problem_id int64 1 978 | question stringlengths 86 2.11k | source stringlengths 19 76 | solution stringlengths 94 14.7k | asymptote_code stringlengths 44 17.8k | solution_image_url stringclasses 16
values |
|---|---|---|---|---|---|
784 | Carl decided to fence in his rectangular garden. He bought $20$ fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly $4$ yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the sho... | 2016 AMC 10B Problem 11 | If the dimensions are $4a\times 4b$, then one side will have $a+1$ posts (including corners) and the other $b+1$ (including corners).
The total number of posts is $2(a+b)=20$.
This diagram represents the number of posts around the garden.
We solve the system $\begin{cases}b+1=2(a+1)\\a+b=10\end{cases}$ to get $a=... | // Block 1
size(7cm);fill((0,0)--(5,0)--(5,7)--(0,7)--cycle,lightgreen);
for(int i=0;i<5;++i)dot((i,0),red);for(int i=0;i<7;++i)dot((5,i),blue);dot((5,7));
draw(arc((0,0),.5,-90,-270)--arc((4,0),.5,90,-90)--cycle,gray+dotted);
draw(arc((5,0),.5,-180,0)--arc((5,6),.5,0,180)--cycle,gray+dotted);
draw((0,-1)--(5,-1),Arrow... | [] |
785 | How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$, the line $y=-0.1$ and the line $x=5.1?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}... | 2016 AMC 10B Problem 14 | The region is a right triangle which contains the following lattice points:
$(0,0); (1,0)\rightarrow(1,3); (2,0)\rightarrow(2,6); (3,0)\rightarrow(3,9); (4,0)\rightarrow(4,12); (5,0)\rightarrow(5,15)$
Squares $1\times 1$:
Suppose that the top-right corner is $(x,y)$, with $2\le x\le 5$. Then to include all other cor... | // Block 1
size(10cm);
for(int i=0;i<6;++i)for(int j=0;j<=3*i;++j)dot((i,j));
draw((0,-1)--(0,16),EndArrow);draw((-1,0)--(6,0),EndArrow);
draw((-.5,-pi/2)--(5.2,5.2*pi),gray);draw((-1,-.1)--(6,-.1)^^(5.1,-1)--(5.1,16),gray);
// Block 2
size(10cm); for(int i=0;i<6;++i)for(int j=0;j<=3*i;++j)dot((i,j)); draw((0,-1)--(0,1... | [] |
786 | All the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a $3\times3$ array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to $18$. What is the number in the center?
$\textbf{(A)}\ 5\qquad\te... | 2016 AMC 10B Problem 15 | Consecutive numbers share an edge. That means that it is possible to walk from $1$ to $9$ by single steps north, south, east, or west. Consequently, the squares in the diagram with different shades have different parity:
But since there are only four even numbers in the set, the five darker squares must contain the odd... | // Block 1
size(4cm);
for(int i=0;i<3;++i)for(int j=0;j<3;++j)filldraw(box((i,j),(i+1,j+1)),gray((i+j)%2*.2+.7));
// Block 2
size(4cm); for(int i=0;i<3;++i)for(int j=0;j<3;++j)filldraw(box((i,j),(i+1,j+1)),gray((i+j)%2*.2+.7)); | [] |
787 | Rectangle $ABCD$ has $AB=5$ and $BC=4$. Point $E$ lies on $\overline{AB}$ so that $EB=1$, point $G$ lies on $\overline{BC}$ so that $CG=1$, and point $F$ lies on $\overline{CD}$ so that $DF=2$. Segments $\overline{AG}$ and $\overline{AC}$ intersect $\overline{EF}$ at $Q$ and $P$, respectively. What is the value of $\fr... | 2016 AMC 10B Problem 19 | First, we will define point $D$ as the origin. Then, we will find the equations of the following three lines: $AG$, $AC$, and $EF$. The slopes of these lines are $-\frac{3}{5}$, $-\frac{4}{5}$, and $2$, respectively. Next, we will find the equations of $AG$, $AC$, and $EF$. They are as follows:
\[AG = f(x) = -\frac... | // Block 1
pair A1=(2,0),A2=(4,4);
pair B1=(0,4),B2=(5,1);
pair C1=(5,0),C2=(0,4);
pair D1=(20/7,0),D2=(20/7,12/7);
pair E1=(40/13,0),E2=(40/13,28/13);
pair F1=(4,0),F2=(4,4);
draw(A1--A2);
draw(B1--B2);
draw(C1--C2);
draw(D1--D2,dashed);
draw(E1--E2,dashed);
draw(F1--F2,dashed);
draw((0,0)--B1--(5,4)--C1--cycle);
dot... | [] |
787 | Rectangle $ABCD$ has $AB=5$ and $BC=4$. Point $E$ lies on $\overline{AB}$ so that $EB=1$, point $G$ lies on $\overline{BC}$ so that $CG=1$, and point $F$ lies on $\overline{CD}$ so that $DF=2$. Segments $\overline{AG}$ and $\overline{AC}$ intersect $\overline{EF}$ at $Q$ and $P$, respectively. What is the value of $\fr... | 2016 AMC 10B Problem 19 | Extend $AG$ to intersect $CD$ at $H$. Letting $x=\overline{HC}$, we have that \[\triangle{HCG}\sim\triangle{HDA}\implies \dfrac{\overline{HC}}{\overline{CG}}=\dfrac{\overline{HD}}{\overline{AD}}\implies \dfrac{x}{1}=\dfrac{x+5}{4}\implies x=\dfrac{5}{3}.\]
Then, notice that $\triangle{AEQ}\sim\triangle{HFQ}$ and $\tri... | // Block 1
pair A1=(2,0),A2=(4,4);
pair B1=(0,4),B2=(5,1);
pair C1=(5,0),C2=(0,4);
pair H = (20/3,0);
draw(A1--A2);
draw(B1--B2);
draw(C1--C2);
draw(B1--H);
draw((0,0)--H);
draw((0,0)--B1--(5,4)--C1--cycle);
dot((20/7,12/7));
dot((3.07692307692,2.15384615384));
label("$Q$",(3.07692307692,2.15384615384),N);
label("$P$"... | [] |
787 | Rectangle $ABCD$ has $AB=5$ and $BC=4$. Point $E$ lies on $\overline{AB}$ so that $EB=1$, point $G$ lies on $\overline{BC}$ so that $CG=1$, and point $F$ lies on $\overline{CD}$ so that $DF=2$. Segments $\overline{AG}$ and $\overline{AC}$ intersect $\overline{EF}$ at $Q$ and $P$, respectively. What is the value of $\fr... | 2016 AMC 10B Problem 19 | We will calculate $\frac{EP}{EF}$ using similar triangle, and $\frac{EQ}{EF}$ using ratio of area of $\triangle AEG$ to $\triangle AFG$.
\[\triangle AEP \sim \triangle CFP, \frac{AE}{CF}=\frac{EP}{FP}, \frac{EP}{FP}=\frac{4}{3}, \frac{EP}{EF}=\frac{4}{7}\]
\[[AEG]=\frac{1}{2} \cdot 4\cdot 3=6\]
\[[AFG]=[ABCD]-[ADF]... | // Block 1
pair A1=(2,0),A2=(4,4);
pair B1=(0,4),B2=(5,1),B3=(4,4);
pair C1=(5,0),C2=(0,4),C3=(2,0);
draw(A1--A2);
draw(B1--B2);
draw(B2--B3);
draw(C1--C2);
draw(C2--C3);
draw(A1--B2);
draw((0,0)--B1--(5,4)--C1--cycle);
dot((20/7,12/7));
dot((3.07692307692,2.15384615384));
label("$Q$",(3.07692307692,2.15384615384),N);
... | [] |
787 | Rectangle $ABCD$ has $AB=5$ and $BC=4$. Point $E$ lies on $\overline{AB}$ so that $EB=1$, point $G$ lies on $\overline{BC}$ so that $CG=1$, and point $F$ lies on $\overline{CD}$ so that $DF=2$. Segments $\overline{AG}$ and $\overline{AC}$ intersect $\overline{EF}$ at $Q$ and $P$, respectively. What is the value of $\fr... | 2016 AMC 10B Problem 19 | We will calculate $\frac{PQ}{QE}$ using the ratio of area of $\triangle APG$ to that of $\triangle AEG$.
\[[ACG]=\frac{1}{2} \cdot 5 \cdot 1 = \frac{5}{2}\]
\[[CAB]=\frac{1}{2} \cdot 5 \cdot 4=10\]
\[\triangle AEP \sim \triangle CFP\]
\[\frac{CP}{AP}=\frac{CF}{AE}=\frac{3}{4}\]
\[\frac{CP}{AC}=\frac{3}{7}\]
\[\frac{... | // Block 1
pair A1=(2,0),A2=(4,4);
pair B1=(0,4),B2=(5,1),B3=(20/7,12/7);
pair C1=(5,0),C2=(0,4);
draw(A1--A2);
draw(B1--B2);
draw(C1--C2);
draw(A2--B2);
draw(B2--B3);
draw((0,0)--B1--(5,4)--C1--cycle);
dot((20/7,12/7));
dot((3.07692307692,2.15384615384));
label("$Q$",(3.07692307692,2.15384615384),N);
label("$P$",(20/... | [] |
788 | A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A’(5,6)$. What distance does the origin $O(0,0)$, move under this transformation?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sq... | 2016 AMC 10B Problem 20 | Using analytic geometry, we find that the center of dilation is at $(-4,-6)$ and the coefficient/factor is $1.5$. Then, we see that the origin is $2\sqrt{13}$ from the center, and will be $1.5 \times 2\sqrt{13} = 3\sqrt{13}$ from it afterwards.
