $5$ balls being put into the bins. For the $4,4,4,4,4$ case, after we canceled the $4,4,4$ out, we have $\binom {8}{4} = 70$ ways to put the $4$ balls inside the bins. Therefore, we have $\frac {56\cdot 20}{70}$ which is equal to $8 \cdot 2 = \boxed{16}$
| 16
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1,927
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_21
| 4
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Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 16$
|
Construct the set $A$ consisting of all possible $3{-}5{-}4{-}4{-}4$ bin configurations, and construct set $B$ consisting of all possible $4{-}4{-}4{-}4{-}4$ configurations. If we let $N$ be the total number of configurations possible, it's clear we want to solve for $\frac{p}{q} = \frac{\frac{|A|}{N}}{\frac{|B|}{N}} = \frac{|A|}{|B|}$
Consider drawing an edge between an element in $A$ and an element in $B$ if it is possible to reach one configuration from the other by moving a single ball (Note this process is reversible.). Let us consider the total number of edges drawn.
For any element in $A$ , we may choose one of the $5$ balls in the $5$ -bin and move it to the $3$ -bin to get a valid element in $B$ . This implies the number of edges is $5|A|$
On the other hand, for any element in $B$ , we may choose one of the $20$ balls and move it to one of the other $4$ -bins to get a valid element in $A$ . This implies the number of edges is $80|B|$
We equate the expressions to get $5|A| = 80|B|$ , from which $\frac{|A|}{|B|} = \frac{80}{5} = \boxed{16}$
| 16
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1,928
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_23
| 2
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For each positive integer $n$ , let $f_1(n)$ be twice the number of positive integer divisors of $n$ , and for $j \ge 2$ , let $f_j(n) = f_1(f_{j-1}(n))$ . For how many values of $n \le 50$ is $f_{50}(n) = 12?$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$
|
First, take note that the maximum possible value of $f_1(n)$ for $1 \le n \le k$ increases as $k$ increases (it is a step function), i.e. it is increasing. Likewise, as $k$ decreases, the maximum possible value of $f_1(n)$ decreases as well. Also, let $f_1(n) = 2d(n)$ where $d(n)$ is the number of divisors of n.
Since $n \le 50$ $f_1(n) <= 20$ . This maximum occurs when $d(n) = 10 \implies n = 2^4 \cdot 3 = 48$ . Next, since $f_1(n) <=20$ $f_1(f_1(n)) \le 12 \implies f_2(n) \le 12$ . This maximum occurs when $d(f_1(n)) = 6 \implies n = 2 \cdot 3^2 = 18, n = 2^2 \cdot 3 = 12$ . Since $f_2(n) \le 12$ $f_1(f_2(n)) \le 12 \implies f_3(n) \le 12$ , once again. This maximum again occurs when $d(f_2(n)) = 6 \implies f_2(n) = 2^2 \cdot 3 = 12$ . Now, suppose for the sake of contradiction that $f_2(n) < 12$ . Then, $f_3(n) < 12$ (since $f_2(n) = 12$ was the only number that would maximize $f_3(n))$ for $f_2(n) \le 12$ ). As a result, since $f_1(n)$ is increasing, and because $12$ is where $f_1$ steps down from a maximum of $6 \cdot 2 = 12$ , we must have that $f_1(f_3(n)) < f_1(12) = 12 \implies f_4(n) < 12$ . We continue applying $f_1$ on both sides (which is possible since $f_1$ is increasing) until we reach $f_50$ , giving us that $f_50(n) < 12$ . However, $f_50(n) = 12$ , which is a contradiction. Thus, $f_2(n) = 12$
Now, let us finally solve for the solutions. $f_2(n) = 12 \implies f_1(f_1(n)) = 12 \implies d(f_1(n)) = 6$ $d(f_1(n)) = 6 \implies f_1(n) = p^2 \cdot q$ where $p$ and $q$ are primes. Since $f_1(n) \le 20$ $f_1(n)$ can only be $12$ $18$ , or $20$ . If $f_1(n) = 12$ , then $d(n) = 6 \implies n = p^5, p^2 \cdot q \implies n \in \{ 12, 18, 20, 28, 32, 44, 45, 50 \}$ , resulting in 8 solutions. If $f_1(n) = 18$ , then $d(n) = 9 \implies n = p^8, p^2 \cdot q^2 \implies n = 36$ , giving us one more solution. Finally, $f_1(n) = 20 \implies d(n) = 10 \implies n = p^9, p^4 \cdot q \implies n = 48$ . Thus, in total, we have $\boxed{10}$ solutions.
| 10
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1,929
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_23
| 3
|
For each positive integer $n$ , let $f_1(n)$ be twice the number of positive integer divisors of $n$ , and for $j \ge 2$ , let $f_j(n) = f_1(f_{j-1}(n))$ . For how many values of $n \le 50$ is $f_{50}(n) = 12?$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$
|
$\textbf{Observation 1}$ $f_1 \left( 12 \right) = 12$
Hence, if $n$ has the property that $f_j \left( n \right) = 12$ for some $j$ , then $f_k \left( n \right) = 12$ for all $k > j$
$\textbf{Observation 2}$ $f_1 \left( 8 \right) = 8$
Hence, if $n$ has the property that $f_j \left( n \right) = 8$ for some $j$ , then $f_k \left( n \right) = 8$ for all $k > j$
$\textbf{Case 1}$ $n = 1$
We have $f_1 \left( n \right) = 2$ $f_2 \left( n \right) = f_1 \left( 2 \right) = 4$ $f_3 \left( n \right) = f_1 \left( 4 \right) = 6$ $f_4 \left( n \right) = f_1 \left( 6 \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
$\textbf{Case 2}$ $n$ is prime.
We have $f_1 \left( n \right) = 4$ $f_2 \left( n \right) = f_1 \left( 4 \right) = 6$ $f_3 \left( n \right) = f_1 \left( 6 \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
$\textbf{Case 3}$ : The prime factorization of $n$ takes the form $p_1^2$
We have $f_1 \left( n \right) = 6$ $f_2 \left( n \right) = f_1 \left( 6 \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
$\textbf{Case 4}$ : The prime factorization of $n$ takes the form $p_1^3$
We have $f_1 \left( n \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
$\textbf{Case 5}$ : The prime factorization of $n$ takes the form $p_1^4$
We have $f_1 \left( n \right) = 10$ $f_2 \left( n \right) = f_1 \left( 10 \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
$\textbf{Case 6}$ : The prime factorization of $n$ takes the form $p_1^5$
We have $f_1 \left( n \right) = 12$ . Hence, Observation 1 implies $f_{50} \left( n \right) = 12$
In this case the only $n$ is $2^5 = 32$
$\textbf{Case 7}$ : The prime factorization of $n$ takes the form $p_1 p_2$
We have $f_1 \left( n \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
$\textbf{Case 8}$ : The prime factorization of $n$ takes the form $p_1 p_2^2$
We have $f_1 \left( n \right) = 12$ . Hence, Observation 1 implies $f_{50} \left( n \right) = 12$
In this case, all $n$ are $12, 18, 20, 28, 44, 45,$ and $50$
$\textbf{Case 9}$ : The prime factorization of $n$ takes the form $p_1 p_2^3$
We have $f_1 \left( n \right) = 16$ $f_2 \left( n \right) = f_1 \left( 16 \right) = 10$ $f_3 \left( n \right) = f_1 \left( 10 \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
$\textbf{Case 10}$ : The prime factorization of $n$ takes the form $p_1 p_2^4$
We have $f_1 \left( n \right) = 20$ $f_2 \left( n \right) = f_1 \left( 20 \right) = 12$ . Hence, Observation 1 implies $f_{50} \left( n \right) = 12$
In this case, the only $n$ is $48$
$\textbf{Case 11}$ : The prime factorization of $n$ takes the form $p_1^2 p_2^2$
We have $f_1 \left( n \right) = 18$ $f_2 \left( n \right) = f_1 \left( 18 \right) = 12$ . Hence, Observation 1 implies $f_{50} \left( n \right) = 12$
In this case, the only $n$ is $36$
$\textbf{Case 12}$ : The prime factorization of $n$ takes the form $p_1 p_2 p_3$
We have $f_1 \left( n \right) = 16$ $f_2 \left( n \right) = f_1 \left( 16 \right) = 10$ $f_3 \left( n \right) = f_2 \left( 10 \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
Putting all cases together, the number of feasible $n \leq 50$ is $\boxed{10}$
| 10
|
1,930
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_24
| 1
|
Each of the $12$ edges of a cube is labeled $0$ or $1$ . Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations and/or reflections. For how many such labelings is the sum of the labels on the edges of each of the $6$ faces of the cube equal to $2$
$\textbf{(A) } 8 \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 20$
|
For simplicity, we will name this cube $ABCDEFGH$ by vertices, as shown below. [asy] /* Made by MRENTHUSIASM */ size(150); pair A, B, C, D, E, F, G, H; A = (0,1); B = (1,1); C = (1,0); D = (0,0); E = (0.3,1.3); F = (1.3,1.3); G = (1.3,0.3); H = (0.3,0.3); draw(A--B--C--D--cycle^^A--E^^B--F^^C--G^^E--F--G); draw(H--D^^H--E^^H--G,dashed); dot("$A$",A,1.5*W,linewidth(4)); dot("$B$",B,1.5*(1,0),linewidth(4)); dot("$C$",C,1.5*SE,linewidth(4)); dot("$D$",D,1.5*SW,linewidth(4)); dot("$E$",E,1.5*NW,linewidth(4)); dot("$F$",F,1.5*NE,linewidth(4)); dot("$G$",G,1.5*NE,linewidth(4)); dot("$H$",H,1.5*NW,linewidth(4)); [/asy] Note that for each face of this cube, two edges are labeled $0,$ and two edges are labeled $1.$ For all twelve edges of this cube, we conclude that six edges are labeled $0,$ and six edges are labeled $1.$
We apply casework to face $ABCD.$ Recall that there are $\binom42=6$ ways to label its edges:
Therefore, we have $4+16=\boxed{20}$ such labelings in total.
| 20
|
1,931
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_24
| 2
|
Each of the $12$ edges of a cube is labeled $0$ or $1$ . Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations and/or reflections. For how many such labelings is the sum of the labels on the edges of each of the $6$ faces of the cube equal to $2$
$\textbf{(A) } 8 \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 20$
|
Since we want the sum of the edges of each face to be $2$ , we need there to be two $1$ s and two $0$ s on each face. Through experimentation, we find that either $2, 4,$ or all of them have $1$ s adjacent to $1$ s and $0$ s adjacent to $0$ on each face. WLOG, let the first face (counterclockwise) be $0,0,1,1$ . In this case we are trying to have all of them be adjacent to each other. First face: $0,0,1,1$ . Second face: $2$ choices: $1,0,0,1$ or $0,0,1,1$ . After that, it is basically forced and everything will fall in to place. Since we assumed WLOG, we need to multiply $2$ by $4$ to get a total of $8$ different arrangements.
Secondly, $4$ of the faces have all of them adjacent and $2$ of the faces do not: WLOG counting counterclockwise, we have $0,0,1,1$ . Then, we choose the other face next to it. There are two cases, which are $0,1,0,1$ and $1,0,1,0$ . Therefore, this subcase has $4$ different arrangements. Then, we can choose the face at front to be $1,0,1,0$ . This has $4$ cases. The sides can either be $0,1,1,0$ or $1,1,0,0$ . Therefore, we have another $8$ cases.
Summing these up, we have $8+4+8 = 20$ . Therefore, our answer is $\boxed{20}$
| 20
|
1,932
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_24
| 3
|
Each of the $12$ edges of a cube is labeled $0$ or $1$ . Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations and/or reflections. For how many such labelings is the sum of the labels on the edges of each of the $6$ faces of the cube equal to $2$
$\textbf{(A) } 8 \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 20$
|
We see that if the $3$ edges connecting to $A$ has two $0$ 's, and one $1$ , it would have the same solutions as if it had two $1$ 's, and one $0$ . The solutions would just be inverted. As case 2.1 and case 2.2.2 are inverses, and case 2.2.1 has case 1 as an inverse, there would not be any additional solutions.
Similarly, if the $3$ edges connecting to $A$ has three $0$ 's, it would be the same as the inverse of case 1, or case 2.2.1, resulting in no new solutions.
Putting all the cases together, we have $4+6+4+6=\boxed{20}$ solutions.
| 20
|
1,933
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_24
| 4
|
Each of the $12$ edges of a cube is labeled $0$ or $1$ . Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations and/or reflections. For how many such labelings is the sum of the labels on the edges of each of the $6$ faces of the cube equal to $2$
$\textbf{(A) } 8 \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 20$
|
The problem states the sum of the labels on the edges of each of the $6$ faces of the cube equal to $2$ . That is, the sum of the labels on the $4$ edges of a face is equal to $2$ . The labels can only be $0$ or $1$ , meaning $2$ edges are labeled $1$ , the other $2$ are labeled $0$
This problem can be approached by Graph Coloring of Graph Theory . Note that each face of the cube connects to $4$ other faces, each with a shared edge. We use the following graph to represent the problem. Each vertex represents a face, each edge represent the cube's edge. Each vertex has $4$ edges connecting to $4$ other vertices. The edges can be colored red or blue, with red as label $0$ , and blue as label $1$ . Each vertex must have $2$ red edges and $2$ blue edges.
$\textbf{Case 1}$ $2$ adjacent red edges from vertex A. There are $4$ ways to choose $2$ red edges adjacent to each other and connect to $2$ vertices with an edge between them as shown below.
