AMC10 2021 A

$(4-2)-(9-3)+(16-4)=2-6+12=8.$ This corresponds to $\boxed{\textbf{(D) } 8}.$

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 $P=\boxed{\textbf{(C)} ~1950}.$

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 $9(1582) = \boxed{\textbf{(D)} ~14{,}238}$.

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 \[5+12+19+26+\cdots+208=\frac{5+208}{2}\cdot30=\boxed{\textbf{(D)} ~3195}.\] Recall that to find the sum of an arithmetic series, we take the average of the first and last terms, then multiply by the number of terms: \[\mathrm{Sum}=\frac{\mathrm{First}+\mathrm{Last}}{2}\cdot\mathrm{Count}.\]

The total score of the class is $8k,$ and the total score of the $12$ quizzes is $12\cdot14=168.$ Therefore, for the remaining quizzes ($k-12$ of them), the total score is $8k-168.$ Their mean score is $\boxed{\textbf{(B)} ~\frac{8k-168}{k-12}}.$

Let $2d$ miles be the distance from the trailhead to the fire tower, where $d>0.$ When Chantal meets Jean, the two have traveled for \[\frac d4 + \frac d2 + \frac d3 = d\left(\frac 14 + \frac 12 + \frac 13\right) =d\left(\frac{3}{12} + \frac{6}{12} + \frac{4}{12}\right)=\frac{13}{12}d\] hours. At that point, Jean has traveled for $d$ miles, so his average speed is $\frac{d}{\frac{13}{12}d}=\boxed{\textbf{(A)} ~\frac{12}{13}}$ miles per hour.

To see that choices (A), (B), (C), and (E) do not follow from the given information, consider the following two snakes that may be part of Tom's collection. One snake is happy but not purple and can both add and subtract. The second is purple but not happy and can neither add nor subtract. Then each of the three bulleted statements is true, but each of these choices is false. To show that answer choice (D) is correct, first observe that the third bulleted statement is equivalent to "Snakes that can add also can subtract." The second bulleted statement is equivalent to "Snakes that can subtract are not purple." The three bulleted statements combined then lead to the conclusion (D) Happy snakes are not purple.

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{\textbf{(E) }75}. \end{align*}

Expanding, we get that the expression is $x^2+2xy+y^2+x^2y^2-2xy+1$ or $x^2+y^2+x^2y^2+1$. By the Trivial Inequality (all squares are nonnegative) the minimum value for this is $\boxed{\textbf{(D)} ~1}$, which can be achieved at $x=y=0$.

By multiplying the entire equation by $3-2=1$, all the terms will simplify by difference of squares, and the final answer is $\boxed{\textbf{(C)} ~3^{128}-2^{128}}$. Additionally, we could also multiply the entire equation (we can let it be equal to $x$) by $2-3=-1$. The terms again simplify by difference of squares. This time, we get $-x=2^{128}-3^{128} \Rightarrow x=3^{128}-2^{128}$. Both solutions yield the same answer. Note: Also notice when you multiply it by the first pair $(2+3)$, it immediately factors. Notice how it "domino" effects to the others, ultimately collapsing into $\boxed{\textbf{(C)} ~3^{128}-2^{128}}$.

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 $\boxed{\textbf{(E)} ~8}.$

Initial Scenario Let the heights of the narrow cone and the wide cone be $h_1$ and $h_2,$ respectively. We have the following table: \[\begin{array}{cccccc} & \textbf{Base Radius} & \textbf{Height} & & \textbf{Volume} & \\ [2ex] \textbf{Narrow Cone} & 3 & h_1 & & \frac13\pi(3)^2h_1=3\pi h_1 & \\ [2ex] \textbf{Wide Cone} & 6 & h_2 & & \hspace{2mm}\frac13\pi(6)^2h_2=12\pi h_2 & \end{array}\] Equating the volumes gives $3\pi h_1=12\pi h_2,$ which simplifies to $\frac{h_1}{h_2}=4.$ Furthermore, by similar triangles: - For the narrow cone, the ratio of the base radius to the height is $\frac{3}{h_1},$ which always remains constant. - For the wide cone, the ratio of the base radius to the height is $\frac{6}{h_2},$ which always remains constant. Two solutions follow from here: Final Scenario For the narrow cone and the wide cone, let their base radii be $3x$ and $6y$ (for some $x,y>1$), respectively. By the similar triangles discussed above, their heights must be $h_1x$ and $h_2y,$ respectively. We have the following table: \[\begin{array}{cccccc} & \textbf{Base Radius} & \textbf{Height} & & \textbf{Volume} & \\ [2ex] \textbf{Narrow Cone} & 3x & h_1x & & \frac13\pi(3x)^2(h_1x)=3\pi h_1 x^3 & \\ [2ex] \textbf{Wide Cone} & 6y & h_2y & & \hspace{2.0625mm}\frac13\pi(6y)^2(h_2y)=12\pi h_2 y^3 & \end{array}\] Recall that $\frac{h_1}{h_2}=4.$ Equating the volumes gives $3\pi h_1 x^3=12\pi h_2 y^3,$ which simplifies to $x^3=y^3,$ or $x=y.$ Finally, the requested ratio is \[\frac{h_1 x - h_1}{h_2 y - h_2}=\frac{h_1 (x-1)}{h_2 (y-1)}=\frac{h_1}{h_2}=\boxed{\textbf{(E) }4:1}.\] Remarks 1. This solution uses the following property of fractions: For unequal positive numbers $a,b,c$ and $d,$ if $\frac ab = \frac cd = k,$ then $\frac{a\pm c}{b\pm d}=\frac{bk\pm dk}{b\pm d}=\frac{(b\pm d)k}{b\pm d}=k.$ 2. This solution shows that, regardless of the shape or the volume of the solid dropped into each cone, the requested ratio stays the same as long as the solid sinks to the bottom and is completely submerged without spilling any liquid.