Thus, it will move $3\sqrt{13} - 2\sqrt{13} = \boxed{\sqrt{13}}$. | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
/* by adihaya */
import graph; size(13cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ rea... | [] |
789 | What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$
$\textbf{(A)}\ \pi+\sqrt{2}\qquad\textbf{(B)}\ \pi+2\qquad\textbf{(C)}\ \pi+2\sqrt{2}\qquad\textbf{(D)}\ 2\pi+\sqrt{2}\qquad\textbf{(E)}\ 2\pi+2\sqrt{2}$ | 2016 AMC 10B Problem 21 | Without loss of generality (WLOG) note that if a point in the first quadrant satisfies the equation, so do its corresponding points in the other three quadrants. Therefore, we can assume that $x, y \ge 0$, which implies that $|x|=x$ and $|y|=y$, and multiply by $4$ at the end.
We can rearrange the equation to get $x^2... | // Block 1
draw((0,-1.5)--(0,1.5),EndArrow);draw((-1.5,0)--(1.5,0),EndArrow);draw((0,1)--(1,0)--(0,-1)--(-1,0)--cycle,dotted);
for(int i=0;i<4;++i){draw(rotate(i*90,(0,0))*arc((1/2,1/2),sqrt(1/2),-45,135));dot(rotate(i*90,(0,0))*(1/2,1/2));}
// Block 2
draw((0,-1.5)--(0,1.5),EndArrow);draw((-1.5,0)--(1.5,0),EndArrow);d... | [] |
789 | What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$
$\textbf{(A)}\ \pi+\sqrt{2}\qquad\textbf{(B)}\ \pi+2\qquad\textbf{(C)}\ \pi+2\sqrt{2}\qquad\textbf{(D)}\ 2\pi+\sqrt{2}\qquad\textbf{(E)}\ 2\pi+2\sqrt{2}$ | 2016 AMC 10B Problem 21 | We solve with cases. The cases are:
Case 1: $x\geq0, y\geq0.$
Case 2: $x\geq0, y<0.$
Case 3: $x<0, y\geq0.$
Case 4: $x<0, y<0.$
We can quickly realize that the whole figure is symmetrical; so when you figure out the first case, you get the first part is $\left(x-\dfrac12\right)^2+\left(x-\dfrac12\right)^2$ you can fig... | // Block 1
size(400);
import TrigMacros;
rr_cartesian_axes(-10,10,-10,10,complexplane=false, usegrid = true);
draw(circle((4,4), 4*1.41421356237));
draw(circle((4,-4), 4*1.41421356237));
draw(circle((-4,4), 4*1.41421356237));
draw(circle((-4,-4), 4*1.41421356237));
// Block 2
size(400);
import TrigMacros;
rr_carte... | [] |
790 | A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$, $B$ beat $C$, and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ ... | 2016 AMC 10B Problem 22 | There are $10 \cdot 2+1=21$ teams. Any of the $\tbinom{21}3=1330$ sets of three teams must either be a fork (in which one team beat both the others) or a cycle:
But we know that every team beat exactly $10$ other teams, so for each possible $A$ at the head of a fork, there are always exactly $\tbinom{10}2$ choices fo... | // Block 1
size(7cm);label("A",(5,5));label("C",(10,0));label("B",(0,0));draw((4,4)--(1,1),EndArrow);draw((6,4)--(9,1),EndArrow);
label("A",(20,5));label("C",(25,0));label("B",(15,0));draw((19,4)--(16,1),EndArrow);draw((16,0)--(24,0),EndArrow);draw((24,1)--(21,4),EndArrow);
// Block 2
size(7cm);label("A",(5,5));label("... | [] |
790 | A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$, $B$ beat $C$, and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ ... | 2016 AMC 10B Problem 22 | This is a Graph Theory problem with directed graph. There are $21$ teams in total. WLOG, pick team $A$, there are $10$ teams that lost to $A$ and $10$ teams that won over $A$. Call the group of teams that lost to $A$ group $L$, and the group of teams that won over $A$ group $W$.
Any team from group $L$ that won a te... | // Block 1
size(7cm);
label("A",(20,5));
label("Group W",(27,0));
label("Group L",(13,0));
draw((19,4)--(16,1),EndArrow);
draw((16,0)--(24,0),EndArrow);
draw((24,1)--(21,4),EndArrow);
// Block 2
size(7cm); label("A",(20,5)); label("Group W",(27,0)); label("Group L",(13,0)); draw((19,4)--(16,1),EndArrow); draw((16,0)--(... | [] |
791 | In regular hexagon $ABCDEF$, points $W$, $X$, $Y$, and $Z$ are chosen on sides $\overline{BC}$, $\overline{CD}$, $\overline{EF}$, and $\overline{FA}$ respectively, so lines $AB$, $ZW$, $YX$, and $ED$ are parallel and equally spaced. What is the ratio of the area of hexagon $WCXYFZ$ to the area of hexagon $ABCDEF$?
$\t... | 2016 AMC 10B Problem 23 | We draw a diagram to make our work easier:
Assume that $AB$ is of length $1$. Therefore, the area of $ABCDEF$ is $\frac{3\sqrt 3}2$. To find the area of $WCXYFZ$, we draw $\overline{CF}$, and find the area of the trapezoids $WCFZ$ and $CXYF$.
From this, we know that $CF=2$. We also know that the combined heig... | // Block 1
pair A,B,C,D,E,F,W,X,Y,Z;
A=(0,0);
B=(1,0);
C=(3/2,sqrt(3)/2);
D=(1,sqrt(3));
E=(0,sqrt(3));
F=(-1/2,sqrt(3)/2);
X=(4/3,2sqrt(3)/3);
W=(4/3,sqrt(3)/3);
Z=(-1/3,sqrt(3)/3);
Y=(-1/3,2sqrt(3)/3);
draw(A--B--C--D--E--F--cycle);
draw(W--Z);
draw(X--Y);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,ESE);
la... | [] |
791 | In regular hexagon $ABCDEF$, points $W$, $X$, $Y$, and $Z$ are chosen on sides $\overline{BC}$, $\overline{CD}$, $\overline{EF}$, and $\overline{FA}$ respectively, so lines $AB$, $ZW$, $YX$, and $ED$ are parallel and equally spaced. What is the ratio of the area of hexagon $WCXYFZ$ to the area of hexagon $ABCDEF$?
$\t... | 2016 AMC 10B Problem 23 | First, like in the first solution, split the large hexagon into 6 equilateral triangles. Each equilateral triangle can be split into three rows of smaller equilateral triangles. The first row will have one triangle, the second three, the third five. Once you have drawn these lines, it's just a matter of counting triang... | // Block 1
pair A,B,C,D,E,F,W,X,Y,Z;
A=(0,0);
B=(1,0);
C=(3/2,sqrt(3)/2);
D=(1,sqrt(3));
E=(0,sqrt(3));
F=(-1/2,sqrt(3)/2);
W=(4/3,2sqrt(3)/3);
X=(4/3,sqrt(3)/3);
Y=(-1/3,sqrt(3)/3);
Z=(-1/3,2sqrt(3)/3);
draw(A--B--C--D--E--F--cycle);
draw(W--Z);
draw(X--Y);
draw(F--C--B--E--D--A);
label("$A$",A,SW);
label("$B$",B,SE)... | [] |
791 | In regular hexagon $ABCDEF$, points $W$, $X$, $Y$, and $Z$ are chosen on sides $\overline{BC}$, $\overline{CD}$, $\overline{EF}$, and $\overline{FA}$ respectively, so lines $AB$, $ZW$, $YX$, and $ED$ are parallel and equally spaced. What is the ratio of the area of hexagon $WCXYFZ$ to the area of hexagon $ABCDEF$?
$\t... | 2016 AMC 10B Problem 23 | Extend $\overline{EF}$ and $\overline{CD}$ to meet at point $G$, as shown in the diagram. Then $\triangle GZW \sim \triangle GFC$. Then $[GZW] = \left(\dfrac53\right)^2[GED]$ and $[GFC] = 2^2[GED]$. Subtracting $[GED]$, we find that $[EDWZ] = \dfrac{16}{9}[GED]$ and $[EDCF] = 3[GED]$. Subtracting again, we find that \[... | // Block 1
pair A,B,C,D,E,F,W,X,Y,Z;
A=(0,0);
B=(1,0);
C=(3/2,sqrt(3)/2);
D=(1,sqrt(3));
E=(0,sqrt(3));
F=(-1/2,sqrt(3)/2);
W=(4/3,2sqrt(3)/3);
X=(4/3,sqrt(3)/3);
Y=(-1/3,sqrt(3)/3);
Z=(-1/3,2sqrt(3)/3);
pair G = (0.5, sqrt(3)*3/2);
draw(A--B--C--D--E--F--cycle);
draw(W--Z);
draw(X--Y);
draw(E--G--D);
draw(F--C);
labe... | [] |
792 | The ratio of the measures of two acute angles is $5:4$, and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?