$\textbf{Case 1.1}$ $2$ adjacent red edges from vertex $A$ form a closed loop with a third red edge. There is only $1$ case as shown below.
$\textbf{Case 1.2}$ $2$ adjacent red edges from vertex $A$ does not form a closed loop with a third red edge. There are $3$ cases as shown below.
In case $1$ , there are total $4 \cdot (1 + 3) = 16$ ways.
$\textbf{Case 2}$ $2$ red edges from vertex $A$ with $1$ blue edge in between. There are $2$ ways to choose $2$ red edges with $1$ blue edge in between.
There are only $2$ cases as shown below.
In case $2$ , there are total $2 \cdot 2 = 4$ ways.
From both case $1$ and case $2$ , there are $16 + 4 = \boxed{20}$ ways in total.
| 20
|
1,934
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_1
| 1
|
What is the value of $1234 + 2341 + 3412 + 4123$
$\textbf{(A)}\: 10000\qquad\textbf{(B)} \: 10010\qquad\textbf{(C)} \: 10110\qquad\textbf{(D)} \: 11000\qquad\textbf{(E)} \: 11110$
|
We see that $1, 2, 3,$ and $4$ each appear in the ones, tens, hundreds, and thousands digit exactly once. Since $1+2+3+4=10$ , we find that the sum is equal to \[10\cdot(1+10+100+1000)=\boxed{11110}.\] Note that it is equally valid to manually add all four numbers together to get the answer.
| 110
|
1,935
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_1
| 2
|
What is the value of $1234 + 2341 + 3412 + 4123$
$\textbf{(A)}\: 10000\qquad\textbf{(B)} \: 10010\qquad\textbf{(C)} \: 10110\qquad\textbf{(D)} \: 11000\qquad\textbf{(E)} \: 11110$
|
We have \[1234 + 2341 + 3412 + 4123 = 1111 \left( 1 + 2 + 3 + 4 \right) = \boxed{11110}.\] ~Steven Chen (www.professorchenedu.com)
| 110
|
1,936
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_1
| 3
|
What is the value of $1234 + 2341 + 3412 + 4123$
$\textbf{(A)}\: 10000\qquad\textbf{(B)} \: 10010\qquad\textbf{(C)} \: 10110\qquad\textbf{(D)} \: 11000\qquad\textbf{(E)} \: 11110$
|
We see that the units digit must be $0$ , since $4+3+2+1$ is $0$ . But every digit from there, will be a $1$ since we have that each time afterwards, we must carry the $1$ from the previous sum. The answer choice that satisfies these conditions is $\boxed{11110}$
| 110
|
1,937
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_1
| 4
|
What is the value of $1234 + 2341 + 3412 + 4123$
$\textbf{(A)}\: 10000\qquad\textbf{(B)} \: 10010\qquad\textbf{(C)} \: 10110\qquad\textbf{(D)} \: 11000\qquad\textbf{(E)} \: 11110$
|
We can simply add the numbers. $1234 + 2341 + 3412 + 4123 = 11110 \implies \boxed{11110}$
| 110
|
1,938
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_2
| 1
|
What is the area of the shaded figure shown below? [asy] size(200); defaultpen(linewidth(0.4)+fontsize(12)); pen s = linewidth(0.8)+fontsize(8); pair O,X,Y; O = origin; X = (6,0); Y = (0,5); fill((1,0)--(3,5)--(5,0)--(3,2)--cycle, palegray+opacity(0.2)); for (int i=1; i<7; ++i) { draw((i,0)--(i,5), gray+dashed); label("${"+string(i)+"}$", (i,0), 2*S); if (i<6) { draw((0,i)--(6,i), gray+dashed); label("${"+string(i)+"}$", (0,i), 2*W); } } label("$0$", O, 2*SW); draw(O--X+(0.35,0), black+1.5, EndArrow(10)); draw(O--Y+(0,0.35), black+1.5, EndArrow(10)); draw((1,0)--(3,5)--(5,0)--(3,2)--(1,0), black+1.5); [/asy]
$\textbf{(A)}\: 4\qquad\textbf{(B)} \: 6\qquad\textbf{(C)} \: 8\qquad\textbf{(D)} \: 10\qquad\textbf{(E)} \: 12$
|
The line of symmetry divides the shaded figure into two congruent triangles, each with base $3$ and height $2.$
Therefore, the area of the shaded figure is \[2\cdot\left(\frac12\cdot3\cdot2\right)=2\cdot3=\boxed{6}.\] ~MRENTHUSIASM ~Wilhelm Z
| 6
|
1,939
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_2
| 2
|
What is the area of the shaded figure shown below? [asy] size(200); defaultpen(linewidth(0.4)+fontsize(12)); pen s = linewidth(0.8)+fontsize(8); pair O,X,Y; O = origin; X = (6,0); Y = (0,5); fill((1,0)--(3,5)--(5,0)--(3,2)--cycle, palegray+opacity(0.2)); for (int i=1; i<7; ++i) { draw((i,0)--(i,5), gray+dashed); label("${"+string(i)+"}$", (i,0), 2*S); if (i<6) { draw((0,i)--(6,i), gray+dashed); label("${"+string(i)+"}$", (0,i), 2*W); } } label("$0$", O, 2*SW); draw(O--X+(0.35,0), black+1.5, EndArrow(10)); draw(O--Y+(0,0.35), black+1.5, EndArrow(10)); draw((1,0)--(3,5)--(5,0)--(3,2)--(1,0), black+1.5); [/asy]
$\textbf{(A)}\: 4\qquad\textbf{(B)} \: 6\qquad\textbf{(C)} \: 8\qquad\textbf{(D)} \: 10\qquad\textbf{(E)} \: 12$
|
To find the area of the shaded figure, we subtract the area of the smaller triangle (base $4$ and height $2$ ) from the area of the larger triangle (base $4$ and height $5$ ): \[\frac12\cdot4\cdot5-\frac12\cdot4\cdot2=10-4=\boxed{6}.\] ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com)
| 6
|
1,940
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_2
| 4
|
What is the area of the shaded figure shown below? [asy] size(200); defaultpen(linewidth(0.4)+fontsize(12)); pen s = linewidth(0.8)+fontsize(8); pair O,X,Y; O = origin; X = (6,0); Y = (0,5); fill((1,0)--(3,5)--(5,0)--(3,2)--cycle, palegray+opacity(0.2)); for (int i=1; i<7; ++i) { draw((i,0)--(i,5), gray+dashed); label("${"+string(i)+"}$", (i,0), 2*S); if (i<6) { draw((0,i)--(6,i), gray+dashed); label("${"+string(i)+"}$", (0,i), 2*W); } } label("$0$", O, 2*SW); draw(O--X+(0.35,0), black+1.5, EndArrow(10)); draw(O--Y+(0,0.35), black+1.5, EndArrow(10)); draw((1,0)--(3,5)--(5,0)--(3,2)--(1,0), black+1.5); [/asy]
$\textbf{(A)}\: 4\qquad\textbf{(B)} \: 6\qquad\textbf{(C)} \: 8\qquad\textbf{(D)} \: 10\qquad\textbf{(E)} \: 12$
|
We have $4$ lattice points in the interior and $6$ lattice points on the boundary. By Pick's Theorem , the area of the shaded figure is \[4+\frac{6}{2}-1 = 4+3-1 = \boxed{6}.\] ~danprathab
| 6
|
1,941
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_3
| 1
|
The expression $\frac{2021}{2020} - \frac{2020}{2021}$ is equal to the fraction $\frac{p}{q}$ in which $p$ and $q$ are positive integers whose greatest common divisor is ${ }1$ . What is $p?$
$(\textbf{A})\: 1\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 2020\qquad(\textbf{D}) \: 2021\qquad(\textbf{E}) \: 4041$
|
We write the given expression as a single fraction: \[\frac{2021}{2020} - \frac{2020}{2021} = \frac{2021\cdot2021-2020\cdot2020}{2020\cdot2021}\] by cross multiplication. Then by factoring the numerator, we get \[\frac{2021\cdot2021-2020\cdot2020}{2020\cdot2021}=\frac{(2021-2020)(2021+2020)}{2020\cdot2021}.\] The question is asking for the numerator, so our answer is $2021+2020=4041,$ giving $\boxed{4041}$
| 41
|
1,942
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_3
| 2
|
The expression $\frac{2021}{2020} - \frac{2020}{2021}$ is equal to the fraction $\frac{p}{q}$ in which $p$ and $q$ are positive integers whose greatest common divisor is ${ }1$ . What is $p?$
$(\textbf{A})\: 1\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 2020\qquad(\textbf{D}) \: 2021\qquad(\textbf{E}) \: 4041$
|
Denote $a = 2020$ . Hence, \begin{align*} \frac{2021}{2020} - \frac{2020}{2021} & = \frac{a + 1}{a} - \frac{a}{a + 1} \\ & = \frac{\left( a + 1 \right)^2 - a^2}{a \left( a + 1 \right)} \\ & = \frac{2 a + 1}{a \left( a + 1 \right)} . \end{align*}
We observe that ${\rm gcd} \left( 2a + 1 , a \right) = 1$ and ${\rm gcd} \left( 2a + 1 , a + 1 \right) = 1$
Hence, ${\rm gcd} \left( 2a + 1 , a \left( a + 1 \right) \right) = 1$
Therefore, $p = 2 a + 1 = 4041$
Therefore, the answer is $\boxed{4041}$
| 41
|
1,943
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_4
| 1
|
At noon on a certain day, Minneapolis is $N$ degrees warmer than St. Louis. At $4{:}00$ the temperature in Minneapolis has fallen by $5$ degrees while the temperature in St. Louis has risen by $3$ degrees, at which time the temperatures in the two cities differ by $2$ degrees. What is the product of all possible values of $N?$
$\textbf{(A)}\: 10\qquad\textbf{(B)} \: 30\qquad\textbf{(C)} \: 60\qquad\textbf{(D)} \: 100\qquad\textbf{(E)} \: 120$
|
At noon on a certain day, let $M$ and $L$ be the temperatures (in degrees) in Minneapolis and St. Louis, respectively. It follows that $M=L+N.$
At $4{:}00,$ we get \begin{align*} |(M-5)-(L+3)| &= 2 \\ |M-L-8| &= 2 \\ |N-8| &= 2. \end{align*} We have two cases:
Together, the product of all possible values of $N$ is $10\cdot6=\boxed{60}.$
| 60
|
1,944
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_4
| 2
|
At noon on a certain day, Minneapolis is $N$ degrees warmer than St. Louis. At $4{:}00$ the temperature in Minneapolis has fallen by $5$ degrees while the temperature in St. Louis has risen by $3$ degrees, at which time the temperatures in the two cities differ by $2$ degrees. What is the product of all possible values of $N?$
$\textbf{(A)}\: 10\qquad\textbf{(B)} \: 30\qquad\textbf{(C)} \: 60\qquad\textbf{(D)} \: 100\qquad\textbf{(E)} \: 120$
|
At noon on a certain day, the difference of temperatures in Minneapolis and St. Louis is $N$ degrees.
At $4{:}00,$ the difference of temperatures in Minneapolis and St. Louis is $N-8$ degrees.
It follows that \[|N-8|=2.\] We continue with the casework in Solution 1 to get the answer $\boxed{60}.$
| 60
|
1,945
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_6
| 1
|
The least positive integer with exactly $2021$ distinct positive divisors can be written in the form $m \cdot 6^k$ , where $m$ and $k$ are integers and $6$ is not a divisor of $m$ . What is $m+k?$
$(\textbf{A})\: 47\qquad(\textbf{B}) \: 58\qquad(\textbf{C}) \: 59\qquad(\textbf{D}) \: 88\qquad(\textbf{E}) \: 90$
|
Let this positive integer be written as $p_1^{e_1}\cdot p_2^{e_2}$ . The number of factors of this number is therefore $(e_1+1) \cdot (e_2+1)$ , and this must equal 2021. The prime factorization of 2021 is $43 \cdot 47$ , so $e_1+1 = 43 \implies e_1=42$ and $e_2+1=47\implies e_2=46$ . To minimize this integer, we set $p_1 = 3$ and $p_2 = 2$ . Then this integer is $3^{42} \cdot 2^{46} = 2^4 \cdot 2^{42} \cdot 3^{42} = 16 \cdot 6^{42}$ .
Now $m=16$ and $k=42$ so $m+k = 16 + 42 = \boxed{58}$
| 58
|
1,946
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_6
| 2
|
The least positive integer with exactly $2021$ distinct positive divisors can be written in the form $m \cdot 6^k$ , where $m$ and $k$ are integers and $6$ is not a divisor of $m$ . What is $m+k?$
$(\textbf{A})\: 47\qquad(\textbf{B}) \: 58\qquad(\textbf{C}) \: 59\qquad(\textbf{D}) \: 88\qquad(\textbf{E}) \: 90$
|
Recall that $6^k$ can be written as $2^k \cdot 3^k$ . Since we want the integer to have $2021$ divisors, we must have it in the form $p_1^{42} \cdot p_2^{46}$ , where $p_1$ and $p_2$ are prime numbers. Therefore, we want $p_1$ to be $3$ and $p_2$ to be $2$ . To make up the remaining $2^4$ , we multiply $2^{42} \cdot 3^{42}$ by $m$ , which is $2^4$ which is $16$ . Therefore, we have $42 + 16 = \boxed{58}$
| 58
|
1,947
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_6
| 3
|
The least positive integer with exactly $2021$ distinct positive divisors can be written in the form $m \cdot 6^k$ , where $m$ and $k$ are integers and $6$ is not a divisor of $m$ . What is $m+k?$
$(\textbf{A})\: 47\qquad(\textbf{B}) \: 58\qquad(\textbf{C}) \: 59\qquad(\textbf{D}) \: 88\qquad(\textbf{E}) \: 90$
|
If a number has prime factorization $p_1^{k_1} p_2^{k_2} \cdots p_m^{k_m}$ , then the number of distinct positive divisors of this number is $\left( k_1 + 1 \right) \left( k_2 + 1 \right) \cdots \left( k_m + 1 \right)$
We have $2021 = 43 \cdot 47$ .