Drawing the tetrahedron out and testing side lengths, we realize that the $\triangle ACD, \triangle ABC,$ and $\triangle ABD$ are right triangles by the Converse of the Pythagorean Theorem. It is now easy to calculate the volume of the tetrahedron using the formula for the volume of a pyramid. If we take $\triangle ADC$ as the base, then $\overline{AB}$ must be the altitude. The volume of tetrahedron $ABCD$ is $\dfrac{1}{3} \cdot \dfrac{3 \cdot 4}{2} \cdot 2=\boxed{\textbf{(C)} ~4}.$

By Vieta's formulas, the sum of the six roots is $10$ and the product of the six roots is $16$. By inspection, we see the roots are $1, 1, 2, 2, 2,$ and $2$, so the function is $(z-1)^2(z-2)^4=(z^2-2z+1)(z^4-8z^3+24z^2-32z+16)$. Therefore, calculating just the $z^3$ terms, we get $B = -32 - 48 - 8 = \boxed{\textbf{(A) }{-}88}$.

Visualizing the two curves, we realize they are both parabolas with the same axis of symmetry. WLOG the first equation is above the second, since order doesn't matter. Then $C>A$ and $B>D$. Therefore the number of ways to choose the four integers is $\tbinom{6}{2}\tbinom{4}{2}=90$, and the answer is $\boxed{\textbf{(C) }90}$.

There are $1+2+..+199+200=\frac{(200)(201)}{2}=20100$ numbers in total. Let the median be $k$. We want to find the median $k$ such that \[\frac{k(k+1)}{2}=20100/2,\] or \[k(k+1)=20100.\] Note that $\sqrt{20100} \approx 142$. Plugging this value in as $k$ gives \[\frac{1}{2}(142)(143)=10153.\] $10153-142<10050$, so $142$ is the $152$nd and $153$rd numbers, and hence, our desired answer. $\fbox{(C) 142}$. Note that we can derive $\sqrt{20100} \approx 142$ through the formula \[\sqrt{n} = \sqrt{a+b} \approx \sqrt{a} + \frac{b}{2\sqrt{a} + 1},\] where $a$ is a perfect square less than or equal to $n$. We set $a$ to $19600$, so $\sqrt{a} = 140$, and $b = 500$. We then have $n \approx 140 + \frac{500}{2(140)+1} \approx 142$. ~approximation by ciceronii Note by Fasolinka (use answer choices): Once you know that the answer is in the 140s range (200,000 is around 14^2 times 10^2) by the approximation, it is highly improbable for the answer to be anything but C.

Angle chasing* reveals that $\triangle BPC\sim\triangle BDA$, therefore \[2=\frac{BD}{BP}=\frac{AB}{BC}=\frac{AB}{43},\] or $AB=86$. Additional angle chasing shows that $\triangle ABO\sim\triangle CDO$, therefore \[2=\frac{AB}{CD}=\frac{BO}{OD}=\frac{BP+11}{BP-11},\] or $BP=33$ and $BD=66$. Since $\triangle ADB$ is right, the Pythagorean theorem implies that \[AD=\sqrt{86^2-66^2}=4\sqrt{190}.\] The answer is $4+190=\boxed{\textbf{(D) }194}$. - Angle Chasing: If we set $\angle DBC = \alpha$, then we know that $\angle DCB = 180^\circ-2\alpha$ because $\triangle DBC$ is isosceles. Then, $\angle BCP = 90^\circ-\alpha$, so $\angle BPC$ is a right angle. Because $\angle BDC = \alpha$ and $\overline{AB}\parallel\overline{DC}$, we conclude that $\angle ABD = \alpha$ too. Lastly, because $\triangle BPC$ and $\triangle BDA$ are both right triangles, they are similar by AA.