$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$ | 2016 AMC 12B Problem 4 | We can visualize the problem like so:
\[5x+1y = 90^\circ = 4x+2y\]
Moving like terms to the same side gets $x = y$, and substituting this back gets $6x = 90^\circ \implies x = \frac{90^\circ}{6} = 15^\circ$, so the sum of the degree measures is $5x + 4x = 9x = 9(15) = \boxed{\textbf{(C)}\ 135}$. ~emerald_block | // Block 1
path b = brace((0,10),(90,10),5);
draw(b);
label("$90^\circ$",b,N);
draw("$5x$",(0,0)--(75,0),N);
draw((75,2.5)--(75,-2.5));
draw("$1y$",(75,0)--(90,0),N);
draw("$4x$",(0,-10)--(60,-10),S);
draw((60,-7.5)--(60,-12.5));
draw("$2y$",(60,-10)--(90,-10),S);
draw((0,5)--(0,-15));
draw((90,5)--(90,-15));
// Blo... | [] |
793 | How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$, the line $y=-0.1$ and the line $x=5.1?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}... | 2016 AMC 12B Problem 11 | Solution by e_power_pi_times_i
Revised by Kinglogic and RJ5303707
(red shows lattice points within the triangle)
If we draw a picture showing the triangle, we see that it would be easier to count the squares vertically and not horizontally. The upper bound is $16$ squares $(y=5.1*\pi)$, and the limit for the $x$-val... | // Block 1
Label l;
l.p=fontsize(8);
xaxis(-1,8,Ticks(l, 1.0));
yaxis(-1,16,Ticks(l, 1.0));
real f(real x)
{
return x * pi;
}
D(graph(f,-1/pi,5.1));
D((5.1,-1)--(5.1,16));
D((-1,-0.1)--(8,-0.1));
for(int x = 0; x < 5.1; ++x) {
for(int y = 0; y < 16; ++y) {
if(x * pi > y) {
D((x,y));
}
}... | [] |
794 | Alice and Bob live $10$ miles apart. One day Alice looks due north from her house and sees an airplane. At the same time Bob looks due west from his house and sees the same airplane. The angle of elevation of the airplane is $30^\circ$ from Alice's position and $60^\circ$ from Bob's position. Which of the following is ... | 2016 AMC 12B Problem 13 | ~credit to mathmaster2012 for original diagram
~BakedPotato66 added/contributed some elements | // Block 1
import graph; usepackage("amsmath"); size(7.2cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -4.3, xmax = 10.1, ymin = -4.44, ymax = 6.3; /* image dime... | [] |
795 | What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$
$\textbf{(A)}\ \pi+\sqrt{2} \qquad\textbf{(B)}\ \pi+2 \qquad\textbf{(C)}\ \pi+2\sqrt{2} \qquad\textbf{(D)}\ 2\pi+\sqrt{2} \qquad\textbf{(E)}\ 2\pi+2\sqrt{2}$ | 2016 AMC 12B Problem 18 | Consider the case when $x \geq 0$, $y \geq 0$.
\[x^2+y^2=x+y\]
\[(x - \frac{1}{2})^2+(y - \frac{1}{2})^2=\frac{1}{2}\]
Notice the circle intersects the axes at points $(0, 1)$ and $(1, 0)$. Find the area of this circle in the first quadrant. The area is made of a semi-circle with radius of $\frac{\sqrt{2}}{2}$ and a t... | draw((0,-1.5)--(0,1.5),EndArrow);draw((-1.5,0)--(1.5,0),EndArrow);draw((0,1)--(1,0)--(0,-1)--(-1,0)--cycle,dotted);draw(arc((1/2,1/2),sqrt(2)/2,135, 315),dotted);for(int i=0;i<4;++i){draw(rotate(i*90,(0,0))*arc((1/2,1/2),sqrt(2)/2,-45,135)); dot(rotate(i*90,(0,0))*(1/2,1/2));} | [] |
796 | A square with side length $x$ is inscribed in a right triangle with sides of length $3$, $4$, and $5$ so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length $y$ is inscribed in another right triangle with sides of length $3$, $4$, and $5$ so that one side of th... | 2017 AMC 10A Problem 21 | Analyze the first right triangle.
Note that $\triangle ABC$ and $\triangle FBE$ are similar, so $\frac{BF}{FE} = \frac{AB}{AC}$. This can be written as $\frac{4-x}{x}=\frac{4}{3}$. Solving, $x = \frac{12}{7}$.
Now we analyze the second triangle.
Similarly, $\triangle A'B'C'$ and $\triangle RB'Q$ are similar, s... | // Block 1
pair A,B,C;
pair D, e, F;
A = (0,0);
B = (4,0);
C = (0,3);
D = (0, 12/7);
e = (12/7 , 12/7);
F = (12/7, 0);
draw(A--B--C--cycle);
draw(D--e--F);
label("$x$", D/2, W);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("$D$", D, W);
label("$E$", e, NE);
label("$F$", F, S);
// Block 2
pair ... | [] |
797 | Sides $\overline{AB}$ and $\overline{AC}$ of equilateral triangle $ABC$ are tangent to a circle at points $B$ and $C$ respectively. What fraction of the area of $\triangle ABC$ lies outside the circle?
$\textbf{(A) } \frac{4\sqrt{3}\pi}{27}-\frac{1}{3}\qquad \textbf{(B) } \frac{\sqrt{3}}{2}-\frac{\pi}{8}\qquad \textbf... | 2017 AMC 10A Problem 22 | Let the radius of the circle be $r$, and let its center be $O$. Since $\overline{AB}$ and $\overline{AC}$ are tangent to circle $O$, then $\angle OBA = \angle OCA = 90^{\circ}$, so $\angle BOC = 120^{\circ}$. Therefore, since $\overline{OB}$ and $\overline{OC}$ are equal to $r$, then (pick your favorite method) $\overl... | // Block 1
real sqrt3 = 1.73205080757;
draw(Circle((4, 4), 4));
draw((4-2*sqrt3,6)--(4,4)--(4+2*sqrt3,6)--(4-2*sqrt3,6)--(4,12)--(4+2*sqrt3,6));
label("A", (4, 12.4));
label("B", (-.3, 6.3));
label("C", (8.3, 6.3));
label("O", (4, 3.4));
// Block 2
real sqrt3 = 1.73205080757; draw(Circle((4, 4), 4)); draw((4-2*sqrt3,6)... | [] |
797 | Sides $\overline{AB}$ and $\overline{AC}$ of equilateral triangle $ABC$ are tangent to a circle at points $B$ and $C$ respectively. What fraction of the area of $\triangle ABC$ lies outside the circle?
$\textbf{(A) } \frac{4\sqrt{3}\pi}{27}-\frac{1}{3}\qquad \textbf{(B) } \frac{\sqrt{3}}{2}-\frac{\pi}{8}\qquad \textbf... | 2017 AMC 10A Problem 22 | (same diagram as Solution 1)
Without the Loss of Generality, let the side length of the triangle be $1$.
Then, the area of the triangle is $\frac{\sqrt{3}}{4}$. We are looking for the area of the portion inside the triangle but outside the circle divided by the area of the triangle. Since $m\angle ABO = 90^{\circ}=m... | // Block 1
real sqrt3 = 1.73205080757;
draw(Circle((4, 4), 4));
draw((4-2*sqrt3,6)--(4,4)--(4+2*sqrt3,6)--(4-2*sqrt3,6)--(4,12)--(4+2*sqrt3,6));
draw((4, 6)--(4, 4));
label("A", (4, 12.4));
label("B", (-.3, 6.3));
label("C", (8.3, 6.3));
label("O", (4, 3.4));
label("D", (4, 6.6));
// Block 2
real sqrt3 = 1.73205080757;... | [] |
798 | Let $f(x) = \sin{x} + 2\cos{x} + 3\tan{x}$, using radian measure for the variable $x$. In what interval does the smallest positive value of $x$ for which $f(x) = 0$ lie?
$\textbf{(A)}\ (0,1) \qquad \textbf{(B)}\ (1, 2) \qquad\textbf{(C)}\ (2, 3) \qquad\textbf{(D)}\ (3, 4) \qquad\textbf{(E)}\ (4,5)$ | 2017 AMC 12A Problem 15 | If you quickly take a moment to sketch the graphs of the three functions, you will see that between $0$ and $\frac{\pi}{2}$ everything is positive, while the positive number created by the sin does not outweigh the negative by the cos and tan function. Upon further examination, it is clear that the positive the tan fun... | // Block 1
Label f;
f.p=fontsize(6);
xaxis(-5,5,Ticks(f, 1.0));
yaxis(-8,8,Ticks(f, 2.0));
real f(real x)
{
return sin(x);
}
draw(graph(f, -5,5));
real g(real x)
{
return 2*cos(x);
}
draw(graph(g, -5,5));
real h(real x)
{
return 3*tan(x);
}
draw(graph(h, -1.2,1.2));
draw(graph(h, 1.94, 4.34));
draw(graph(h, -4.34, -... | [] |
799 | In the figure below, semicircles with centers at $A$ and $B$ and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter $JK$. The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered a... | 2017 AMC 12A Problem 16 | Connect the centers of the tangent circles! (call the center of the large circle $C$)
Notice that we don't even need the circles anymore; thus, draw triangle $\Delta ABP$ with cevian $PC$:
and use Stewart's Theorem:
\[AB \cdot AC \cdot BC + AB \cdot {CP}^2 = AC \cdot {BP}^2 + BC \cdot {AP}^2\]
From what we lear... | // Block 1
size(5cm);
draw(arc((0,0),3,0,180));
draw(arc((2,0),1,0,180));
draw(arc((-1,0),2,0,180));
draw((-3,0)--(3,0));
pair P = (9/7,12/7);
pair A = (-1,0);
pair C = (0,0);
pair B = (2,0);
draw(circle(P,6/7));
dot((-1,0)); dot((2,0)); dot((0,0)); dot(P);
draw((-1,0)--P);
draw((2,0)--P);
draw((0,0)--(9/5,12/5));
labe... | [] |
799 | In the figure below, semicircles with centers at $A$ and $B$ and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter $JK$. The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered a... | 2017 AMC 12A Problem 16 | Like the solution above, connecting the centers of the circles results in triangle $\Delta ABP$ with cevian $PC$. The two triangles $\Delta APC$ and $\Delta ABP$ share angle $A$, which means we can use Law of Cosines to set up a system of 2 equations that solve for $r$ respectively:
$(2 + r)^2 + 1^2 - 2(2 + r)(1)\cos ... | // Block 1
size(5cm);
draw(arc((0,0),3,0,180));
draw(arc((2,0),1,0,180));
draw(arc((-1,0),2,0,180));
draw((-3,0)--(3,0));
pair P = (9/7,12/7);
draw(circle(P,6/7));
dot((-1,0)); dot((2,0)); dot((0,0)); dot(P);
draw((-1,0)--P);
draw((2,0)--P);
draw((0,0)--(9/5,12/5));
// Block 2
size(5cm); draw(arc((0,0),3,0,180)); draw(... | [] |
799 | In the figure below, semicircles with centers at $A$ and $B$ and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter $JK$. The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered a... | 2017 AMC 12A Problem 16 | Let $C$ be the center of the largest semicircle and $r$ be the radius of $\circ P$. We know that $AC = 1$, $CB = 2$, $AP = r + 2$, $BP = r + 1$, and $CP = 3 - r$. Notice that $\Delta ACP$ and $\Delta CBP$ are bounded by the same two parallel lines, so these triangles have the same heights. Because the bases of thes... | // Block 1
size(7.5cm);
draw(arc((0,0),3,0,180));
draw(arc((2,0),1,0,180));
draw(arc((-1,0),2,0,180));
draw((-3,0)--(3,0));
pair P = (9/7,12/7);
pair A = (-1,0);
pair C = (0,0);
pair B = (2,0);
draw(circle(P,6/7));
dot((-1,0)); dot((2,0)); dot((0,0)); dot(P);
draw((-1,0)--P);
draw((-2.45,12/7)--P);
draw((2.45,12/7)--P)... | [] |
799 | In the figure below, semicircles with centers at $A$ and $B$ and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter $JK$. The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered a... | 2017 AMC 12A Problem 16 | Let $C$, the center of the large semicircle, to be at $(0, 0)$, and $P$ to be at $(h, k)$.