Hence, if a number $N$ has 2021 distinct positive divisors, then $N$ takes one of the following forms: $p_1^{2020}$ $p_1^{42} p_2^{46}$
Therefore, the smallest $N$ is $3^{42} 2^{46} = 2^4 \cdot 6^{42} = 16 \cdot 6^{42}$
Therefore, the answer is $\boxed{58}$
| 58
|
1,948
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_7
| 2
|
Call a fraction $\frac{a}{b}$ , not necessarily in the simplest form, special if $a$ and $b$ are positive integers whose sum is $15$ . How many distinct integers can be written as the sum of two, not necessarily different, special fractions?
$\textbf{(A)}\ 9 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$
|
Let $a=15-b,$ so the special fraction is \[\frac ab = \frac{15-b}{b} = \frac{15}{b}-1.\] We can ignore the $-1$ part and only focus on $\frac{15}{b}.$
The integers are $\frac{15}{1},\frac{15}{3},\frac{15}{5},$ which are $15,5,3,$ respectively. We get $30,20,18,10,8,6$ from this group of numbers.
The halves are $\frac{15}{2},\frac{15}{6},\frac{15}{10},$ which are $7\frac12,2\frac12,1\frac12,$ respectively. We get $15,10,9,5,4,3$ from this group of numbers.
The quarters are $\frac{15}{4},\frac{15}{12},$ which are $3\frac34,1\frac14,$ respectively. We get $5$ from this group of numbers.
Note that $10$ and $5$ each appear twice. Therefore, the answer is $\boxed{11}.$
| 11
|
1,949
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_7
| 3
|
Call a fraction $\frac{a}{b}$ , not necessarily in the simplest form, special if $a$ and $b$ are positive integers whose sum is $15$ . How many distinct integers can be written as the sum of two, not necessarily different, special fractions?
$\textbf{(A)}\ 9 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$
|
We split this up into two cases:
Case 1: integer + integer
The whole numbers we have are $\frac{10}{5}$ (or $2$ ), $\frac{12}{3}$ (or $4$ ), and $\frac{14}{1}$ (or $14$ ). There are $\dbinom{3}{2}=3$ ways to choose different-numbered pairs and $3$ ways to choose the same-numbered pairs. So, $3+3=6$
Case 2: fraction + fraction
The fractions we have are $\frac{5}{10}$ (or $\frac{1}{2}$ ), $\frac{9}{6}$ (or $\frac{3}{2}$ ), and $\frac{13}{2}$ . Similarly, there are $\dbinom{3}{2}=3$ ways to choose different-numbered pairs and $3$ ways to choose the same-numbered pairs. So, $3+3=6$
Thus, $6+6=12$
So now you would just go ahead and innocently choose $\textbf{(D) }12$ , right? No! We overcounted $8$ , as $\frac{13}{2}+\frac96=\frac{12}{3}+\frac{12}{3}=8$ . Therefore, the correct answer is actually $12-1=\boxed{11}$
| 11
|
1,950
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_8
| 1
|
The greatest prime number that is a divisor of $16384$ is $2$ because $16384 = 2^{14}$ . What is the sum of the digits of the greatest prime number that is a divisor of $16383$
$\textbf{(A)} \: 3\qquad\textbf{(B)} \: 7\qquad\textbf{(C)} \: 10\qquad\textbf{(D)} \: 16\qquad\textbf{(E)} \: 22$
|
We have \begin{align*} 16383 & = 2^{14} - 1 \\ & = \left( 2^7 + 1 \right) \left( 2^7 - 1 \right) \\ & = 129 \cdot 127 \\ \end{align*}
Since $129$ is composite, $127$ is the largest prime which can divide $16383$ . The sum of $127$ 's digits is $1+2+7=\boxed{10}$
| 10
|
1,951
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_10
| 1
|
Forty slips of paper numbered $1$ to $40$ are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, "I can't tell who has the larger number." Then Bob says, "I know who has the larger number." Alice says, "You do? Is your number prime?" Bob replies, "Yes." Alice says, "In that case, if I multiply your number by $100$ and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?
$\textbf{(A) }27\qquad\textbf{(B) }37\qquad\textbf{(C) }47\qquad\textbf{(D) }57\qquad\textbf{(E) }67$
|
Let Alice have the number A, Bob B.
When Alice says that she can't tell who has the larger number, it means that $A$ cannot equal $1$ . Therefore, it makes sense that Bob has $2$ because he now knows that Alice has the larger number. $2$ is also prime. The last statement means that $200+A$ is a perfect square. The three squares in the range $200-300$ are $225$ $256$ , and $289$ . So, $A$ could equal $25$ $56$ , or $89$ , so $A+B$ is $27$ $58$ , or $91$ , of only $\boxed{27}$ is an answer choice.
| 27
|
1,952
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_10
| 2
|
Forty slips of paper numbered $1$ to $40$ are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, "I can't tell who has the larger number." Then Bob says, "I know who has the larger number." Alice says, "You do? Is your number prime?" Bob replies, "Yes." Alice says, "In that case, if I multiply your number by $100$ and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?
$\textbf{(A) }27\qquad\textbf{(B) }37\qquad\textbf{(C) }47\qquad\textbf{(D) }57\qquad\textbf{(E) }67$
|
Denote by $A$ and $B$ the numbers drawn by Alice and Bob, respectively.
Alice's sentence “I can't tell who has the larger number.” implies $A \in \left\{ 2 , \cdots , 39 \right\}$
Bob's sentence “I know who has the larger number.” implies $B \in \left\{ 1 , 2 , 39, 40 \right\}$
Their subsequent conversation that $B$ is prime implies $B = 2$
Then, Alice's next sentence “In that case, if I multiply your number by 100 and add my number, the result is a perfect square.” implies $200 + A$ is a perfect square.
Hence, $A = 25$
Therefore, the answer is $\boxed{27}$
| 27
|
1,953
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_10
| 3
|
Forty slips of paper numbered $1$ to $40$ are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, "I can't tell who has the larger number." Then Bob says, "I know who has the larger number." Alice says, "You do? Is your number prime?" Bob replies, "Yes." Alice says, "In that case, if I multiply your number by $100$ and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?
$\textbf{(A) }27\qquad\textbf{(B) }37\qquad\textbf{(C) }47\qquad\textbf{(D) }57\qquad\textbf{(E) }67$
|
We see that $225$ is one such square that works. Bob gets $2$ and Alice gets $25$ which is valid. Thus, $2 + 25 = 27.$ So $\boxed{27}$ is our answer.
| 27
|
1,954
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_15
| 1
|
In square $ABCD$ , points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$ , with $BR = 6$ and $PR = 7$ . What is the area of the square?
[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]
$\textbf{(A) }85\qquad\textbf{(B) }93\qquad\textbf{(C) }100\qquad\textbf{(D) }117\qquad\textbf{(E) }125$
|
Note that $\triangle APB \cong \triangle BQC.$ Then, it follows that $\overline{PB} \cong \overline{QC}.$ Thus, $QC = PB = PR + RB = 7 + 6 = 13.$ Define $x$ to be the length of side $CR,$ then $RQ = 13-x.$ Because $\overline{BR}$ is the altitude of the triangle, we can use the property that $QR \cdot RC = BR^2.$ Substituting the given lengths, we have \[(13-x) \cdot x = 36.\] Solving, gives $RQ = 4$ and $RC = 9.$ We eliminate the possibility of $x=4$ because $RC > QR.$ Thus, the side length of the square, by Pythagorean Theorem, is \[\sqrt{9^2 +6^2} = \sqrt{81+36} = \sqrt{117}.\] Thus, the area of the square is $(\sqrt{117})^2 = 117,$ so the answer is $\boxed{117}.$
| 117
|
1,955
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_15
| 2
|
In square $ABCD$ , points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$ , with $BR = 6$ and $PR = 7$ . What is the area of the square?
[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]
$\textbf{(A) }85\qquad\textbf{(B) }93\qquad\textbf{(C) }100\qquad\textbf{(D) }117\qquad\textbf{(E) }125$
|
As above, note that $\bigtriangleup BPA \cong \bigtriangleup CQB$ , which means that $QC = 13$ . In addition, note that $BR$ is the altitude of a right triangle to its hypotenuse, so $\bigtriangleup BQR \sim \bigtriangleup CBR \sim \bigtriangleup CQB$ . Let the side length of the square be $x$ ; using similarity side ratios of $\bigtriangleup BQR$ to $\bigtriangleup CQB$ , we get \[\frac{6}{x} = \frac{QB}{13} \implies QB \cdot x = 78\] Note that $QB^2 + x^2 = 13^2 = 169$ by the Pythagorean theorem, so we can use the expansion $(a+b)^2 = a^2+2ab+b^2$ to produce two equations and two variables;
\[(QB + x)^2 = QB^2 + 2QB\cdot x + x^2 \implies (QB+x)^2 = 169 + 2 \cdot 78 \implies QB+x = \sqrt{13(13)+13(12)} = \sqrt{13 \cdot 25} = 5\sqrt{13}\] \[(QB-x)^2 = QB^2 - 2QB\cdot x + x^2 \implies (QB - x)^2 = 169 - 2\cdot 78 \implies \pm(QB-x) = \sqrt{13(13) - 13(12)}\]
Since $QB-x$ is negative, it doesn't make sense in the context of this problem, so we go with \[x-QB = \sqrt{13(13) - 13(12)} = \sqrt{13 \cdot 1} = \sqrt{13}\]
We want $x^2$ , so we want to find $x$ . Adding the first equation to the second, we get \[2x = 6\sqrt{13} \implies x = 3\sqrt{13}\]
Then $x^2$ $(3\sqrt{13}^2) = 9 \cdot 13 = 117 = \boxed{117}$
| 117
|
1,956
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_15
| 3
|
In square $ABCD$ , points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$ , with $BR = 6$ and $PR = 7$ . What is the area of the square?
[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]
$\textbf{(A) }85\qquad\textbf{(B) }93\qquad\textbf{(C) }100\qquad\textbf{(D) }117\qquad\textbf{(E) }125$
|
We have that $\triangle CRB \sim \triangle BAP.$ Thus, $\frac{\overline{CB}}{\overline{CR}} = \frac{\overline{PB}}{\overline{AB}}$ . Now, let the side length of the square be $s.$ Then, by the Pythagorean theorem, $CR = \sqrt{x^2-36}.$ Plugging all of this information in, we get \[\frac{s}{\sqrt{s^2-36}} = \frac{13}{s}.\] Simplifying gives \[s^2=13\sqrt{s^2-36},\] Squaring both sides gives \[s^4 = 169s^2- 169\cdot 36 \implies s^4-169s^2 + 169\cdot 36 = 0.\] We now set $s^2=t,$ and get the equation $t^2-169t + 169\cdot 36 = 0.$ From here, notice we want to solve for $t$ , as it is precisely $s^2,$ or the area of the square. So we use the Quadratic formula , and though it may seem bashy, we hope for a nice cancellation of terms. \[t = \frac{169\pm\sqrt{169^2-4\cdot 36 \cdot 169}}{2}.\] It seems scary, but factoring $169$ from the square root gives us \[t = \frac{169\pm \sqrt{169 \cdot (169-144)}}{2} = \frac{169 \pm \sqrt{169 \cdot 25}}{2} = \frac{169 \pm 13\cdot 5}{2} = \frac{169\pm 65}{2},\] giving us the solutions \[t=52, 117.\] We instantly see that $t=52$ is way too small to be an area of this square ( $52$ isn't even an answer choice, so you can skip this step if out of time) because then the side length would be $2\sqrt{13}$ and then, even the largest line you can draw inside the square (the diagonal) is $2\sqrt{26},$ which is less than $13$ (line $PB$ ) And thus, $t$ must be $117$ , and our answer is $\boxed{117}.$ $\blacksquare$
| 117
|
1,957
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_15
| 4
|
In square $ABCD$ , points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$ , with $BR = 6$ and $PR = 7$ . What is the area of the square?
[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]
$\textbf{(A) }85\qquad\textbf{(B) }93\qquad\textbf{(C) }100\qquad\textbf{(D) }117\qquad\textbf{(E) }125$
|
Denote $a = RC$ . Now tilt your head to the right and view $R, \overrightarrow{RB}$ and $\overrightarrow{RC}$ as the origin, $x$ -axis and $y$ -axis, respectively. In particular, we have points $B(6,0), C(0,a), P(-7,0)$ . Note that side length of the square $ABCD$ is $BC = \sqrt{a^2 + 36}$ . Also equation of line $BC$ is \[\underbrace{\frac{x}{6} + \frac{y}{a} = 1}_{\text{intercepts form}} \quad \implies \quad ax + 6y - 6a = 0.\] Because the distance from $P(-7,0)$ to line $\color[rgb]{0,0.4,0.65}BC: ax + 6y - 6a = 0$ is also the side length $\sqrt{a^2 + 36}$ , we can apply the point-line distance formula to get \[\frac{|a\cdot(-7) + 6 \cdot 0 - 6a|}{\sqrt{a^2 + 36}} = {\sqrt{a^2 + 36}}\] which reduces to $|13a| = a^2 + 36$ . Since $a$ is positive, the last equations factors as $a^2 - 13a + 36 = (a-4)(a-9) = 0$ . Now judging from the figure, we learn that $a > RB = 6$ . So $a = 9$ .