From the answer choices, note that \begin{align*} f(25)&=f\left(\frac{25}{11}\cdot11\right) \\ &=f\left(\frac{25}{11}\right)+f(11) \\ &=f\left(\frac{25}{11}\right)+11. \end{align*} On the other hand, we have \begin{align*} f(25)&=f(5\cdot5) \\ &=f(5)+f(5) \\ &=5+5 \\ &=10. \end{align*} Equating the expressions for $f(25)$ produces \[f\left(\frac{25}{11}\right)+11=10,\] from which $f\left(\frac{25}{11}\right)=-1.$ Therefore, the answer is $\boxed{\textbf{(E) }\frac{25}{11}}.$ Remark Similarly, we can find the outputs of $f$ at the inputs of the other answer choices: \begin{alignat*}{10} &\textbf{(A)} \qquad && f\left(\frac{17}{32}\right) \quad && = \quad && 7 \\ &\textbf{(B)} \qquad && f\left(\frac{11}{16}\right) \quad && = \quad && 3 \\ &\textbf{(C)} \qquad && f\left(\frac{7}{9}\right) \quad && = \quad && 1 \\ &\textbf{(D)} \qquad && f\left(\frac{7}{6}\right) \quad && = \quad && 2 \end{alignat*} Alternatively, refer to Solutions 2 and 4 for the full processes.

In order to attack this problem, we can use casework on the sign of $|x-y|$ and $|x+y|$. Case 1: $|x-y|=x-y, |x+y|=x+y$ Substituting and simplifying, we have $x^2-6x+y^2=0$, i.e. $(x-3)^2+y^2=3^2$, which gives us a circle of radius $3$ centered at $(3,0)$. Case 2: $|x-y|=y-x, |x+y|=x+y$ Substituting and simplifying again, we have $x^2+y^2-6y=0$, i.e. $x^2+(y-3)^2=3^2$. This gives us a circle of radius $3$ centered at $(0,3)$. Case 3: $|x-y|=x-y, |x+y|=-x-y$ Doing the same process as before, we have $x^2+y^2+6y=0$, i.e. $x^2+(y+3)^2=3^2$. This gives us a circle of radius $3$ centered at $(0,-3)$. Case 4: $|x-y|=y-x, |x+y|=-x-y$ One last time: we have $x^2+y^2+6x=0$, i.e. $(x+3)^2+y^2=3^2$. This gives us a circle of radius $3$ centered at $(-3,0)$. After combining all the cases and drawing them on the Cartesian Plane, this is what the diagram looks like: Now, the area of the shaded region is just a square with side length $6$ with four semicircles of radius $3$. The area is $6\cdot6+4\cdot \frac{9\pi}{2} = 36+18\pi$. The answer is $36+18$ which is $\boxed{\textbf{(E) }54}$

By symmetry with respect to $3,$ note that $(x_1,x_2,x_3,x_4,x_5)$ is a valid sequence if and only if $(6-x_1,6-x_2,6-x_3,6-x_4,6-x_5)$ is a valid sequence. We enumerate the valid sequences that start with $1,2,31,$ or $32,$ as shown below: There are $16$ valid sequences that start with $1,2,31,$ or $32.$ By symmetry, there are $16$ valid sequences that start with $5,4,35,$ or $34.$ So, the answer is $16+16=\boxed{\textbf{(D)} ~32}.$