Therefore $A$ is at $(-1, 0)$ and $B$ is at $(2, 0)$.
Let the radius of circle $P$ be $r$.
Using Distance Formula, we get the following system of three equations:
\[h^2+k^2=(3-r)^2, (h+1)^2+k^2=(r+2)^2, (h-2)^2+k^2=(r+1)^2\... | // Block 1
size(5cm);
draw(arc((0,0),3,0,180));
draw(arc((2,0),1,0,180));
draw(arc((-1,0),2,0,180));
draw((-3,0)--(3,0));
pair P = (9/7,12/7);
pair A = (-1,0);
pair C = (0,0);
pair B = (2,0);
draw(circle(P,6/7));
dot((-1,0)); dot((2,0)); dot((0,0)); dot(P);
draw((-1,0)--P);
draw((2,0)--P);
draw((0,0)--(9/5,12/5));
labe... | [] |
799 | In the figure below, semicircles with centers at $A$ and $B$ and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter $JK$. The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered a... | 2017 AMC 12A Problem 16 | Let $\Omega$ be a circle with radius of $6$ and centered at the left corner of the semi-circle (O) with radius $3$. Extend the three semicircles to full circles. Label the resulting four circles as shown in the diagram:
$C_1$ has radius $3$, $C_2$ has radius $2$, and $C_3$ has radius $1$. We want to find the radius ... | // Block 1
size(5cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7); draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(6, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); ... | [] |
800 | A square with side length $x$ is inscribed in a right triangle with sides of length $3$, $4$, and $5$ so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length $y$ is inscribed in another right triangle with sides of length $3$, $4$, and $5$ so that one side of th... | 2017 AMC 12A Problem 19 | Analyze the first right triangle.
Note that $\triangle ABC$ and $\triangle FBE$ are similar, so $\frac{BF}{FE} = \frac{AB}{AC}$. This can be written as $\frac{4-x}{x}=\frac{4}{3}$. Solving, $x = \frac{12}{7}$.
Now we analyze the second triangle.
Similarly, $\triangle A'B'C'$ and $\triangle RB'Q$ are similar, s... | // Block 1
pair A,B,C;
pair D, e, F;
A = (0,0);
B = (4,0);
C = (0,3);
D = (0, 12/7);
e = (12/7 , 12/7);
F = (12/7, 0);
draw(A--B--C--cycle);
draw(D--e--F);
label("$x$", D/2, W);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("$D$", D, W);
label("$E$", e, NE);
label("$F$", F, S);
// Block 2
pair ... | [] |
801 | Points $A(11, 9)$ and $B(2, -3)$ are vertices of $\triangle ABC$ with $AB=AC$. The altitude from $A$ meets the opposite side at $D(-1, 3)$. What are the coordinates of point $C$?
$\textbf{(A)}\ (-8, 9)\qquad\textbf{(B)}\ (-4, 8)\qquad\textbf{(C)}\ (-4, 9)\qquad\textbf{(D)}\ (-2, 3)\qquad\textbf{(E)}\ (-1, 0)$ | 2017 AMC 10B Problem 8 | Since $AB = AC$, then $\triangle ABC$ is isosceles, so $BD = CD$. Therefore, the coordinates of $C$ are $(-1 - 3, 3 + 6) = \boxed{\textbf{(C) } (-4,9)}$. | // Block 1
pair A,B,C,D;
A=(11,9);
B=(2,-3);
C=(-4,9);
D=(-1,3);
draw(A--B--C--cycle);
draw(A--D);
draw(rightanglemark(A,D,B,30));
label("$A$",A,E);
label("$B$",B,S);
label("$D$",D,W);
label("$C$",C,N);
// Block 2
pair A,B,C,D; A=(11,9); B=(2,-3); C=(-4,9); D=(-1,3); draw(A--B--C--cycle); draw(A--D); draw(rightanglemar... | [] |
801 | Points $A(11, 9)$ and $B(2, -3)$ are vertices of $\triangle ABC$ with $AB=AC$. The altitude from $A$ meets the opposite side at $D(-1, 3)$. What are the coordinates of point $C$?
$\textbf{(A)}\ (-8, 9)\qquad\textbf{(B)}\ (-4, 8)\qquad\textbf{(C)}\ (-4, 9)\qquad\textbf{(D)}\ (-2, 3)\qquad\textbf{(E)}\ (-1, 0)$ | 2017 AMC 10B Problem 8 | As stated in solution 1, the triangle is isosceles.
This means that $D(-1, 3)$ is the midpoint of $B(2, -3)$ and $C(x,y)$. So $\frac{x+2}{2}$ $= -1$ and so $x = -4$. Similarly for $y$, we have $\frac{y-3}{2}$ $=3$ and so $y=9$. So our final answer is $\boxed{\textbf{(C) } (-4,9)}$.
- youtube.com/indianmathguy | // Block 1
pair A,B,C,D;
A=(11,9);
B=(2,-3);
C=(-4,9);
D=(-1,3);
draw(A--B--C--cycle);
draw(A--D);
draw(rightanglemark(A,D,B));
label("$A$",A,E);
label("$B$",B,S);
label("$D$",D,W);
label("$C$",C,N);
// Block 2
pair A,B,C,D; A=(11,9); B=(2,-3); C=(-4,9); D=(-1,3); draw(A--B--C--cycle); draw(A--D); draw(rightanglemark(A... | [] |
802 | Rectangle $ABCD$ has $AB=3$ and $BC=4$. Point $E$ is the foot of the perpendicular from $B$ to diagonal $\overline{AC}$. What is the area of $\triangle AED$?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{42}{25}\qquad\textbf{(C)}\ \frac{28}{15}\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{54}{25}$ | 2017 AMC 10B Problem 15 | First, note that $AC=5$ because $ABC$ is a right triangle. In addition, we have $AB\cdot BC=2[ABC]=AC\cdot BE$, so $BE=\frac{12}{5}$. Using similar triangles within $ABC$, we get that $AE=\frac{9}{5}$ and $CE=\frac{16}{5}$.
Let $F$ be the foot of the perpendicular from $E$ to $AB$. Since $EF$ and $BC$ are parallel, $\... | // Block 1
pair A,B,C,D,E;
A=(0,4);
B=(3,4);
C=(3,0);
D=(0,0);
draw(A--B--C--D--cycle);
label("$A$",A,N);
label("$B$",B,N);
label("$C$",C,S);
label("$D$",D,S);
E=foot(B,A,C);
draw(E--B);
draw(A--C);
draw(rightanglemark(B,E,C));
label("$E$",E,N);
draw(D--E);
label("$3$",A--B,N);
label("$4$",B--C,E);
// Block 2
pair A,B,... | [] |
803 | In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
$\textbf{(A)}\ 6\qqua... | 2017 AMC 10B Problem 18 | Note that the green disk has two possibilities; in a corner or on the side. WLOG, we can arrange these as
Take the first case. Now, we must pick two of the five remaining circles to fill in the red. There are $\dbinom{5}{2}=10$ of these. However, due to reflection we must divide this by two. But, in two of these cas... | // Block 1
filldraw(circle((0,0),1),green);
draw(circle((2,0),1));
draw(circle((4,0),1));
draw(circle((1,sqrt(3)),1));
draw(circle((3,sqrt(3)),1));
draw(circle((2,2sqrt(3)),1));
draw(circle((8,0),1));
filldraw(circle((10,0),1),green);
draw(circle((12,0),1));
draw(circle((9,sqrt(3)),1));
draw(circle((11,sqrt(3)),1));
d... | [] |
804 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$. Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$, and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$. What is the ratio of the... | 2017 AMC 10B Problem 19 | This looks like an easy application of Routh's Theorem, except we are only given information about the ratios of the cevians, not the side lengths of $\triangle A'B'C'$.
Let's figure those out. Extend $A'A$ and $B'B$ to meet $B'C'$ and $A'C'$ at $D$ and $E$, respectively (Note we only need to look at 2 of 3 cevians ... | // Block 1
size(250);
real a = 1;
pair A = (0, 0), B = rotate(60)*A + (a, 0), C = rotate(120)*B + (a, 0);
pair shiftVector = (8, 0); // define a shift vector to the right
// calculate extended points
pair A_prime = A + 3*(A - C);
pair B_prime = B + 3*(B - A);
pair C_prime = C + 3*(C - B);
pair D = extension(A, A_prime... | [] |
805 | In $\triangle ABC$, $AB=6$, $AC=8$, $BC=10$, and $D$ is the midpoint of $\overline{BC}$. What is the sum of the radii of the circles inscribed in $\triangle ADB$ and $\triangle ADC$?