Therefore, the area of the square $ABCD$ is $BC^2 = RC^2 + RB^2 = a^2 + 6^2 = \boxed{117}$
| 117
|
1,958
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_15
| 5
|
In square $ABCD$ , points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$ , with $BR = 6$ and $PR = 7$ . What is the area of the square?
[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]
$\textbf{(A) }85\qquad\textbf{(B) }93\qquad\textbf{(C) }100\qquad\textbf{(D) }117\qquad\textbf{(E) }125$
|
Denote $\angle PBA = \alpha$ .
Because $\angle QRB = \angle QBC = 90^\circ$ $\angle BCQ = \alpha$
Hence, $AB = BP \cos \angle PBA = 13 \cos \alpha$ $BC = \frac{BR}{\sin \angle BCQ} = \frac{6}{\sin \alpha}$
Because $ABCD$ is a square, $AB = BC$ .
Hence, $13 \cos \alpha = \frac{6}{\sin \alpha}$
Therefore, \begin{align*} \sin 2 \alpha & = 2 \sin \alpha \cos \alpha \\ & = \frac{12}{13} . \end{align*}
Thus, $\cos 2 \alpha = \pm \frac{5}{13}$
$\textbf{Case 1}$ $\cos 2 \alpha = \frac{5}{13}$
Thus, $\cos \alpha = \sqrt{\frac{1 + \cos 2 \alpha}{2}} = \frac{3}{\sqrt{13}}$
Hence, $AB = 13 \cos \alpha = 3 \sqrt{13}$
Therefore, ${\rm Area} \ ABCD = AB^2 = 117$
$\textbf{Case 2}$ $\cos 2 \alpha = - \frac{5}{13}$
Thus, $\cos \alpha = \sqrt{\frac{1 + \cos 2 \alpha}{2}} = \frac{2}{\sqrt{13}}$
Hence, $AB = 13 \cos \alpha = 2 \sqrt{13}$
However, we observe $BQ = \frac{BR}{\cos \alpha} = 3 \sqrt{13} > AB$ .
Therefore, in this case, point $Q$ is not on the segment $AB$
Therefore, this case is infeasible.
Putting all cases together, the answer is $\boxed{117}$
| 117
|
1,959
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_15
| 6
|
In square $ABCD$ , points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$ , with $BR = 6$ and $PR = 7$ . What is the area of the square?
[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]
$\textbf{(A) }85\qquad\textbf{(B) }93\qquad\textbf{(C) }100\qquad\textbf{(D) }117\qquad\textbf{(E) }125$
|
Note that if we connect points $P$ and $C$ , we get a triangle with height $RC$ and length $13$ . This triangle has an area of $\frac {1}{2}$ the square. We can now use answer choices to our advantage!
Answer choice A: If $BC$ was $\sqrt {85}$ $RC$ would be $7$ . The triangle would therefore have an area of $\frac {91}{2}$ which is not half of the area of the square. Therefore, A is wrong.
Answer choice B: If $BC$ was $\sqrt {93}$ $RC$ would be $\sqrt {57}$ . This is obviously wrong.
Answer choice C: If $BC$ was $10$ , we would have that $RC$ is $8$ . The area of the triangle would be $52$ , which is not half the area of the square. Therefore, C is wrong.
Answer choice D: If $BC$ was $\sqrt {117}$ , that would mean that $RC$ is $9$ . The area of the triangle would therefore be $\frac {117}{2}$ which IS half the area of the square. Therefore, our answer is $\boxed{117}$
| 117
|
1,960
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_18
| 1
|
Three identical square sheets of paper each with side length $6{ }$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$ -sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a-b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a+b+c?$
$(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147$
|
First note the useful fact that if $R$ is the circumradius of a dodecagon ( $12$ -gon) the area of the figure is $3R^2.$ If we connect the vertices of the $3$ squares we get a dodecagon. The radius of circumcircle of the dodecagon is simply half the diagonal of the square, which is $3\sqrt{2}.$ Thus the area of the dodecagon is $3 \cdot (3\sqrt{2})^2 = 3 \cdot 18 = 54.$ But, the problem asked for the area of the combined figure which was made from the rotated squares. This area is the area of the dodecagon, which was found, subtracting the $12$ isosceles triangles, which are formed when connecting the vertices of the squares to created the dodecagon. To find this area, we need to know the base of the isosceles triangle, call this $x.$ Then, we can use the Law of Cosines on the triangle that is formed from the two vertices of the square and the center of the square. After computing, we get that $x = 3\sqrt{3} -3.$ Realize that the $12$ isosceles are congruent with an angle measure of $120^{\circ},$ this means that we can create $4$ congruent equilateral triangles with side length of $3\sqrt3 - 3.$ The area of the equilateral triangle is $\frac{\sqrt{3}}{4} \cdot (3\sqrt{3} -3)^2 = \frac{\sqrt{3}}{4} \cdot (36 - 18\sqrt{3}) = \frac{36\sqrt{3} - 54}{4}.$ Thus, the area of all the twelve small equilateral traingles are $4 \cdot \frac{36\sqrt{3} - 54}{4} = 36\sqrt{3} - 54$ . Thus, the requested area is $54 - (36\sqrt{3} - 54) = 108 - 36\sqrt{3}.$ Thus, $a+b+c = 108 + 36 + 3 = 147,$ so the answer is $\boxed{147}.$
| 147
|
1,961
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_18
| 2
|
Three identical square sheets of paper each with side length $6{ }$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$ -sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a-b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a+b+c?$
$(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147$
|
To make things simpler, let's take only the original sheet and the 30 degree rotated sheet. Then the diagram is this;
[asy] size(10cm,0); path p = box((0,0), (1,1)); draw(p, black + linewidth(2.0pt)); draw(rotate(30,(1/2,1/2))*p,black + linewidth(2.0pt)); /*Rotate 60 degrees*/ [/asy]
The area of this diagram is the original square plus the area of the four triangles that 'jut' out of the square. Because the square is rotated $30^{\circ}$ , each triangle is a 30-60-90 triangle. Similarly, the triangles that are bounded on the inside of the original square outside of the rotated square are also congruent 30-60-90 triangles. Noting this, we can do some labelling:
[asy] size(10cm,0); path p = box((0,0), (1,1)); draw(p, black + linewidth(2.0pt)); draw(rotate(30,(1/2,1/2))*p,black + linewidth(2.0pt)); /*Rotate 60 degrees*/ label("$y$",(0.1,-0.05)); label("$x$",(0.4,0.05)); label("$y\sqrt{3}$",(0.8,-0.05)); label("$\frac{x}{2}$",(0.22,-0.12)); label("$\frac{x\sqrt{3}}{2}$",(0.5,-0.15)); label("$2y$",(0.8,0.15)); label("$y$",(1.05,0.1)); label("$\frac{x}{2}$",(1.12,0.22)); [/asy]
Since the side lengths of the squares must be the same, and they are both 6, we have a system of equations; \[y+x+y\sqrt{3} = 6\] \[\frac{x\sqrt{3}}{2} + 2y + \frac{x}{2} = 6\]
We solve this to get $x = 6-2\sqrt{3}$ and $y = 3-\sqrt{3}$
The area of each triangle is $\frac{x}{2} \cdot \frac{x\sqrt{3}}{2} \cdot \frac{1}{2} = 6\sqrt{3} - 9$ by plugging in $x$
The rotated 60 degree square is the same thing as rotating it 30 degrees counterclockwise, so it's triangles that jut out of the square will be congruent to the triangles we have found, and therefore they will have the same area.
Unfortunately, when drawing all three squares, we see the two triangles overlap; take the very top for example.
[asy] import olympiad; size(10cm); pair A,B,C,D,E,F,G; A = (0.211,0); B=(0.3657,0); C = (0.63397,0); D = (0.789,0); E = (0.31666,0.1823); F=(0.5,0.077); G=(0.68334,0.1823); draw((0,0)--(1,0),black+linewidth(2pt)); draw(A--E--C--cycle); draw(B--D--G--cycle); label("$A$",A,S); label("$B$",B,S); label("$C$",C,S); label("$D$",D,S); label("$E$",E,N); label("$F$",F,N); label("$G$",G,N); [/asy]
The area of this shape is twice the area of each of the triangles that we have already found minus the area of the small triangle that is overlapped by the two by PIE. Now we only need to find the area of $\bigtriangleup BCF$
$\angle GBD \cong \angle ECA \cong 30^{\circ}$ and by symmetry $\bigtriangleup BCF$ is isosceles, so it is a 30-30-120 triangle. If we draw a perpendicular, we split it into two 30-60-90 triangles;
[asy] import olympiad; size(10cm); pair A,B,C,D,E,F,G; A = (0.211,0); B=(0.3657,0); C = (0.63397,0); D = (0.789,0); E = (0.31666,0.1823); F=(0.5,0.077); G=(0.68334,0.1823); draw((0,0)--(1,0),black+linewidth(2pt)); draw(A--E--C--cycle); draw(B--D--G--cycle); draw(F--(0.5,0)); label("$A$",A,S); label("$B$",B,S); label("$C$",C,S); label("$D$",D,S); label("$E$",E,N); label("$F$",F,N); label("$G$",G,N); label("$H$",(0.5,0),S); [/asy]
By symmetry, the distance from A to the edge of the square is equal to the distance from D to the edge of the square is equal to $y$ . AC = BD = $x$ , and the side length of the square is 6, so we use PIE to obtain \[x+x-BC = 6-y-y \implies BC = 12 - 6\sqrt{3}\] To find the height of $\bigtriangleup BFC$ , we see that $HC = \frac{BC}{2} = 6-3\sqrt{3}$ . Then by 30-60-90 triangles, $HF = \frac{HC}{\sqrt{3}} = 2\sqrt{3} - 3$ . Finally, the area of $\bigtriangleup BFC = \frac{BC \cdot HF}{2} = 21\sqrt{3}-36$
Putting it all together, the area of the entire diagram is the area of the square plus four of these triangle-triangle intersections. The area of these intersections by PIE is $2 \cdot [ACE] - [BFC] = 12\sqrt{3}-18-(21\sqrt{3}-36) = 18-9\sqrt{3}$ .
Therefore the total area is $36 + 4 \cdot(18-9\sqrt{3}) = 36 + 72 - 36 \sqrt{3} = 108 - 36\sqrt{3}$
Thus $a + b + c = 108+36+3 = 147 = \boxed{147}$
| 147
|
1,962
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_18
| 3
|
Three identical square sheets of paper each with side length $6{ }$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$ -sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a-b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a+b+c?$
$(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147$
|
As shown in Image:2021_AMC_12B_(Nov)_Problem_15,_sol.png , all 12 vertices of three squares form a regular dodecagon (12-gon).
Denote by $O$ the center of this dodecagon.
Hence, $\angle AOB = \frac{360^\circ}{12} = 30^\circ$
Because the length of a side of a square is 6, $AO = 3 \sqrt{2}$
Hence, $AB = 2 AO \sin \frac{\angle AOB}{2} = 3 \left( \sqrt{3} - 1 \right)$
We notice that $\angle MAB = \angle MBA = 30^\circ$ .
Hence, $AM = \frac{AB}{2\cos \angle MAB} = 3 - \sqrt{3}$
Therefore, the area of the region that three squares cover is \begin{align*} & \ [ABCDEFGHIJKL] - 12[MAB] \\ & = 12 [OAB] - 12 [MAB] \\ & = 12 \cdot \frac{1}{2} OA \cdot OB \sin \angle AOB - 12 \cdot \frac{1}{2} MA \cdot MB \sin \angle AMB \\ & = 6 OA^2 \sin \angle AOB - 6 MA^2 \sin \angle AMB \\ & = 108 - 36\sqrt{3} \end{align*}
Therefore, the answer is $\boxed{147}$
| 147
|
1,963
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_18
| 4
|
Three identical square sheets of paper each with side length $6{ }$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$ -sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a-b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a+b+c?$
$(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147$
|
Let $O$ be the center of the polygon, $A$ be the bottom right corner of the first square, $C$ be the next vertex to the left of $A$ , and $M$ be the midpoint between $A$ and $B$ , where $B$ is the bottom left corner of the first square. Note that because there are three $90^{\circ}$ squares separated by $\frac{90^{\circ}}{3} = 30^{\circ}$ , each side of the 24-sided polygon is equal in length, meaning to calculate the area of the whole polygon, we find the area of $\bigtriangleup OAC$ and multiply by 24.