Let $P,Q,R,X,Y,$ and $Z$ be the intersections $\overleftrightarrow{AB}\cap\overleftrightarrow{CD},\overleftrightarrow{CD}\cap\overleftrightarrow{EF},\overleftrightarrow{EF}\cap\overleftrightarrow{AB},\overleftrightarrow{BC}\cap\overleftrightarrow{DE},\overleftrightarrow{DE}\cap\overleftrightarrow{FA},$ and $\overleftrightarrow{FA}\cap\overleftrightarrow{BC},$ respectively. The sum of the interior angles of any hexagon is $720^\circ.$ Since hexagon $ABCDEF$ is equiangular, each of its interior angles is $720^\circ\div6=120^\circ.$ By angle chasing, we conclude that the interior angles of $\triangle PBC,\triangle QDE,\triangle RFA,\triangle XCD,\triangle YEF,$ and $\triangle ZAB$ are all $60^\circ.$ Therefore, these triangles are all equilateral triangles, from which $\triangle PQR$ and $\triangle XYZ$ are both equilateral triangles. We are given that \begin{alignat*}{8} [PQR]&=\frac{\sqrt{3}}{4}\cdot PQ^2&&=192\sqrt3, \\ [XYZ]&=\frac{\sqrt{3}}{4}\cdot YZ^2&&=324\sqrt3, \end{alignat*} so we get $PQ=16\sqrt3$ and $YZ=36,$ respectively. By equilateral triangles and segment addition, we find the perimeter of hexagon $ABCDEF:$ \begin{align*} AB+BC+CD+DE+EF+FA&=AZ+PC+CD+DQ+YF+FA \\ &=(YF+FA+AZ)+(PC+CD+DQ) \\ &=YZ+PQ \\ &=36+16\sqrt{3}. \end{align*} Finally, the answer is $36+16+3=\boxed{\textbf{(C)} ~55}.$

Suppose the roommate took sheets $a$ through $b$, or equivalently, page numbers $2a-1$ through $2b$. Because there are $(2b-2a+2)$ numbers taken, \[\frac{(2a-1+2b)(2b-2a+2)}{2}+19(50-(2b-2a+2))=\frac{50\cdot51}{2}\] \[\implies (2a+2b-39)(b-a+1)=\frac{50\cdot13}{2}=25\cdot13.\] The first possible solution that comes to mind is if $2a+2b-39=25, b-a+1=13 \implies a+b=32, b-a=12$, which indeed works, giving $b=22$ and $a=10$. The answer is $22-10+1=\boxed{\textbf{(B)} ~13}$.

We will use complementary counting. First, the frog can go left with probability $\frac14$. We observe symmetry, so our final answer will be multiplied by 4 for the 4 directions, and since $4\cdot \frac14=1$, we will ignore the leading probability. From the left, she either goes left to another edge ($\frac14$) or back to the center ($\frac14$). Time for some casework. $\textbf{Case 1:}$ She goes back to the center. Now, she can go in any 4 directions, and then has 2 options from that edge. This gives $\frac12$. --End case 1 $\textbf{Case 2:}$ She goes to another edge (rightmost). Subcase 1: She goes back to the left edge. She now has 2 places to go, giving $\frac12$ Subcase 2: She goes to the center. Now any move works. $\frac14 \cdot \frac12 + \frac14 \cdot 1=\frac18 + \frac 14=\frac38$ for this case. --End case 2 She goes back to the center in Case 1 with probability $\frac14$, and to the right edge with probability $\frac14$ So, our answer is $\frac14 \cdot \frac12 + \frac14 \cdot \frac38=\frac14 (\frac12+\frac38)=\frac14 \cdot \frac78 = \frac7{32}$ But, don't forget complementary counting. So, we get $1-\frac7{32}=\frac{25}{32} \implies \boxed{D}$.

The cases for $(x+ay)^2 = 4a^2$ are $x+ay = \pm2a,$ or two parallel lines. We rearrange each case and construct the table below: The cases for $(ax-y)^2 = a^2$ are $ax-y=\pm a,$ or two parallel lines. We rearrange each case and construct the table below: Since the slopes of intersecting lines $(1)\cap(1*), (1)\cap(2*), (2)\cap(1*),$ and $(2)\cap(2*)$ are negative reciprocals, we get four right angles, from which the quadrilateral is a rectangle. Two solutions follow from here: Recall that for constants $A,B,C_1$ and $C_2,$ the distance $d$ between parallel lines $\begin{cases} Ax+By+C_1=0 \\ Ax+By+C_2=0 \end{cases}$ is \[d=\frac{\left|C_2-C_1\right|}{\sqrt{A^2+B^2}}.\] From this formula: - The distance between lines $(1)$ and $(2)$ is $\frac{4a}{\sqrt{1+a^2}},$ the length of this rectangle. - The distance between lines $(1*)$ and $(2*)$ is $\frac{2a}{\sqrt{a^2+1}},$ the width of this rectangle. The area we seek is \[\frac{4a}{\sqrt{1+a^2}}\cdot\frac{2a}{\sqrt{a^2+1}}=\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.\]

Call the different colors A,B,C. There are $3!=6$ ways to rearrange these colors to these three letters, so $6$ must be multiplied after the letters are permuted in the grid. WLOG assume that A is in the center. This means that there are $4+2=6$ ways to arrange A,B, and C in the grid, and there are $6$ ways to rearrange the colors. Therefore, there are $6\cdot6=36$ ways in total, which is $\boxed{\textbf{(E)} ~36}$.