$\textbf{(A)}\ \sqrt{5}\qquad\textbf{(B)}\ \frac{11}{4}\qquad\textbf{(C)}\ 2\sqrt{2}\qquad\textbf{(D)}\ \frac{17}{6}\qquad\textbf{(E)}\ ... | 2017 AMC 10B Problem 21 | We have
Let $x$ be the radius of circle $F$, and let $y$ be the radius of circle $J$. We want to find $x+y$.
We form 6 kites: $GAKF$, $HFKD$, $GFHB$, $EJIA$, $LJIC$, and $JEDL$.
Since $G$ and $I$ are the midpoints of $\overline{AB}$ and $\overline{AC}$, respectively, this means that $BG = AG = \frac{6}{2} = 3$, and $... | // Block 1
draw((0,0)--(8,0));
draw((0,0)--(0,6));
draw((8,0)--(0,6));
draw((0,0)--(4,3));
label("A",(0,0),W);
label("B",(0,6),N);
label("C",(8,0),E);
label("D",(4,3),NE);
label("H",(2.3,4.2),NE);
label("K",(2.3,1.8),S);
draw(circle((1.54,3),1.49));
draw(circle((4,1.35),1.33));
dot((4,1.35));
dot((1.54,3));
label("F",(... | [] |
806 | The diameter $\overline{AB}$ of a circle of radius $2$ is extended to a point $D$ outside the circle so that $BD=3$. Point $E$ is chosen so that $ED=5$ and line $ED$ is perpendicular to line $AD$. Segment $\overline{AE}$ intersects the circle at a point $C$ between $A$ and $E$. What is the area of $\triangle ABC$?
$\t... | 2017 AMC 10B Problem 22 | Notice that $ADE$ and $ABC$ are right triangles. Then $AE = \sqrt{7^2+5^2} = \sqrt{74}$. $\sin{DAE} = \frac{5}{\sqrt{74}} = \sin{BAE} = \sin{BAC} = \frac{BC}{4}$, so $BC = \frac{20}{\sqrt{74}}$. We also find that $AC = \frac{28}{\sqrt{74}}$ (You can also use power of point ~MATHWIZARD2010), and thus the area of $ABC$ i... | // Block 1
size(10cm);
pair A, B, C, D, E, O;
A = (-2,0);
B = (2,0);
C = (2*cos(1.24),2*sin(1.24));
D = (5,0);
E = (5,5);
O = (A+B)/2;
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(O);
draw(Circle((A+B)/2,2));
draw(A--D--E--C--A);
draw(C--B);
draw(rightanglemark(A,C,B,5));
draw(rightanglemark(A,D,E,5));
label("$A$",A,W);... | [] |
807 | The diameter $AB$ of a circle of radius $2$ is extended to a point $D$ outside the circle so that $BD=3$. Point $E$ is chosen so that $ED=5$ and line $ED$ is perpendicular to line $AD$. Segment $AE$ intersects the circle at a point $C$ between $A$ and $E$. What is the area of $\triangle ABC$?
$\textbf{(A)}\ \frac{120... | 2017 AMC 12B Problem 18 | Let $O$ be the center of the circle. Note that $EC + CA = EA = \sqrt{AD^2 + DE^2} = \sqrt{(2+2+3)^2 + 5^2} = \sqrt{74}$. However, by Power of a Point, $(EC)(EC + CA) = EO^2 - R^2 = (2+3)^2 + 5^2 - 2^2 = 25 + 25 - 4 = 46 \implies EC = \frac{46}{\sqrt{74}}$, so $AC = \sqrt{74} - \frac{46}{\sqrt{74}} = \frac{28}{\sqrt{74}... | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(8.865514650638614cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen ... | [] |
808 | All of the triangles in the diagram below are similar to isosceles triangle $ABC$, in which $AB=AC$. Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$. What is the area of trapezoid $DBCE$?
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad... | 2018 AMC 10A Problem 9 | Notice $\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]$.
Let the base of the small triangles of area 1 be $x$, then the base length of $\Delta ADE=4x$. Notice, $\left(\frac{DE}{BC}\right)^2=\frac{1}{40}\implies \frac{x}{BC}=\frac{1}{\sqrt{40}}$, then $4x=\frac{4BC}{\sqrt{40}}\implies \big[ADE\big]=\left(\frac{4}{\sqrt{40}}... | // Block 1
unitsize(5);
dot((0,0));
dot((60,0));
dot((50,10));
dot((10,10));
dot((30,30));
draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0));
draw((10,10)--(50,10));
label("$B$",(0,0),SW);
label("$C$",(60,0),SE);
label("$E$",(50,10),E);
label("$D$",(10,10),W);
label("$A$",(30,30),N);
draw((10,10)--(15,15)--(20,10)-... | [] |
809 | How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations?
\begin{align*} x+3y&=3 \\ \big||x|-|y|\big|&=1 \end{align*}
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$ | 2018 AMC 10A Problem 12 | We can solve this by graphing the equations. The second equation looks challenging to graph, but start by graphing it in the first quadrant only (which is easy since the inner absolute value signs can be ignored), then simply reflect that graph into the other quadrants.
The graph looks something like this:
Now, it be... | // Block 1
draw((-3,0)--(3,0), Arrows);
draw((0,-3)--(0,3), Arrows);
draw((2,3)--(0,1)--(-2,3), blue);
draw((-3,2)--(-1,0)--(-3,-2), blue);
draw((-2,-3)--(0,-1)--(2,-3), blue);
draw((3,-2)--(1,0)--(3,2), blue);
draw((-3,2)--(3,0), red);
dot((-3,2));
dot((3/2,1/2));
dot((0,1));
// Block 2
draw((-3,0)--(3,0), Arrows); dr... | [] |
810 | A paper triangle with sides of lengths $3,4,$ and $5$ inches, as shown, is folded so that point $A$ falls on point $B$. What is the length in inches of the crease?
$\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad ... | 2018 AMC 10A Problem 13 | First, we need to realize that the crease line is just the perpendicular bisector of side $AB$, the hypotenuse of right triangle $\triangle ABC$. Call the midpoint of $AB$ point $D$. Draw this line and call the intersection point with $AC$ as $E$. Now, $\triangle ACB$ is similar to $\triangle ADE$ by $AA$ similarity. S... | // Block 1
draw((0,0)--(4,0)--(4,3)--(0,0)); label("$A$", (0,0), SW); label("$B$", (4,3), NE); label("$C$", (4,0), SE); label("$D$", (2,1.5), NW); label("$E$", (3.125,0), S);
draw ((2,1.5)--(3.125,0),linewidth(1));
draw(rightanglemark((0,0),(2,1.5),(3.125,0)));
// Block 2
draw((0,0)--(4,0)--(4,3)--(0,0)); label("$A$"... | [] |
810 | A paper triangle with sides of lengths $3,4,$ and $5$ inches, as shown, is folded so that point $A$ falls on point $B$. What is the length in inches of the crease?
$\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad ... | 2018 AMC 10A Problem 13 | Since $\triangle ABC$ is a right triangle, we can see that the slope of line $AB$ is $\frac{BC}{AC}$ = $\frac{3}{4}$. We know that if we fold $\triangle ABC$ so that point $A$ meets point $B$ the crease line will be perpendicular to $AB$ and we also know that the slopes of perpendicular lines are negative reciprocals o... | // Block 1
pen dotstyle = black;
draw((0,0)--(4,0)--(4,3)--(0,0));
fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray);
dot((0,0),dotstyle);
label("$A$", (0.03153837092244126,0.07822624343603715), SW);
dot((4,0),dotstyle);
label("$C$", (4.028913881471271,0.07822624343603715), SE);
dot((4,3),dotstyle);
label("$B$", (4... | [] |
810 | A paper triangle with sides of lengths $3,4,$ and $5$ inches, as shown, is folded so that point $A$ falls on point $B$. What is the length in inches of the crease?
$\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad ... | 2018 AMC 10A Problem 13 | Like in Solution 3, we can make use of coordinate geometry to solve this problem. Because the length of the crease is constant no matter the positioning of the triangle, reorient the triangle so it has C at the origin and B and A at $(0,3)$ and $(4,0)$ respectively.
We know that each point in the crease is equidista... | // Block 1
draw((0,0)--(0,3)--(4,0)--(0,0));
label("$C$", (0,0), SW); label("$B$", (0,3), NW); label("$A$", (4,0), SE);
draw ((2,1.5)--(0.875,0),blue+linewidth(1));
label("$(\frac{7}{8},0)$", (0.875,0), S); label("$(2,\frac{3}{2})$", (2,1.5), NE);
draw(rightanglemark((4,0),(2,1.5),(0.875,0)));
// Block 2
draw((0,0)--(0... | [] |
811 | Two circles of radius $5$ are externally tangent to each other and are internally tangent to a circle of radius $13$ at points $A$ and $B$, as shown in the diagram. The distance $AB$ can be written in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?
$\textbf{(A) } ... | 2018 AMC 10A Problem 15 | Let the center of the surrounding circle be $X$. The circle that is tangent at point $A$ will have point $Y$ as the center. Similarly, the circle that is tangent at point $B$ will have point $Z$ as the center. Connect $AB$, $YZ$, $XA$, and $XB$. Now observe that $\triangle XYZ$ is similar to $\triangle XAB$ by SAS.