To find $[\bigtriangleup OAC]$ , we already know the height $\overline{OM}$ is the sidelength of the square over $2$ , or $\frac{6}{2}=3$ , so we just need the length of the base $\overline{AC}$ . Notice that $\bigtriangleup OCM$ is a $30-60-90$ triangle since $\angle COM = \frac{360^{\circ}}{12} = 30^{\circ}$ , so $\overline{CM} = \frac{\overline{OM}}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$ . Then $\overline{AC} = \overline{AM} - \overline{CM} = \frac{6}{2} - \sqrt{3} = 3 - \sqrt{3}$ , so \begin{align*} & [\bigtriangleup AOC] = \frac{1}{2} \cdot \overline{OM} \cdot \overline{AC} \\ & = \frac{1}{2} (3)(3 - \sqrt{3}) \\ & = \frac{9 - 3\sqrt{3}}{2} \end{align*}
Then the whole area of the polygon is $\frac{9 - 3\sqrt{3}}{2} \cdot 24 = 108 - 36\sqrt{3}$ . The desired solution is then $108 + 36 + 3 = 147$ , so the answer is $\boxed{147}$
| 147
|
1,964
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_19
| 1
|
Let $N$ be the positive integer $7777\ldots777$ , a $313$ -digit number where each digit is a $7$ . Let $f(r)$ be the leading digit of the $r{ }$ th root of $N$ . What is \[f(2) + f(3) + f(4) + f(5)+ f(6)?\] $(\textbf{A})\: 8\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 11\qquad(\textbf{D}) \: 22\qquad(\textbf{E}) \: 29$
|
We can rewrite $N$ as $\frac{7}{9}\cdot 9999\ldots999 = \frac{7}{9}\cdot(10^{313}-1)$ .
When approximating values, as we will shortly do, the minus one will become negligible so we can ignore it.
When we take the power of ten out of the square root, we’ll be multiplying by another power of ten, so the leading digit will not change. Thus the leading digit of $f(r)$ will be equal to the leading digit of $\sqrt[r]{\frac{7}{9} \cdot 10^{313(\mod r)}}$
Then $f(2)$ is the first digit of $\sqrt{\frac{7}{9}\cdot(10)} = \sqrt{\frac{70}{9}} = \sqrt{7.\ldots} \approx 2$
$f(3) - \sqrt[3]{\frac{7}{9} \cdot 10} = \sqrt[3]{\frac{70}{9}} = \sqrt[3]{7.\ldots} \approx 1$
$f(4) - \sqrt[4]{\frac{7}{9} \cdot 10} = \sqrt[4]{\frac{70}{9}} = \sqrt[4]{7.\ldots} \approx 1$
$f(5) - \sqrt[5]{\frac{7}{9} \cdot 1000} = \sqrt[5]{\frac{7000}{9}} = \sqrt[5]{777.\ldots} \approx 3$
$f(6) - \sqrt[6]{\frac{7}{9} \cdot 10} = \sqrt[6]{\frac{70}{9}} = \sqrt[6]{7.\ldots} \approx 1$
The final answer is therefore $2+1+1+3+1 = \boxed{8}$
| 8
|
1,965
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_19
| 2
|
Let $N$ be the positive integer $7777\ldots777$ , a $313$ -digit number where each digit is a $7$ . Let $f(r)$ be the leading digit of the $r{ }$ th root of $N$ . What is \[f(2) + f(3) + f(4) + f(5)+ f(6)?\] $(\textbf{A})\: 8\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 11\qquad(\textbf{D}) \: 22\qquad(\textbf{E}) \: 29$
|
For notation purposes, let $x$ be the number $777 \ldots 777$ with $313$ digits, and let $B(n)$ be the leading digit of $n$ . As an example, $B(x) = 7$ , because $x = 777 \ldots 777$ , and the first digit of that is $7$
Notice that \[B(\sqrt{\frac{n}{100}}) = B(\sqrt{n})\] for all numbers $n \geq 100$ ; this is because $\sqrt{\frac{n}{100}} = \frac{\sqrt{n}}{10}$ , and dividing by $10$ does not affect the leading digit of a number. Similarly, \[B(\sqrt[3]{\frac{n}{1000}}) = B(\sqrt[3]{n}).\] In general, for positive integers $k$ and real numbers $n > 10^{k}$ , it is true that \[B(\sqrt[k]{\frac{n}{10^{k}}}) = B(\sqrt[k]{n}).\] Behind all this complex notation, all that we're really saying is that the first digit of something like $\sqrt[3]{123456789}$ has the same first digit as $\sqrt[3]{123456.789}$ and $\sqrt[3]{123.456789}$
The problem asks for \[B(\sqrt[2]{x}) + B(\sqrt[3]{x}) + B(\sqrt[4]{x}) + B(\sqrt[5]{x}) + B(\sqrt[6]{x}).\]
From our previous observation, we know that \[B(\sqrt[2]{x}) = B(\sqrt[2]{\frac{x}{100}} = B(\sqrt[2]{\frac{x}{10,000}} = B(\sqrt[2]{\frac{x}{1,000,000}} = \ldots .\] Therefore, $B(\sqrt[2]{x}) = B(\sqrt[2]{7.777 \dots})$ . We can evaluate $B(\sqrt[2]{7.777 \dots})$ , the leading digit of $\sqrt[2]{7.777 \dots}$ , to be $2$ . Therefore, $f(2) = 2$
Similarly, we have \[B(\sqrt[3]{x}) = B(\sqrt[3]{\frac{x}{1,000}} = B(\sqrt[3]{\frac{x}{1,000,000}} = B(\sqrt[3]{\frac{x}{1,000,000,000}} = \ldots .\] Therefore, $B(\sqrt[3]{x}) = B(\sqrt[3]{7.777 \ldots})$ . We know $B(\sqrt[3]{7.777 \ldots}) = 1$ , so $f(3) = 1$
Next, \[B(\sqrt[4]{x}) = B(\sqrt[4]{7.777 \ldots})\] and $B(\sqrt[4]{7.777 \ldots}) = 1$ , so $f(4) = 1$
We also have \[B(\sqrt[5]{x}) = B(\sqrt[5]{777.777 \ldots})\] and $B(\sqrt[5]{777.777 \ldots}) = 3$ , so $f(5) = 3$
Finally, \[B(\sqrt[6]{x}) = B(\sqrt[6]{7.777 \ldots})\] and $B(\sqrt[4]{7.777 \ldots}) = 1$ , so $f(6) = 1$
We have that $f(2)+f(3)+f(4)+f(5)+f(6) = 2+1+1+3+1 = \boxed{8}$
| 8
|
1,966
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_19
| 3
|
Let $N$ be the positive integer $7777\ldots777$ , a $313$ -digit number where each digit is a $7$ . Let $f(r)$ be the leading digit of the $r{ }$ th root of $N$ . What is \[f(2) + f(3) + f(4) + f(5)+ f(6)?\] $(\textbf{A})\: 8\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 11\qquad(\textbf{D}) \: 22\qquad(\textbf{E}) \: 29$
|
Since $7777..7$ is a $313$ digit number and $\sqrt {7}$ is around $2.5$ , we have $f(2)$ is $2$ $f(3)$ is the same story, so $f(3)$ is $1$ . It is the same as $f(4)$ as well, so $f(4)$ is also $1$ . However, $313$ is $3$ mod $5$ , so we need to take the 5th root of $777$ , which is between $3$ and $4$ , and therefore, $f(5)$ is $3$ $f(6)$ is the same as $f(4)$ , since it is $1$ more than a multiple of $6$ . Therefore, we have $2+1+1+3+1$ which is $\boxed{8}$
| 8
|
1,967
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_19
| 4
|
Let $N$ be the positive integer $7777\ldots777$ , a $313$ -digit number where each digit is a $7$ . Let $f(r)$ be the leading digit of the $r{ }$ th root of $N$ . What is \[f(2) + f(3) + f(4) + f(5)+ f(6)?\] $(\textbf{A})\: 8\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 11\qquad(\textbf{D}) \: 22\qquad(\textbf{E}) \: 29$
|
First, we compute $f \left( 2 \right)$
Because $N > 4 \cdot 10^{312}$ $\sqrt{N} > 2 \cdot 10^{156}$ .
Because $N < 9 \cdot 10^{312}$ $\sqrt{N} < 3 \cdot 10^{156}$
Therefore, $f \left( 2 \right) = 2$
Second, we compute $f \left( 3 \right)$
Because $N > 1 \cdot 10^{312}$ $\sqrt[3]{N} > 1 \cdot 10^{104}$ .
Because $N < 8 \cdot 10^{312}$ $\sqrt[3]{N} < 2 \cdot 10^{104}$
Therefore, $f \left( 3 \right) = 1$
Third, we compute $f \left( 4 \right)$
Because $N > 1 \cdot 10^{312}$ $\sqrt[4]{N} > 1 \cdot 10^{78}$ .
Because $N < 16 \cdot 10^{312}$ $\sqrt[4]{N} < 2 \cdot 10^{78}$
Therefore, $f \left( 4 \right) = 1$
Fourth, we compute $f \left( 5 \right)$
Because $N > 3^5 \cdot 10^{310}$ $\sqrt[5]{N} > 3 \cdot 10^{62}$ .
Because $N < 4^5 \cdot 10^{310}$ $\sqrt[5]{N} < 4 \cdot 10^{62}$
Therefore, $f \left( 5 \right) = 3$
Fifth, we compute $f \left( 6 \right)$
Because $N > 1 \cdot 10^{312}$ $\sqrt[6]{N} > 1 \cdot 10^{52}$ .
Because $N < 2^6 \cdot 10^{312}$ $\sqrt[6]{N} < 2 \cdot 10^{52}$
Therefore, $f \left( 6 \right) = 1$
Therefore, \begin{align*} f \left( 2 \right) + f \left( 3 \right) + f \left( 4 \right) + f \left( 5 \right) + f \left( 6 \right) & = 2 + 1 + 1 + 3 + 1 \\ & = 8 . \end{align*}
Therefore, the answer is $\boxed{8}$
| 8
|
1,968
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_21
| 1
|
Regular polygons with $5,6,7,$ and $8$ sides are inscribed in the same circle. No two of the polygons share a vertex, and no three of their sides intersect at a common point. At how many points inside the circle do two of their sides intersect?
$(\textbf{A})\: 52\qquad(\textbf{B}) \: 56\qquad(\textbf{C}) \: 60\qquad(\textbf{D}) \: 64\qquad(\textbf{E}) \: 68$
|
Imagine we have $2$ regular polygons with $m$ and $n$ sides and $m>n$ inscribed in a circle without sharing a vertex. We see that each side of the polygon with $n$ sides (the polygon with fewer sides) will be intersected twice.
(We can see this because to have a vertex of the $m$ -gon on an arc subtended by a side of the $n$ -gon, there will be one intersection to “enter” the arc and one to “exit” the arc. ~KingRavi)
This means that we will end up with $2$ times the number of sides in the polygon with fewer sides.
If we have polygons with $5,$ $6,$ $7,$ and $8$ sides, we need to consider each possible pair of polygons and count their intersections.
Throughout $6$ of these pairs, the $5$ -sided polygon has the least number of sides $3$ times, the $6$ -sided polygon has the least number of sides $2$ times, and the $7$ -sided polygon has the least number of sides $1$ time.