Wr... | // Block 1
draw(circle((0,0),13));
draw(circle((5,-6.25),5));
draw(circle((-5,-6.25),5));
label("$A$", (-8.125,-10.15), S);
label("$B$", (8.125,-10.15), S);
draw((0,0)--(-8.125,-10.15));
draw((0,0)--(8.125,-10.15));
draw((-5,-6.25)--(5,-6.25));
draw((-8.125,-10.15)--(8.125,-10.15));
label("$X$", (0,0), N);
label("$Y$",... | [] |
812 | Right triangle $ABC$ has leg lengths $AB=20$ and $BC=21$. Including $\overline{AB}$ and $\overline{BC}$, how many line segments with integer length can be drawn from vertex $B$ to a point on hypotenuse $\overline{AC}$?
$\textbf{(A) }5 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }13 \qquad \textbf{(... | 2018 AMC 10A Problem 16 | As the problem has no diagram, we draw a diagram. The hypotenuse has length $29$. Let $P$ be the foot of the altitude from $B$ to $AC$. Note that $BP$ is the shortest possible length of any segment. Writing the area of the triangle in two ways, we can solve for $BP=\dfrac{20\cdot 21}{29}$, which is between $14$ and $1... | // Block 1
unitsize(4);
pair A, B, C, E, P;
A=(-20, 0);
B=origin;
C=(0,21);
E=(-21, 20);
P=extension(B,E, A, C);
draw(A--B--C--cycle);
draw(B--P);
dot("$A$", A, SW);
dot("$B$", B, SE);
dot("$C$", C, NE);
dot("$P$", P, NW);
// Block 2
unitsize(4); pair A, B, C, E, P; A=(-20, 0); B=origin; C=(0,21); E=(-21, 20); P=extens... | [] |
812 | Right triangle $ABC$ has leg lengths $AB=20$ and $BC=21$. Including $\overline{AB}$ and $\overline{BC}$, how many line segments with integer length can be drawn from vertex $B$ to a point on hypotenuse $\overline{AC}$?
$\textbf{(A) }5 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }13 \qquad \textbf{(... | 2018 AMC 10A Problem 16 | Note that if a circle with an integer radius $r$ centered at vertex $B$ intersects hypotenuse $\overline{AB}$, the lines drawn from $B$ to the points of intersection are integer lengths. As in the previous solution, the shortest distance $14<\overline{BP}<15$. As a result, a circle of $14$ will not reach the hypotenuse... | // Block 1
unitsize(4);
pair A, B, C, E, P;
A=(-20, 0);
B=origin;
C=(0,21);
E=(-21, 20);
P=extension(B,E, A, C);
draw(A--B--C--cycle);
draw(B--P);
dot("$A$", A, SW);
dot("$B$", B, SE);
dot("$C$", C, NE);
dot("$P$", P, S);
draw(arc((0,0),21, 90, 180));
draw(arc((0,0),20, 90, 180));
draw(arc((0,0),19, 90, 180));
draw(arc... | [] |
813 | A scanning code consists of a $7 \times 7$ grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of $49$ squares. A scanning code is called $\textit{symmetric}$ if its look does not change when the entire square is rotated by a m... | 2018 AMC 10A Problem 20 | Imagine folding the scanning code along its lines of symmetry. There will be $10$ regions which you have control over coloring. Since we must subtract off $2$ cases for the all-black and all-white cases, the answer is $2^{10}-2=\boxed{\textbf{(B) } 1022.}$ | // Block 1
size(100pt);
draw((1,0)--(8,0),linewidth(0.5));
draw((1,2)--(6,2),linewidth(0.5));
draw((1,4)--(4,4),linewidth(0.5));
draw((1,6)--(2,6),linewidth(0.5));
draw((2,6)--(2,0),linewidth(0.5));
draw((4,4)--(4,0),linewidth(0.5));
draw((6,2)--(6,0),linewidth(0.5));
draw((1,0)--(1,7),dashed+linewidth(0.5));
draw((1,7... | [] |
814 | Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$-plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 ... | 2018 AMC 10A Problem 21 | Looking at a graph, it is obvious that the two curves intersect at $(0, -a)$. We also see that if the parabola goes "in" the circle, then by going out of it (as it will), it will intersect five times. This is impossible. Thus, we only look for cases where the parabola becomes externally tangent to the circle.
We have... | // Block 1
Label f;
f.p=fontsize(6);
xaxis(-2,2,Ticks(f, 0.2));
yaxis(-2,2,Ticks(f, 0.2));
real g(real x)
{
return x^2-1;
}
draw(graph(g, 1.7, -1.7));
real h(real x)
{
return sqrt(1-x^2);
}
draw(graph(h, 1, -1));
real j(real x)
{
return -sqrt(1-x^2);
}
draw(graph(j, 1, -1));
// Block 2
Label f; f.p=font... | [] |
814 | Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$-plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 ... | 2018 AMC 10A Problem 21 | Now, let's graph these two equations. We want the blue parabola to be inside this red circle.
Then we substitute $y$ into the first equation to get $x^2+(x^2-a)^2=a^2$. Expanding, we get $x^4-2ax^2+x^2=0$. Factoring out the $x^2$, we get $x^2(x^2-2a+1)=0$. Then we find that $x=0$ or $x=\pm\sqrt{2a-1}$. Therefore, $2a-... | // Block 1
import graph;
size(6cm);
draw((0,0)--(0,10),EndArrow);
draw((0,0)--(0,-10),EndArrow);
draw((0,0)--(10,0),EndArrow);
draw((0,0)--(-10,0),EndArrow);
Label f;
f.p=fontsize(6);
xaxis(-10,10);
yaxis(-10,10);
real f(real x)
{
return x^2-5;
}
draw(graph(f,-4,4),blue+linewidth(1));
draw(circle((0,0),5),red);
dot(s... | [] |
815 | Let $a, b, c,$ and $d$ be positive integers such that $\gcd(a, b)=24$, $\gcd(b, c)=36$, $\gcd(c, d)=54$, and $70<\gcd(d, a)<100$. Which of the following must be a divisor of $a$? (gcd means greatest common factor)
$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} ... | 2018 AMC 10A Problem 22 | The relationship of $a$, $b$, $c$, and $d$ is shown in the above diagram. $gcd(a,d)=2 \cdot 3 \cdot k$.
$70 < 6k < 100$, $12 \le k \le 16$, $k=\boxed{\textbf{(D) }13}$
Note that $gcd(b,c)=36$ is not required to solve the problem.
~isabelchen | // Block 1
//Variable Declarations
defaultpen(0.45);
size(200pt);
fontsize(15pt);
pair X, Y, Z, W;
real R;
path quad;
//Variable Definitions
R = 1;
X = R*dir(45);
Y = R*dir(135);
Z = R*dir(-135);
W = R*dir(-45);
quad = X--Y--Z--W--cycle;
//Diagram
draw(quad);
label("$b$",X,NE);
label("$a=2^3 \cdot 3 \cdot p$",Y,NW);
... | [] |
816 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortes... | 2018 AMC 10A Problem 23 | Note that the hypotenuse of the field is $5,$ and the area of the field is $6.$ Let $x$ be the side-length of square $S.$
We partition the field into a red triangle, a yellow triangle, and a green triangle, as shown below:
Let the brackets denote areas. By area addition, we set up an equation for $x:$
\begin{align*} ... | // Block 1
/* Edited by MRENTHUSIASM */
size(180);
pair A, B, C, D, F;
A = origin;
B = (4,0);
C = (0,3);
D = (2/7,2/7);
F = foot(D,B,C);
fill(A--D--C--cycle, red);
fill(A--D--B--cycle, yellow);
fill(B--D--C--cycle, green);
draw(A--B--C--cycle);
label("$5$", midpoint(B--C), NE);
label("$4$", midpoint(A--B), S);
label("$... | [] |
816 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortes... | 2018 AMC 10A Problem 23 | Let the square have side length $x$. Connect the upper-right vertex of square $S$ with the two vertices of the triangle's hypotenuse. This divides the triangle in several regions whose areas must add up to the area of the whole triangle, which is $6$.
Square $S$ has area $x^2$, and the two thin triangle regions have a... | // Block 1
/* Edited by MRENTHUSIASM */
size(180);
pair A, B, C, D, F;
A = origin;
B = (4,0);
C = (0,3);
D = (2/7,2/7);
F = foot(D,B,C);
fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray);
draw(A--B--C--cycle);
label("$5$", midpoint(B--C), NE);
label("$4$", midpoint(A--B), S);
label("$3$", midpoint(A--C), W);
label("$2$", ... | [] |
816 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortes... | 2018 AMC 10A Problem 23 | On the diagram above, find two smaller triangles similar to the large one with side lengths $3$, $4$, and $5$; consequently, the segments with length $\frac{5}{2}$ and $\frac{10}{3}$.
With $\ell$ being the side length of the square, we need to find an expression for $\ell$. Using the hypotenuse, we can see that $\frac... | // Block 1
/* Edited by MRENTHUSIASM */
size(180);
pair A, B, C, D, F;
A = origin;
B = (4,0);
C = (0,3);
D = (2/7,2/7);
F = foot(D,B,C);
fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray);
draw(A--B--C--cycle);
label("$4$", midpoint(A--B), S);
label("$3$", midpoint(A--C), W);
label("$2$", midpoint(D--F), SE);
label("$S$", ... | [] |
816 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortes... | 2018 AMC 10A Problem 23 | Let $AD=x$. Note that $\triangle DEF$ is a $3{-}4{-}5$ triangle, so $EF=\frac{5}{4}x$ and $FD=\frac{3}{4}x$. $BF=BD+FD=4-x+\frac{3}{4}x=4-\frac{1}{4}x$. We know that $GE$ is $2$ from the problem so $GF=2+\frac{5}{4}x$. $\triangle FGB$ is also a $3{-}4{-}5$ triangle with $GF:BF=3:5$. We now have $3\left(4-\frac{1}{4}x\r... | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(180);
real labelscalefactor = 1.5; /* changes label-to-point distance */
// pen dps = linewidth(0.5) + fontsize(10);
// defaultpen(dps); /* default pen style */
// pen dotstyl... | [] |
816 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortes... | 2018 AMC 10A Problem 23 | Denote $A,B,C$ to be the three vertices of the triangular field. Also denote $A,M,D,N$ to be the vertices of the square $S$. Let $X$ be on $BC$ such that $AC\parallel DX$ and $Y$ be on $BC$ such that $AB\parallel DY$. Let $P$ and $Q$ be the foot of the altitudes from $X$ to $AC$ and from $Y$ to $AB$ respectively.