Therefore the number of intersections is $2\cdot(3\cdot5+2\cdot6+1\cdot7)=\boxed{68}$
| 68
|
1,969
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_22
| 1
|
For each integer $n\geq 2$ , let $S_n$ be the sum of all products $jk$ , where $j$ and $k$ are integers and $1\leq j<k\leq n$ . What is the sum of the 10 least values of $n$ such that $S_n$ is divisible by $3$
$\textbf{(A)}\ 196\qquad\textbf{(B)}\ 197\qquad\textbf{(C)}\ 198\qquad\textbf{(D)}\ 199\qquad\textbf{(E)}\ 200$
|
To get from $S_n$ to $S_{n+1}$ , we add $1(n+1)+2(n+1)+\cdots +n(n+1)=(1+2+\cdots +n)(n+1)=\frac{n(n+1)^2}{2}$
Now, we can look at the different values of $n$ mod $3$ . For $n\equiv 0\pmod{3}$ and $n\equiv 2\pmod{3}$ , then we have $\frac{n(n+1)^2}{2}\equiv 0\pmod{3}$ . However, for $n\equiv 1\pmod{3}$ , we have \[\frac{1\cdot {2}^2}{2}\equiv 2\pmod{3}.\]
Clearly, $S_2\equiv 2\pmod{3}.$ Using the above result, we have $S_5\equiv 1\pmod{3}$ , and $S_8$ $S_9$ , and $S_{10}$ are all divisible by $3$ . After $3\cdot 3=9$ , we have $S_{17}$ $S_{18}$ , and $S_{19}$ all divisible by $3$ , as well as $S_{26}, S_{27}, S_{28}$ , and $S_{35}$ . Thus, our answer is $8+9+10+17+18+19+26+27+28+35=27+54+81+35=162+35=\boxed{197} .$ -BorealBear
| 197
|
1,970
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_22
| 2
|
For each integer $n\geq 2$ , let $S_n$ be the sum of all products $jk$ , where $j$ and $k$ are integers and $1\leq j<k\leq n$ . What is the sum of the 10 least values of $n$ such that $S_n$ is divisible by $3$
$\textbf{(A)}\ 196\qquad\textbf{(B)}\ 197\qquad\textbf{(C)}\ 198\qquad\textbf{(D)}\ 199\qquad\textbf{(E)}\ 200$
|
Since we have a wonky function, we start by trying out some small cases and see what happens. If $j$ is $1$ and $k$ is $2$ , then there is one case. We have $2$ mod $3$ for this case. If $N$ is $3$ , we have $1 \cdot 2 + 1 \cdot 3 + 2 \cdot 3$ which is still $2$ mod $3$ . If $N$ is $4$ , we have to add $1 \cdot 4 + 2 \cdot 4 + 3 \cdot 4$ which is a multiple of $3$ , meaning that we are still at $2$ mod $3$ . If we try a few more cases, we find that when $N$ is $8$ , we get a multiple of $3$ . When $N$ is $9$ , we are adding $0$ mod $3$ , and therefore, we are still at a multiple of $3$
When $N$ is $10$ , then we get $0$ mod $3$ $10(1+2+3+...+9)$ which is $10$ times a multiple of $3$ . Therefore, we have another multiple of $3$ . When $N$ is $11$ , so we have $2$ mod $3$ . So, every time we have $-1$ mod $9$ $0$ mod $9$ , and $1$ mod $9$ , we always have a multiple of $3$ . Think about it: When $N$ is $1$ , it will have to be $0 \cdot 1$ , so it is a multiple of $3$ . Therefore, our numbers are $8, 9, 10, 17, 18, 19, 26, 27, 28, 35$ . Adding the numbers up, we get $\boxed{197}$
| 197
|
1,971
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_22
| 3
|
For each integer $n\geq 2$ , let $S_n$ be the sum of all products $jk$ , where $j$ and $k$ are integers and $1\leq j<k\leq n$ . What is the sum of the 10 least values of $n$ such that $S_n$ is divisible by $3$
$\textbf{(A)}\ 196\qquad\textbf{(B)}\ 197\qquad\textbf{(C)}\ 198\qquad\textbf{(D)}\ 199\qquad\textbf{(E)}\ 200$
|
Denote $A_{n, <} = \left\{ \left( j , k \right) : 1 \leq j < k \leq n \right\}$ $A_{n, >} = \left\{ \left( j , k \right) : 1 \leq k < j \leq n \right\}$ and $A_{n, =} = \left\{ \left( j , k \right) : 1 \leq j = k \leq n \right\}$
Hence, $\sum_{\left( j , k \right) \in A_{n,<}} jk = \sum_{\left( j , k \right) \in A_{n,>}} jk = S_n$
Therefore, \begin{align*} S_n & = \frac{1}{2} \left( \sum_{\left( j , k \right) \in A_{n,<}} jk = \sum_{\left( j , k \right) \in A_{n,>}} jk \right) \\ & = \frac{1}{2} \left( \sum_{1 \leq j, k \leq n} jk - {\left( j , k \right) \in A_{n,=}} jk \right) \\ & = \frac{1}{2} \left( \sum_{j=1}^n \sum_{k=1}^n jk - \sum_{j=1}^n j^2 \right) \\ & = \frac{1}{2} \left( \frac{n^2 \left( n + 1 \right)^2}{4} - \frac{n \left( n + 1 \right) \left( 2 n + 1 \right) }{6} \right) \\ & = \frac{\left( n - 1 \right) n \left( n + 1 \right) \left( 3 n + 2 \right)}{24} . \end{align*}
Hence, $S_n$ is divisible by 3 if and only if $\left( n - 1 \right) n \left( n + 1 \right) \left( 3 n + 2 \right)$ is divisible by $24 \cdot 3 = 8 \cdot 9$
First, $\left( n - 1 \right) n \left( n + 1 \right) \left( 3 n + 2 \right)$ is always divisible by 8. Otherwise, $S_n$ is not even an integer.
Second, we find conditions for $n$ , such that $\left( n - 1 \right) n \left( n + 1 \right) \left( 3 n + 2 \right)$ is divisible by 9.
Because $3 n + 2$ is not divisible by 3, it cannot be divisible by 9.
Hence, we need to find conditions for $n$ , such that $\left( n - 1 \right) n \left( n + 1 \right)$ is divisible by 9.
This holds of $n \equiv 0, \pm 1 \pmod{9}$
Therefore, the 10 least values of $n$ such that $\left( n - 1 \right) n \left( n + 1 \right)$ is divisible by 9 (equivalently, $S_n$ is divisible by 3) are 8, 9, 10, 17, 18, 19, 26, 27, 28, 35.
Their sum is 197.
Therefore, the answer is $\boxed{197}$
| 197
|
1,972
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_24
| 1
|
A cube is constructed from $4$ white unit cubes and $4$ blue unit cubes. How many different ways are there to construct the $2 \times 2 \times 2$ cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.)
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 11$
|
This problem is about the relationships between the white unit cubes and the blue unit cubes, which can be solved by Graph Theory . We use a Planar Graph to represent the larger cube. Each vertex of the planar graph represents a unit cube. Each edge of the planar graph represents a shared face between $2$ neighboring unit cubes. Each face of the planar graph represents a face of the larger cube.
Now the problem becomes a Graph Coloring problem of how many ways to assign $4$ vertices blue and $4$ vertices white with Topological Equivalence . For example, in Figure $(1)$ , as long as the $4$ blue vertices belong to the same planar graph face, the different planar graphs are considered to be topological equivalent by rotating the larger cube.
Topology.jpg
Here is how the $4$ blue unit cubes are arranged:
In Figure $(1)$ $4$ blue unit cubes are on the same layer (horizontal or vertical).
In Figure $(2)$ $4$ blue unit cubes are in $T$ shape.
In Figure $(3)$ and $(4)$ $4$ blue unit cubes are in $S$ shape.
In Figure $(5)$ $3$ blue unit cubes are in $L$ shape, and the other is isolated without a shared face.
In Figure $(6)$ $2$ pairs of neighboring blue unit cubes are isolated from each other without a shared face.
In Figure $(7)$ $4$ blue unit cubes are isolated from each other without a shared face.
So the answer is $\boxed{7}$
| 7
|
1,973
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_24
| 2
|
A cube is constructed from $4$ white unit cubes and $4$ blue unit cubes. How many different ways are there to construct the $2 \times 2 \times 2$ cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.)
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 11$
|
Let’s split the cube into two layers; a bottom and top. Note that there must be four of each color, so however many number of one color are in the bottom, there will be four minus that number of the color on the top. We do casework on the color distribution of the bottom layer.
Case 1: 4, 0
In this case, there is only one possibility for the top layer - all of the other color - $\binom{4}{4}$ . Therefore there is 1 construction from this case.
Case 2: 3, 1
In this case, the top layer has four possibilities, because there are four different ways to arrange it so that it also has a 3, 1 color distribution - $\binom{4}{3}$ . Therefore there are 4 constructions from this case.
Case 3: 2, 2
In this case, the top layer has six possibilities of arrangement - $\binom{4}{2}$ . However, having adjacent colors one way can be rotated to having adjacent colors any other way, so there is only one construction for the adjacent colors subcase and similarly, only one for the diagonal color subcase. Therefore the total number of constructions for this case is 2.
The total number of constructions for the cube is thus $1+4+2=7=\boxed{7}$
| 7
|
1,974
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_24
| 3
|
A cube is constructed from $4$ white unit cubes and $4$ blue unit cubes. How many different ways are there to construct the $2 \times 2 \times 2$ cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.)
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 11$
|
Divide the $2 \times 2 \times 2$ cube into two layers, say, front and back. Any possible construction can be rotated such that the front layer has the same or greater number of white cubes than blue cubes, so we only need to count the number of cases given that is true.
Therefore, our answer is $6+1+0=\boxed{7}$
| 7
|
1,975
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_24
| 4
|
A cube is constructed from $4$ white unit cubes and $4$ blue unit cubes. How many different ways are there to construct the $2 \times 2 \times 2$ cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.)
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 11$
|
Burnside lemma is used to counting number of orbit where the element on the same orbit can be achieved by the defined operator, naming rotation, reflection and etc.
The fact for Burnside lemma are
1. the sum of stablizer on the same orbit equals to the # of operators;
2. the sum of stablizer can be counted as $fix(g)$
3. the sum of the $fix(g)/|G|$ equals the # of orbit.
Let's start with defining the operator for a cube,
1. $\textbf{e (identity)}$
For identity, there are $\frac{8!}{4!4!} = 70$
2. ${\bf r^{1}, r^{2}, r^{3}}$ to be the rotation axis along three pair of opposite face,
each contains $r^{i}_{90}, r^{i}_{180}, r^{i}_{270}$ where $i= 1, 2, 3$
$fix(r^{i}_{90}) = fix(r^{i}_{270}) = 2\cdot1 = 2$
$fix(r^{i}_{180}) = \frac{4!}{2!\cdot2!} = 6$
therefore $fix(\bf r^{i}) = 2+2+6 = 10$ , and $fix(\bf r^{1})+fix(\bf r^{2})+fix(\bf r^{3}) = 30$
3. ${\bf r^{4}, r^{5}, r^{6}, r^{7}}$ to the rotation axis along four cube diagnals.
each contains $r^{i}_{120}, r^{i}_{240}$ where $i= 4, 5, 6, 7$
$fix(r^{i}_{120}) = fix(r^{i}_{240}) = 2\cdot1\cdot2\cdot1 = 4$
therefore $fix(\bf r^{i}) = 4+4 = 8$ , and $fix(\bf r^{4})+fix(\bf r^{5})+fix(\bf r^{6})+fix(\bf r^{7}) = 32$
4. ${\bf r^{8}, r^{9}, r^{10}, r^{11}, r^{12}, r^{13}}$ to be the rotation axis along 6 pairs of diagnally opposite sides
each contains $r^{i}_{180}$ where $i= 8, 9, 10, 11, 12, 13$
$fix(r^{i}_{180}) = \frac{4!}{2!\cdot2!} = 6$
therefore $fix(\bf r^{8})+fix(\bf r^{9})+fix(\bf r^{10})+fix(\bf r^{11})+fix(\bf r^{12})+fix(\bf r^{13}) = 36$
5. The total number of operators are
$|G| = 1 + 3\cdot3 + 4\cdot2 + 6\cdot1 = 24$
Based on 1, 2, 3, 4 the total number of stablizer is $70 + 30 + 32 + 36 = 168$
therefore the number of orbit $= \frac{168}{G=24} = \boxed{7}$
| 7
|
1,976
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_24
| 5
|
A cube is constructed from $4$ white unit cubes and $4$ blue unit cubes. How many different ways are there to construct the $2 \times 2 \times 2$ cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.)
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 11$
|
Since rotations of a single pattern are considered indistinguishable, we can assume that the forward upper right corner of the 2-by-2-by-2 cube is a blue cube (since we can always rotate the big cube to place a blue cube in that spot).
Once we've assigned this cube to be blue, we note that 3 1-by-1-by-1 cubes share a side with it, 3 1-by-1-by-1 cubes share a corner with it, and 1 1-by-1-by-1 cube does not touch the assigned cube at all, from the perspective of someone who can only see the cube's faces. We'll call the first 3 "adjacent", the second 3 "cornering", and the last one "opposite."
We can use a little bit of intuition to confirm that due to the rotation condition, we should treat all adjacents as indistinguishable, all cornerings as indistinguishable, and of course the opposite one is unique from all the others. Thus, we can list out like so (keeping in mind that there are 3 adjacents, 3 cornerings, and 1 opposite, and that we're choosing the positions of the remaining 3 blue cubes):
OAA, OAC, OCC, CCC, CAA, CCA, AAA.
This gives the answer to be $\boxed{7}$
| 7
|
1,977
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_25
| 1
|
A rectangle with side lengths $1{ }$ and $3,$ a square with side length $1,$ and a rectangle $R$ are inscribed inside a larger square as shown. The sum of all possible values for the area of $R$ can be written in the form $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n?$ [asy] size(8cm); draw((0,0)--(10,0)); draw((0,0)--(0,10)); draw((10,0)--(10,10)); draw((0,10)--(10,10)); draw((1,6)--(0,9)); draw((0,9)--(3,10)); draw((3,10)--(4,7)); draw((4,7)--(1,6)); draw((0,3)--(1,6)); draw((1,6)--(10,3)); draw((10,3)--(9,0)); draw((9,0)--(0,3)); draw((6,13/3)--(10,22/3)); draw((10,22/3)--(8,10)); draw((8,10)--(4,7)); draw((4,7)--(6,13/3)); label("$3$",(9/2,3/2),N); label("$3$",(11/2,9/2),S); label("$1$",(1/2,9/2),E); label("$1$",(19/2,3/2),W); label("$1$",(1/2,15/2),E); label("$1$",(3/2,19/2),S); label("$1$",(5/2,13/2),N); label("$1$",(7/2,17/2),W); label("$R$",(7,43/6),W); [/asy] $(\textbf{A})\: 14\qquad(\textbf{B}) \: 23\qquad(\textbf{C}) \: 46\qquad(\textbf{D}) \: 59\qquad(\textbf{E}) \: 67$
|
We use Image:2021_AMC_10B_(Nov)_Problem_25,_sol.png to facilitate our analysis.
Denote $\angle AFE = \theta$ . Thus, $\angle FIB = \angle CEF = \angle EKG = \angle KLC = \theta$
Hence, \begin{align*} AB & = AF + FB \\ & = EF \cos \angle EFA + IF \sin \angle FIB \\ & = 3 \cos \theta + \sin \theta , \end{align*} and \begin{align*} AC & = AE + EK + KC \\ & = EF \sin \angle EFA + EG \cos \angle CEG + KG \cos \angle EKG + KL \sin \angle CLK \\ & = 3 \sin \theta + \cos \theta + \cos \theta + \sin \theta \\ & = 2 \cos \theta + 4 \sin \theta . \end{align*}
Because $AB = AC$ $3 \cos \theta + \sin \theta = 2 \cos \theta + 4 \sin \theta$ .
Hence, $\tan \theta = \frac{1}{3}$ .