Note... | // Block 1
size(240);
pair A, B, C, D, F, X, Y, P, Q, M, N;
A = origin; label(A, "$A$", SW);
B = (4,0); label(B, "$B$", S);
C = (0,3); label(C, "$C$", W);
D = (2/7,2/7); label(D, "$D$", NE);
F = foot(D,B,C); label(F, "$F$", NE);
X = (2/7,39/14); label(X, "$X$", NE, red);
Y = (76/21,2/7); label(Y, "$Y$", NE, red);
P = f... | [] |
817 | All of the triangles in the diagram below are similar to isosceles triangle $ABC$, in which $AB=AC$. Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$. What is the area of trapezoid $DBCE$?
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad... | 2018 AMC 12A Problem 8 | Notice $\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]$.
Let the base of the small triangles of area 1 be $x$, then the base length of $\Delta ADE=4x$. Notice, $\left(\frac{DE}{BC}\right)^2=\frac{1}{40}\implies \frac{x}{BC}=\frac{1}{\sqrt{40}}$, then $4x=\frac{4BC}{\sqrt{40}}\implies \big[ADE\big]=\left(\frac{4}{\sqrt{40}}... | unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(... | [] |
818 | How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations?
\begin{align*} x+3y&=3 \\ \big||x|-|y|\big|&=1 \end{align*}
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$ | 2018 AMC 12A Problem 10 | We can solve this by graphing the equations. The second equation looks challenging to graph, but start by graphing it in the first quadrant only (which is easy since the inner absolute value signs can be ignored), then simply reflect that graph into the other quadrants.
The graph looks something like this:
Now, it be... | draw((-3,0)--(3,0), Arrows); draw((0,-3)--(0,3), Arrows); draw((2,3)--(0,1)--(-2,3), blue); draw((-3,2)--(-1,0)--(-3,-2), blue); draw((-2,-3)--(0,-1)--(2,-3), blue); draw((3,-2)--(1,0)--(3,2), blue); draw((-3,2)--(3,0), red); dot((-3,2)); dot((3/2,1/2)); dot((0,1)); | [] |
819 | A paper triangle with sides of lengths $3,4,$ and $5$ inches, as shown, is folded so that point $A$ falls on point $B$. What is the length in inches of the crease?
$\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad ... | 2018 AMC 12A Problem 11 | First, we need to realize that the crease line is just the perpendicular bisector of side $AB$, the hypotenuse of right triangle $\triangle ABC$. Call the midpoint of $AB$ point $D$. Draw this line and call the intersection point with $AC$ as $E$. Now, $\triangle ACB$ is similar to $\triangle ADE$ by $AA$ similarity. S... | draw((0,0)--(4,0)--(4,3)--(0,0)); label("$A$", (0,0), SW); label("$B$", (4,3), NE); label("$C$", (4,0), SE); label("$D$", (2,1.5), NW); label("$E$", (3.125,0), S); draw ((2,1.5)--(3.125,0),linewidth(1)); draw(rightanglemark((0,0),(2,1.5),(3.125,0))); | [] |
819 | A paper triangle with sides of lengths $3,4,$ and $5$ inches, as shown, is folded so that point $A$ falls on point $B$. What is the length in inches of the crease?
$\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad ... | 2018 AMC 12A Problem 11 | Since $\triangle ABC$ is a right triangle, we can see that the slope of line $AB$ is $\frac{BC}{AC}$ = $\frac{3}{4}$. We know that if we fold $\triangle ABC$ so that point $A$ meets point $B$ the crease line will be perpendicular to $AB$ and we also know that the slopes of perpendicular lines are negative reciprocals o... | pen dotstyle = black; draw((0,0)--(4,0)--(4,3)--(0,0)); fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray); dot((0,0),dotstyle); label("$A$", (0.03153837092244126,0.07822624343603715), SW); dot((4,0),dotstyle); label("$C$", (4.028913881471271,0.07822624343603715), SE); dot((4,3),dotstyle); label("$B$", (4.0289138814... | [] |
819 | A paper triangle with sides of lengths $3,4,$ and $5$ inches, as shown, is folded so that point $A$ falls on point $B$. What is the length in inches of the crease?
$\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad ... | 2018 AMC 12A Problem 11 | Like in Solution 3, we can make use of coordinate geometry to solve this problem. Because the length of the crease is constant no matter the positioning of the triangle, reorient the triangle so it has C at the origin and B and A at $(0,3)$ and $(4,0)$ respectively.
We know that each point in the crease is equidista... | draw((0,0)--(0,3)--(4,0)--(0,0)); label("$C$", (0,0), SW); label("$B$", (0,3), NW); label("$A$", (4,0), SE); draw ((2,1.5)--(0.875,0),blue+linewidth(1)); label("$(\frac{7}{8},0)$", (0.875,0), S); label("$(2,\frac{3}{2})$", (2,1.5), NE); draw(rightanglemark((4,0),(2,1.5),(0.875,0))); | [] |
820 | A scanning code consists of a $7 \times 7$ grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of $49$ squares. A scanning code is called $\textit{symmetric}$ if its look does not change when the entire square is rotated by a m... | 2018 AMC 12A Problem 15 | Imagine folding the scanning code along its lines of symmetry. There will be $10$ regions which you have control over coloring. Since we must subtract off $2$ cases for the all-black and all-white cases, the answer is $2^{10}-2=\boxed{\textbf{(B) } 1022.}$ | size(100pt); draw((1,0)--(8,0),linewidth(0.5)); draw((1,2)--(6,2),linewidth(0.5)); draw((1,4)--(4,4),linewidth(0.5)); draw((1,6)--(2,6),linewidth(0.5)); draw((2,6)--(2,0),linewidth(0.5)); draw((4,4)--(4,0),linewidth(0.5)); draw((6,2)--(6,0),linewidth(0.5)); draw((1,0)--(1,7),dashed+linewidth(0.5)); draw((1,7)--(8,0),da... | [] |
821 | Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$-plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 ... | 2018 AMC 12A Problem 16 | Looking at a graph, it is obvious that the two curves intersect at $(0, -a)$. We also see that if the parabola goes "in" the circle, then by going out of it (as it will), it will intersect five times. This is impossible. Thus, we only look for cases where the parabola becomes externally tangent to the circle.
We have... | Label f; f.p=fontsize(6); xaxis(-2,2,Ticks(f, 0.2)); yaxis(-2,2,Ticks(f, 0.2)); real g(real x) { return x^2-1; } draw(graph(g, 1.7, -1.7)); real h(real x) { return sqrt(1-x^2); } draw(graph(h, 1, -1)); real j(real x) { return -sqrt(1-x^2); } draw(graph(j, 1, -1)); | [] |
821 | Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$-plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 ... | 2018 AMC 12A Problem 16 | Now, let's graph these two equations. We want the blue parabola to be inside this red circle.
Then we substitute $y$ into the first equation to get $x^2+(x^2-a)^2=a^2$. Expanding, we get $x^4-2ax^2+x^2=0$. Factoring out the $x^2$, we get $x^2(x^2-2a+1)=0$. Then we find that $x=0$ or $x=\pm\sqrt{2a-1}$. Therefore, $2a-... | import graph; size(6cm); draw((0,0)--(0,10),EndArrow); draw((0,0)--(0,-10),EndArrow); draw((0,0)--(10,0),EndArrow); draw((0,0)--(-10,0),EndArrow); Label f; f.p=fontsize(6); xaxis(-10,10); yaxis(-10,10); real f(real x) { return x^2-5; } draw(graph(f,-4,4),blue+linewidth(1)); draw(circle((0,0),5),red); dot(scale(.7)*"$... | [] |
822 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortes... | 2018 AMC 12A Problem 17 | Note that the hypotenuse of the field is $5,$ and the area of the field is $6.$ Let $x$ be the side-length of square $S.$
We partition the field into a red triangle, a yellow triangle, and a green triangle, as shown below:
Let the brackets denote areas. By area addition, we set up an equation for $x:$
\begin{align*} ... | /* Edited by MRENTHUSIASM */ size(180); pair A, B, C, D, F; A = origin; B = (4,0); C = (0,3); D = (2/7,2/7); F = foot(D,B,C); fill(A--D--C--cycle, red); fill(A--D--B--cycle, yellow); fill(B--D--C--cycle, green); draw(A--B--C--cycle); label("$5$", midpoint(B--C), NE); label("$4$", midpoint(A--B), S); label("$3$", midpoi... | [] |
822 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortes... | 2018 AMC 12A Problem 17 | Let the square have side length $x$. Connect the upper-right vertex of square $S$ with the two vertices of the triangle's hypotenuse. This divides the triangle in several regions whose areas must add up to the area of the whole triangle, which is $6$.
Square $S$ has area $x^2$, and the two thin triangle regions have a... | /* Edited by MRENTHUSIASM */ size(180); pair A, B, C, D, F; A = origin; B = (4,0); C = (0,3); D = (2/7,2/7); F = foot(D,B,C); fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray); draw(A--B--C--cycle); label("$5$", midpoint(B--C), NE); label("$4$", midpoint(A--B), S); label("$3$", midpoint(A--C), W); label("$2$", midpoint(D-... | [] |
822 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortes... | 2018 AMC 12A Problem 17 | On the diagram above, find two smaller triangles similar to the large one with side lengths $3$, $4$, and $5$; consequently, the segments with length $\frac{5}{2}$ and $\frac{10}{3}$.