Hence, $\sin \theta = \frac{1}{\sqrt{10}}$ and $\cos \theta = \frac{3}{\sqrt{10}}$
Hence, $AB = AC = BD = CD = \sqrt{10}$
Now, we put the graph to a coordinate plane by setting point $A$ as the origin, putting $AB$ in the $x$ -axis and $AC$ on the $y$ -axis.
Hence, $A = \left( 0 , 0 \right)$ $B = \left( \sqrt{10} , 0 \right)$ $C = \left( 0 , \sqrt{10} \right)$ $D = \left( \sqrt{10} , \sqrt{10} \right)$ $E = \left( 0 , \frac{3}{\sqrt{10}} \right)$ $F = \left( \frac{9}{\sqrt{10}} , 0 \right)$ $G = \left( \frac{1}{\sqrt{10}} , \frac{6}{\sqrt{10}} \right)$ $H = \left( \frac{4}{\sqrt{10}} , \frac{7}{\sqrt{10}} \right)$ $I = \left( \sqrt{10} , \frac{3}{\sqrt{10}} \right)$
Denote $P = \left( \frac{10 - u}{\sqrt{10}} , \sqrt{10} \right)$ $Q = \left( \sqrt{10} , \frac{10 - v}{\sqrt{10}} \right)$
Because $HPQJ$ is a rectangle, $HP \perp PQ$ . Hence, $m_{HP} m_{PQ} = -1$ .
We have $m_{HP} = \frac{3}{6 - u}$ and $m_{PQ} = - \frac{v}{u}$ .
Hence, \[ \frac{3}{6 - u} \cdot \left( - \frac{v}{u} \right) = - 1 . \hspace{1cm} (1) \]
Because $HPQJ$ is a rectangle, $x_J + x_P = x_H + x_Q$ and $y_J + y_P = y_H + y_Q$ .
Hence, $J = \left( \frac{4 + u}{\sqrt{10}} , \frac{7 - v}{\sqrt{10}} \right)$
The equation of line $GI$ is \begin{align*} y & = \frac{\frac{3}{\sqrt{10}} - \frac{6}{\sqrt{10}}}{\sqrt{10} - \frac{1}{\sqrt{10}}} \left( x - \frac{1}{\sqrt{10}} \right) + \frac{6}{\sqrt{10}} \\ & = - \frac{x}{3} + \frac{19}{3 \sqrt{10}} . \end{align*} Because point $J$ is on line $GI$ , plugging the coordinates of $J$ into the equation of line $GI$ , we get \[ \frac{7 - v}{\sqrt{10}} = - \frac{\frac{4 + u}{\sqrt{10}}}{3} + \frac{19}{3 \sqrt{10}} . \hspace{1cm} (2) \]
By solving Equations (1) and (2), we get $\left( u , v \right) = \left( 2 , \frac{8}{3} \right)$ or $\left( 3 , 3 \right)$
Case 1: $\left( u , v \right) = \left( 2 , \frac{8}{3} \right)$
We have $P = \left( \frac{8}{\sqrt{10}} , \sqrt{10} \right)$ and $Q = \left( \sqrt{10} , \frac{22}{3 \sqrt{10}} \right)$ .
Thus, $HP = \frac{5}{\sqrt{10}}$ and $PQ = \frac{10}{3\sqrt{10}}$
Therefore, ${\rm Area} \ R = HP \cdot PQ = \frac{5}{3}$
Case 2: $\left( u , v \right) = \left( 3 , 3 \right)$
We have $P = \left( \frac{7}{\sqrt{10}} , \sqrt{10} \right)$ and $Q = \left( \sqrt{10} , \frac{7}{ \sqrt{10}} \right)$ .
Thus, $HP = \frac{3 \sqrt{2}}{\sqrt{10}}$ and $PQ = \frac{3 \sqrt{2}}{\sqrt{10}}$
Therefore, ${\rm Area} \ R = HP \cdot PQ = \frac{9}{5}$
Putting these two cases together, the sum of all possible values of the area of $R$ is $\frac{5}{3} + \frac{9}{5} = \frac{52}{15}$
Therefore, the answer is $\boxed{67}$
| 67
|
1,978
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_25
| 2
|
A rectangle with side lengths $1{ }$ and $3,$ a square with side length $1,$ and a rectangle $R$ are inscribed inside a larger square as shown. The sum of all possible values for the area of $R$ can be written in the form $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n?$ [asy] size(8cm); draw((0,0)--(10,0)); draw((0,0)--(0,10)); draw((10,0)--(10,10)); draw((0,10)--(10,10)); draw((1,6)--(0,9)); draw((0,9)--(3,10)); draw((3,10)--(4,7)); draw((4,7)--(1,6)); draw((0,3)--(1,6)); draw((1,6)--(10,3)); draw((10,3)--(9,0)); draw((9,0)--(0,3)); draw((6,13/3)--(10,22/3)); draw((10,22/3)--(8,10)); draw((8,10)--(4,7)); draw((4,7)--(6,13/3)); label("$3$",(9/2,3/2),N); label("$3$",(11/2,9/2),S); label("$1$",(1/2,9/2),E); label("$1$",(19/2,3/2),W); label("$1$",(1/2,15/2),E); label("$1$",(3/2,19/2),S); label("$1$",(5/2,13/2),N); label("$1$",(7/2,17/2),W); label("$R$",(7,43/6),W); [/asy] $(\textbf{A})\: 14\qquad(\textbf{B}) \: 23\qquad(\textbf{C}) \: 46\qquad(\textbf{D}) \: 59\qquad(\textbf{E}) \: 67$
|
[asy] size(8cm); label("D",(0,0),SW); label("A",(0,10),NW); label("B",(10,10),NE); label("C",(10,0),SE); label("H",(1,5.7),SE); label("O",(0,6),W); label("I",(0,3),W); draw((0,0)--(10,0)); draw((0,0)--(0,10)); draw((10,0)--(10,10)); draw((0,10)--(10,10)); draw((1,6)--(0,9)); draw((0,9)--(3,10)); draw((3,10)--(4,7)); draw((4,7)--(1,6)); draw((0,3)--(1,6)); draw((1,6)--(10,3)); draw((10,3)--(9,0)); draw((9,0)--(0,3)); draw((6,13/3)--(10,22/3)); draw((10,22/3)--(8,10)); draw((8,10)--(4,7)); draw((4,7)--(6,13/3)); draw((1,6)--(0,6)); label("$3$",(9/2,3/2),N); label("$3$",(11/2,9/2),S); label("$1$",(1/2,9/2),E); label("$1$",(19/2,3/2),W); label("$1$",(1/2,15/2),E); draw((0.3,6)--(0.3,5.7)--(0,5.7)); label("$1$",(3/2,19/2),S); label("$1$",(5/2,13/2),N); label("$1$",(7/2,17/2),W); label("$R$",(7,43/6),W); label("E",(0,9),W); label("F",(3,10),N); label("P",(4,10),N); label("L",(8,10),N); draw((4,7)--(4,10)); draw((4,9.7)--(4.3,9.7)--(4.3,10)); label("G",(3.9,6.8),S); label("N",(6.2,4.1),S); label("M",(10,22/3),E); label("Q",(10,13/3),E); draw((10,13/3)--(6,13/3)); draw((9.7,13/3)--(9.7,4.033333333333333333333333)--(10,4.03333333333333333333333333)); label("K",(10,3),E); label("J",(9,0),S); [/asy] We will scale every number up by a factor of $\sqrt{10}.$ This implies our final area will be $\frac{1}{10}$ of the answer we receive.
We have \[FAE\sim EOH \sim IOH\sim JDI\sim KCJ \sim NQK\sim GPF.\] Let $AE=a$ and $FA=b.$ We have \[FP=AE=OH=JC=\frac13ID=a\] and \[PG=AF=EO=OI=KC=\frac13 DJ=b\] As $ABCD$ is a square, we have $AD=DC$ or \[a+2b+3a=3b+a \Rightarrow 3a=b.\] Since $a^2+b^2=10,$ we have \[a=1,b=3.\] We have $\triangle GPL\cong \triangle MQN$ which implies \[MQ=GP=3.\] Denote $QK=x.$ As $\triangle NQK\sim \triangle JDI,$ we have $NQ=3x.$
We have \begin{align*}BM&=BC-(CK+QK+MQ)\\ &=4-x.\end{align*}
In addition, \begin{align*}LB&=AB-(AF+FP+PL)\\&=6-3x.\end{align*}
Since $\triangle LBM\sim \triangle MQN,$ we have \[\frac{LB}{BM}=\frac{MQ}{QN}\Rightarrow \frac{6-3x}{4-x}=\frac{3}{3x}=\frac{1}{x}.\] Simplifying we have \[3x^2-7x+4=0 \Rightarrow x=\frac43, 1.\] We have \begin{align*}[GLMN]&=MN\cdot LM\\ &= 3\sqrt{x^2+1}\cdot \sqrt{10x^2-44x+52}.\end{align*}
Plugging in $x=1,$ we have $[GLMN]=18.$
Plugging in $x=\frac43,$ we have $[GLMN]=\frac{50}{3}.$
The sum of the two possible $R$ s is \[\frac{1}{10}\cdot\frac{104}{3}=\frac{52}{15}.\] Hence, $52+15=\boxed{67}.$
| 67
|
1,979
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_1
| 1
|
What is the value of \[(2^2-2)-(3^2-3)+(4^2-4)\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 12$
|
$(4-2)-(9-3)+(16-4)=2-6+12=8.$ This corresponds to $\boxed{8}.$
| 8
|
1,980
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_1
| 2
|
What is the value of \[(2^2-2)-(3^2-3)+(4^2-4)\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 12$
|
We have \begin{align*} \left(2^2-2\right)-\left(3^2-3\right)+\left(4^2-4\right) &= 2(2-1)-3(3-1)+4(4-1) \\ &= 2(1)-3(2)+4(3) \\ &= 2-6+12 \\ &= \boxed{8} ~MRENTHUSIASM
| 8
|
1,981
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_1
| 3
|
What is the value of \[(2^2-2)-(3^2-3)+(4^2-4)\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 12$
|
We have \begin{align*} (2^2-2)-(3^2-3)+(4^2-4) &= 2^2+4^2-3^2-2+3-4 \\ &=2^2+(4-3)(4+3)-3 \\ &=2^2+7-3=2^2+4 \\ &=4\cdot 2 \\ &=\boxed{8} PureSwag
| 8
|
1,982
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_1
| 4
|
What is the value of \[(2^2-2)-(3^2-3)+(4^2-4)\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 12$
|
Using the difference of squares, we have \[(2-\sqrt{2})(2+\sqrt{2}) - (3-\sqrt{3})(3+\sqrt{3}) + (4-2)(4+2).\] Knowing that $\sqrt{2} \approx 1.41$ and $\sqrt{3} \approx 1.73,$ we get \[(2-\sqrt{2})(2+\sqrt{2}) - (3-\sqrt{3})(3+\sqrt{3}) + (4-2)(4+2) \approx 0.59\cdot 3.41 -1.26\cdot 4.73 + 2 \cdot 6 =8.0521,\] which is closest to $\boxed{8}.$
| 8
|
1,983
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_2
| 1
|
Portia's high school has $3$ times as many students as Lara's high school. The two high schools have a total of $2600$ students. How many students does Portia's high school have?
$\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050$
|
The following system of equations can be formed with $P$ representing the number of students in Portia's high school and $L$ representing the number of students in Lara's high school: \begin{align*} P&=3L, \\ P+L&=2600. \end{align*} Substituting $P=3L$ gives $4L=2600.$ Solving for $L$ gives $L=650.$ Since we need to find $P,$ we multiply $650$ by $3$ to get $P=\boxed{1950}.$
| 950
|
1,984
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_2
| 2
|
Portia's high school has $3$ times as many students as Lara's high school. The two high schools have a total of $2600$ students. How many students does Portia's high school have?
$\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050$
|
Suppose Lara's high school has $x$ students, so Portia's high school has $3x$ students. We have $x+3x=2600,$ or $4x=2600.$ The answer is \[3x=2600\cdot\frac 34=650\cdot3=\boxed{1950}.\]
| 950
|
1,985
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_2
| 3
|
Portia's high school has $3$ times as many students as Lara's high school. The two high schools have a total of $2600$ students. How many students does Portia's high school have?
$\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050$
|
Clearly, $2600$ is $4$ times the number of students in Lara's high school. Therefore, Lara's high school has $2600\div4=650$ students, and Portia's high school has $650\cdot3=\boxed{1950}$ students.
| 950
|
1,986
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_2
| 4
|
Portia's high school has $3$ times as many students as Lara's high school. The two high schools have a total of $2600$ students. How many students does Portia's high school have?
$\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050$
|
The number of students in Portia's high school must be a multiple of $3.$ This eliminates $\textbf{(B)},\textbf{(D)},$ and $\textbf{(E)}.$ Since $\textbf{(A)}$ is too small (as it is clear that $600+\frac{600}{3}<2600$ ), we are left with $\boxed{1950}.$
| 950
|
1,987
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_3
| 1
|
The sum of two natural numbers is $17402$ . One of the two numbers is divisible by $10$ . If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(E)} ~15426$
|
The units digit of a multiple of $10$ will always be $0$ . We add a $0$ whenever we multiply by $10$ . So, removing the units digit is equal to dividing by $10$
Let the smaller number (the one we get after removing the units digit) be $a$ . This means the bigger number would be $10a$
We know the sum is $10a+a = 11a$ so $11a=17402$ . So $a=1582$ . The difference is $10a-a = 9a$ . So, the answer is $9(1582) = \boxed{14238}$
| 238
|
1,988
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_3
| 2
|
The sum of two natural numbers is $17402$ . One of the two numbers is divisible by $10$ . If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(E)} ~15426$
|
Since the unit's place of a multiple of $10$ is $0$ , the other integer must end with a $2$ , for both integers sum up to a number ending in a $2$ . Thus, the unit's place of the difference must be $10-2=8$ , and the only answer choice that ends with an $8$ is $\boxed{14238}$
| 238
|
1,989
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_3
| 3
|
The sum of two natural numbers is $17402$ . One of the two numbers is divisible by $10$ . If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(E)} ~15426$
|
Let the larger number be $\underline{ABCD0}.$ It follows that the smaller number is $\underline{ABCD}.$ Adding vertically, we have \[\begin{array}{cccccc} & A & B & C & D & 0 \\ +\quad & & A & B & C & D \\ \hline & & & & & \\ [-2.5ex] & 1 & 7 & 4 & 0 & 2 \\ \end{array}\] Working from right to left, we get \[D=2\implies C=8 \implies B=5 \implies A=1.\] The larger number is $15820$ and the smaller number is $1582.$ Their difference is $15820-1582=\boxed{14238}.$
| 238
|
1,990
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_3
| 4
|
The sum of two natural numbers is $17402$ . One of the two numbers is divisible by $10$ . If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(E)} ~15426$
|
We know that the larger number has a units digit of $0$ since it is divisible by 10. If $D$ is the ten's digit of the larger number, then $D$ is the units digit of the smaller number. Since the sum of the natural numbers has a unit's digit of $2$ $D=2$
The units digit of the larger number is $0$ and the units digit of the smaller number is $2$ , so the positive difference between the numbers is 8. There is only one answer choice that has this units digit, and that is $\boxed{14238}.$
| 238
|
1,991
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_4
| 1
|
A cart rolls down a hill, travelling $5$ inches the first second and accelerating so that during each successive $1$ -second time interval, it travels $7$ inches more than during the previous $1$ -second interval. The cart takes $30$ seconds to reach the bottom of the hill. How far, in inches, does it travel?
$\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242$
|
Since \[\mathrm{Distance}=\mathrm{Speed}\cdot\mathrm{Time},\] we seek the sum \[5\cdot1+12\cdot1+19\cdot1+26\cdot1+\cdots=5+12+19+26+\cdots,\] in which there are $30$ terms.
The last term is $5+7\cdot(30-1)=208.$ Therefore, the requested sum is \[5+12+19+26+\cdots+208=\frac{5+208}{2}\cdot30=\boxed{3195}.\] ~MRENTHUSIASM
| 195
|
1,992
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_4
| 3
|
A cart rolls down a hill, travelling $5$ inches the first second and accelerating so that during each successive $1$ -second time interval, it travels $7$ inches more than during the previous $1$ -second interval. The cart takes $30$ seconds to reach the bottom of the hill. How far, in inches, does it travel?
$\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242$
|
From the $30$ -term sum \[5+12+19+26+\cdots\] in Solution 1, taking modulo $10$ gives \[5+12+19+26+\cdots \equiv 3\cdot(5+2+9+6+3+0+7+4+1+8) = 3\cdot45\equiv5 \pmod{10}.\] The only answer choices congruent to $5$ modulo $10$ are $\textbf{(A)}$ and $\textbf{(D)}.$ By a quick estimation, $\textbf{(A)}$ is too small, leaving us with $\boxed{3195}.$
| 195
|
1,993
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_4
| 4
|
A cart rolls down a hill, travelling $5$ inches the first second and accelerating so that during each successive $1$ -second time interval, it travels $7$ inches more than during the previous $1$ -second interval. The cart takes $30$ seconds to reach the bottom of the hill. How far, in inches, does it travel?
$\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242$
|
This problem can be solved by physics method. This method is perhaps the quickest too and shows the beauty of the problem. The average speed increases $7 \ \text{in/s}$ per second. So, the acceleration $a=7 \ \text{in/s\textsuperscript{2}}.$ The average speed of the first second is $5 \ \text{in/s}.$ We can know the initial velocity $v_0=5-0.5\cdot7=1.5.$ The displacement at $t=30$ is \[s=\frac{1}{2}at^2+v_0t=\frac{1}{2}\cdot7\cdot30^2+1.5\cdot30= \boxed{3195}.\] ~Bran_Qin
| 195
|
1,994
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_8
| 1
|
When a student multiplied the number $66$ by the repeating decimal, \[\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},\] where $a$ and $b$ are digits, he did not notice the notation and just multiplied $66$ times $\underline{1}.\underline{a} \ \underline{b}.$ Later he found that his answer is $0.5$ less than the correct answer. What is the $2$ -digit number $\underline{a} \ \underline{b}?$
$\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75$
|
We are given that $66\Bigl(\underline{1}.\overline{\underline{a} \ \underline{b}}\Bigr)-0.5=66\Bigl(\underline{1}.\underline{a} \ \underline{b}\Bigr),$ from which \begin{align*} 66\Bigl(\underline{1}.\overline{\underline{a} \ \underline{b}}\Bigr)-66\Bigl(\underline{1}.\underline{a} \ \underline{b}\Bigr)&=0.5 \\ 66\Bigl(\underline{1}.\overline{\underline{a} \ \underline{b}} - \underline{1}.\underline{a} \ \underline{b}\Bigr)&=0.5 \\ 66\Bigl(\underline{0}.\underline{0} \ \underline{0} \ \overline{\underline{a} \ \underline{b}}\Bigr)&=0.5 \\ 66\left(\frac{1}{100}\cdot\underline{0}.\overline{\underline{a} \ \underline{b}}\right)&=\frac12 \\ \underline{0}.\overline{\underline{a} \ \underline{b}}&=\frac{25}{33} \\ \underline{0}.\overline{\underline{a} \ \underline{b}}&=0.\overline{75} \\ \underline{a} \ \underline{b}&=\boxed{75} ~MRENTHUSIASM
| 75
|
1,995
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_8
| 2
|
When a student multiplied the number $66$ by the repeating decimal, \[\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},\] where $a$ and $b$ are digits, he did not notice the notation and just multiplied $66$ times $\underline{1}.\underline{a} \ \underline{b}.$ Later he found that his answer is $0.5$ less than the correct answer. What is the $2$ -digit number $\underline{a} \ \underline{b}?$
$\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75$
|
It is known that $\underline{0}.\overline{\underline{a} \ \underline{b}}=\frac{\underline{a} \ \underline{b}}{99}$ and $\underline{0}.\underline{a} \ \underline{b}=\frac{\underline{a} \ \underline{b}}{100}.$
Let $x=\underline{a} \ \underline{b}.$ We have \[66\biggl(1+\frac{x}{99}\biggr)-66\biggl(1+\frac{x}{100}\biggr)=0.5.\] Expanding and simplifying give $\frac{x}{150}=0.5,$ so $x=\boxed{75}.$
| 75
|
1,996
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_8
| 3
|
When a student multiplied the number $66$ by the repeating decimal, \[\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},\] where $a$ and $b$ are digits, he did not notice the notation and just multiplied $66$ times $\underline{1}.\underline{a} \ \underline{b}.$ Later he found that his answer is $0.5$ less than the correct answer. What is the $2$ -digit number $\underline{a} \ \underline{b}?$
$\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75$
|
We have \[66 \cdot \left(1 + \frac{10a+b}{100}\right) + \frac{1}{2} = 66 \cdot \left(1+ \frac{10a+b}{99}\right).\] Expanding both sides, we have \[66 + \frac{33(10a+b)}{50} + \frac{1}{2} = 66 + \frac{2(10a+b)}{3}.\] Subtracting $66$ from both sides, we have \[\frac{33(10a+b)}{50} + \frac{1}{2} = \frac{2(10a+b)}{3}.\] Multiplying both sides by $50 \cdot 3 = 150,$ we have \[99(10a+b) + 75 = 100(10a+b).\] Thus, the answer is $10a+b = \boxed{75}.$
| 75
|
1,997
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_9
| 3
|
What is the least possible value of $(xy-1)^2+(x+y)^2$ for real numbers $x$ and $y$
$\textbf{(A)} ~0\qquad\textbf{(B)} ~\frac{1}{4}\qquad\textbf{(C)} ~\frac{1}{2} \qquad\textbf{(D)} ~1 \qquad\textbf{(E)} ~2$
|
Like solution 1, expand and simplify the original equation to $x^2+y^2+x^2y^2+1$ and let $f(x, y) = x^2+y^2+x^2y^2+1$ . To find local extrema, find where $\nabla f(x, y) = \boldsymbol{0}$ . First, find the first partial derivative with respect to x and y and find where they are $0$ \[\frac{\partial f}{\partial x} = 2x + 2xy^{2} = 2x(1 + y^{2}) = 0 \implies x = 0\] \[\frac{\partial f}{\partial y} = 2y + 2yx^{2} = 2y(1 + x^{2}) = 0 \implies y = 0\] Thus, there is a local extremum at $(0, 0)$ . Because this is the only extremum, we can assume that this is a minimum because the problem asks for the minimum (though this can also be proven using the partial second derivative test) and the global minimum since it's the only minimum, meaning $f(0, 0)$ is the minimum of $f(x, y)$ . Plugging $(0, 0)$ into $f(x, y)$ , we find 1 $\implies \boxed{1}$
| 1
|
1,998
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_11
| 1
|
For which of the following integers $b$ is the base- $b$ number $2021_b - 221_b$ not divisible by $3$
$\textbf{(A)} ~3 \qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~6\qquad\textbf{(D)} ~7\qquad\textbf{(E)} ~8$
|
We have \begin{align*} 2021_b - 221_b &= (2021_b - 21_b) - (221_b - 21_b) \\ &= 2000_b - 200_b \\ &= 2b^3 - 2b^2 \\ &= 2b^2(b-1), \end{align*} which is divisible by $3$ unless $b\equiv2\pmod{3}.$ The only choice congruent to $2$ modulo $3$ is $\boxed{8}.$
| 8
|
1,999
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_11
| 2
|
For which of the following integers $b$ is the base- $b$ number $2021_b - 221_b$ not divisible by $3$
$\textbf{(A)} ~3 \qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~6\qquad\textbf{(D)} ~7\qquad\textbf{(E)} ~8$
|
Vertically subtracting $2021_b - 221_b,$ we see that the ones place becomes $0,$ and so does the $b^1$ place. Then, we perform a carry (make sure the carry is in base $b$ ). Let $b-2 = A.$ Then, we have our final number as \[1A00_b.\] Now, when expanding, we see that this number is simply $b^3 - (b - 2)^2.$
Now, notice that the final number will only be congruent to \[b^3-(b-2)^2\equiv0\pmod{3}.\] If either $b\equiv0\pmod{3},$ or if $b\equiv1\pmod{3}$ (because note that $(b - 2)^2$ would become $\equiv1\pmod{3},$ and $b^3$ would become $\equiv1\pmod{3}$ as well, and therefore the final expression would become $1-1\equiv0\pmod{3}.$ Therefore, $b$ must be $\equiv2\pmod{3}.$ Among the answers, only $8$ is $\equiv2\pmod{3},$ and therefore our answer is $\boxed{8}.$
| 8
|
2,000
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_12
| 1
|
Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are $3$ cm and $6$ cm. Into each cone is dropped a spherical marble of radius $1$ cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?
[asy] size(350); defaultpen(linewidth(0.8)); real h1 = 10, r = 3.1, s=0.75; pair P = (r,h1), Q = (-r,h1), Pp = s * P, Qp = s * Q; path e = ellipse((0,h1),r,0.9), ep = ellipse((0,h1*s),r*s,0.9); draw(ellipse(origin,r*(s-0.1),0.8)); fill(ep,gray(0.8)); fill(origin--Pp--Qp--cycle,gray(0.8)); draw((-r,h1)--(0,0)--(r,h1)^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(0,Qp.y),Arrows(size=8)); draw(origin--(0,12),linetype("4 4")); draw(origin--(r*(s-0.1),0)); label("$3$",(-0.9,h1*s),N,fontsize(10)); real h2 = 7.5, r = 6, s=0.6, d = 14; pair P = (d+r-0.05,h2-0.15), Q = (d-r+0.05,h2-0.15), Pp = s * P + (1-s)*(d,0), Qp = s * Q + (1-s)*(d,0); path e = ellipse((d,h2),r,1), ep = ellipse((d,h2*s+0.09),r*s,1); draw(ellipse((d,0),r*(s-0.1),0.8)); fill(ep,gray(0.8)); fill((d,0)--Pp--Qp--cycle,gray(0.8)); draw(P--(d,0)--Q^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(d,Qp.y),Arrows(size=8)); draw((d,0)--(d,10),linetype("4 4")); draw((d,0)--(d+r*(s-0.1),0)); label("$6$",(d-r/4,h2*s-0.06),N,fontsize(10)); [/asy]
$\textbf{(A) }1:1 \qquad \textbf{(B) }47:43 \qquad \textbf{(C) }2:1 \qquad \textbf{(D) }40:13 \qquad \textbf{(E) }4:1$
|
The heights of the cones are not given, so suppose the heights are very large (i.e. tending towards infinity) in order to approximate the cones as cylinders with base radii $3$ and $6$ and infinitely large height. Then the base area of the wide cylinder is $4$ times that of the narrow cylinder. Since we are dropping a ball of the same volume into each cylinder, the water level in the narrow cone/cylinder should rise $\boxed{4}$ times as much.
| 4
|
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