With $\ell$ being the side length of the square, we need to find an expression for $\ell$. Using the hypotenuse, we can see that $\frac... | /* Edited by MRENTHUSIASM */ size(180); pair A, B, C, D, F; A = origin; B = (4,0); C = (0,3); D = (2/7,2/7); F = foot(D,B,C); fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray); draw(A--B--C--cycle); label("$4$", midpoint(A--B), S); label("$3$", midpoint(A--C), W); label("$2$", midpoint(D--F), SE); label("$S$", midpoint(A-... | [] |
822 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortes... | 2018 AMC 12A Problem 17 | Let $AD=x$. Note that $\triangle DEF$ is a $3{-}4{-}5$ triangle, so $EF=\frac{5}{4}x$ and $FD=\frac{3}{4}x$. $BF=BD+FD=4-x+\frac{3}{4}x=4-\frac{1}{4}x$. We know that $GE$ is $2$ from the problem so $GF=2+\frac{5}{4}x$. $\triangle FGB$ is also a $3{-}4{-}5$ triangle with $GF:BF=3:5$. We now have $3\left(4-\frac{1}{4}x\r... | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(180); real labelscalefactor = 1.5; /* changes label-to-point distance */ // pen dps = linewidth(0.5) + fontsize(10); // defaultpen(dps); /* default pen style */ // pen dotstyle = black; ... | [] |
822 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortes... | 2018 AMC 12A Problem 17 | Denote $A,B,C$ to be the three vertices of the triangular field. Also denote $A,M,D,N$ to be the vertices of the square $S$. Let $X$ be on $BC$ such that $AC\parallel DX$ and $Y$ be on $BC$ such that $AB\parallel DY$. Let $P$ and $Q$ be the foot of the altitudes from $X$ to $AC$ and from $Y$ to $AB$ respectively.
Note... | size(240); pair A, B, C, D, F, X, Y, P, Q, M, N; A = origin; label(A, "$A$", SW); B = (4,0); label(B, "$B$", S); C = (0,3); label(C, "$C$", W); D = (2/7,2/7); label(D, "$D$", NE); F = foot(D,B,C); label(F, "$F$", NE); X = (2/7,39/14); label(X, "$X$", NE, red); Y = (76/21,2/7); label(Y, "$Y$", NE, red); P = foot(X,A,C);... | [] |
823 | The solutions to the equations $z^2=4+4\sqrt{15}i$ and $z^2=2+2\sqrt 3i,$ where $i=\sqrt{-1},$ form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form $p\sqrt q-r\sqrt s,$ where $p,$ $q,$ $r,$ and $s$ are positive integers and neither $q$ nor $s$ is divisible... | 2018 AMC 12A Problem 22 | Let $z_1$ and $z_2$ be the solutions to the equation $z^2=4+4\sqrt{15}i,$ and $z_3$ and $z_4$ be the solutions to the equation $z^2=2+2\sqrt 3i.$ Clearly, $z_1$ and $z_2$ are opposite complex numbers, so are $z_3$ and $z_4.$ This solution refers to the results of De Moivre's Theorem in Solution 2.
From Solution 2, let... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
int xMin = -5;
int xMax = 5;
int yMin = -5;
int yMax = 5;
int numRays = 24;
//Draws a polar grid that goes out to a number of circles
//equal to big, with numRays specifying the number of rays:
void polarGrid(int big, int numRays)
{
for (int i = 1; i < big+1; ++i... | [] |
824 | In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=56^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute ang... | 2018 AMC 12A Problem 23 | We will refer to the Diagram section. In this solution, all angle measures are in degrees.
We rotate $\triangle PUM$ by $180^\circ$ about $M$ to obtain $\triangle AU'M.$ Let $H$ be the intersection of $\overline{PA}$ and $\overline{GU'},$ as shown below.
Note that $\triangle GU'A$ is an isosceles triangle with $GA=U'... | // Block 1
/* Made by MRENTHUSIASM */
size(375);
pair P, A, T, U, G, M, N, U1, H;
P = origin;
A = (10,0);
U = intersectionpoint(Circle(P,1),P--P+2*dir(36));
G = intersectionpoint(Circle(A,1),A--A+2*dir(180-56));
T = extension(P,U,A,G);
M = midpoint(P--A);
N = midpoint(U--G);
U1 = rotate(180,M)*U;
H = intersectionpoint... | [] |
825 | Alice, Bob, and Carol play a game in which each of them chooses a real number between $0$ and $1.$ The winner of the game is the one whose number is between the numbers chosen by the other two players. Alice announces that she will choose her number uniformly at random from all the numbers between $0$ and $1,$ and Bob ... | 2018 AMC 12A Problem 24 | Let $a,b,$ and $c$ be the numbers that Alice, Bob, and Carol choose, respectively.
Based on the value of $c,$ we construct the following table:
\[\begin{array}{c|c|c} & & \\ [-2ex] \textbf{Case} & \textbf{Conditions for }\boldsymbol{a}\textbf{ and }\boldsymbol{b} & \textbf{Carol's Probability of Winning} \\ [0.5ex] \h... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
real f(real x) { return x; }
real g(real x) { return -12x^2+13x-3; }
real h(real x) { return 1-x; }
draw((1/2,0)--(1/2,1.25),dashed);
draw((2/3,0)--(2/3,1.25),dashed);
draw(graph(f,0,1/2),red);
draw(graph(g,1/2,2/3),red);
draw(graph(h,2/3,1),red);
real xMin = -0.25;... | [] |
826 | Line segment $\overline{AB}$ is a diameter of a circle with $AB = 24$. Point $C$, not equal to $A$ or $B$, lies on the circle. As point $C$ moves around the circle, the centroid (center of mass) of $\triangle ABC$ traces out a closed curve missing two points. To the nearest positive integer, what is the area of the reg... | 2018 AMC 10B Problem 12 | By the Inscribed Angle Theorem, $\triangle ABC$ is a right triangle with $\angle C=90^{\circ}.$ So, its circumcenter is the midpoint of $\overline{AB},$ and its median from $C$ is half as long as $\overline{AB}.$ For each $\triangle ABC,$ let $O$ and $G$ be its circumcenter and centroid, respectively. It follows that $... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
pair O, A, B, C1, C2, G1, G2, M1, M2;
O = (0,0);
A = (-12,0);
B = (12,0);
C1 = (36/5,48/5);
C2 = (-96/17,-180/17);
G1 = O + 1/3 * C1;
G2 = O + 1/3 * C2;
M1 = (4,0);
M2 = (-4,0);
draw(Circle(O,12));
draw(Circle(O,4),red);
dot("$O$", O, (3/5,-4/5), linewidth(4.5));
dot("... | [] |
827 | A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the side... | 2018 AMC 10B Problem 15 | Consider one-quarter of the image (the wrapping paper is divided up into $4$ congruent squares). The length of each dotted line is $h$. The area of the rectangle that is $w$ by $h$ is $wh$. The combined figure of the two triangles with base $h$ is a square with $h$ as its diagonal. Using the Pythagorean Theorem, each s... | // Block 1
/* Edited by MRENTHUSIASM */
size(180pt);
defaultpen(fontsize(10pt));
fill(((0,3)--(-3,3)--(-3,0)--(0,0)--cycle),lightgrey);
dot((-3,3));
label("$A$",(-3,3),NW);
draw((-3,0)--(-3,3)--(0,3),linewidth(.5));
draw((0,2)--(-1,3)--(-3,1)--(-2,0),dashed+linewidth(.5));
draw((0,2)--(-2,0),linewidth(.5));
draw((0,3)-... | [] |
827 | A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the side... | 2018 AMC 10B Problem 15 | The sheet of paper is made out of the surface area of the box plus the sum of the four yellow triangles, as shown below.
The surface area is $2w^2 + 2wh + 2wh$ which equals $2w^2 + 4wh$.The four triangles each have a height and a base of $h$, so they each have an area of $\frac{h^2}{2}$. There are four of them, so ... | // Block 1
/* Edited by MRENTHUSIASM */
size(180pt);
defaultpen(fontsize(10pt));
fill(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),lightgrey);
fill((2,0)--(3,1)--(3,-1)--cycle,yellow);
fill((-2,0)--(-3,1)--(-3,-1)--cycle,yellow);
fill((0,2)--(1,3)--(-1,3)--cycle,yellow);
fill((0,-2)--(1,-3)--(-1,-3)--cycle,yellow);
draw(((3... | [] |
828 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A... | 2018 AMC 10B Problem 24 | The desired area (hexagon $MPNQOR$) consists of an equilateral triangle ($\triangle MNO$) and three right triangles ($\triangle MPN,\triangle NQO,$ and $\triangle ORM$).
Notice that $\overline {AD}$ (not shown) and $\overline {BC}$ are parallel. $\overline {XY}$ divides transversals $\overline {AB}$ and $\overline {CD... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
draw(polygon(6));
pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R;
A = dir(120);
B = dir(60);
C = dir(0);
D = dir(300);
E = dir(240);
F = dir(180);
X = midpoint(A--B);
Y = midpoint(C--D);
Z = midpoint(E--F);
M = intersectionpoint(A--E,X--Z);
N = intersectionpoint(A--C,X... | [] |
828 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A... | 2018 AMC 10B Problem 24 | Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of $3$ isosceles trapezoids (namely $AXZF,CYXB,$ and $EZYD$) and $3$ right triangles (namely $\triangle XPN,\triangle YQO,$ and $\triangle ZRM$).
Finding the trapezoid's area, we know that one base of each tr... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
draw(polygon(6));
pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R;
A = dir(120);
B = dir(60);
C = dir(0);
D = dir(300);
E = dir(240);
F = dir(180);
X = midpoint(A--B);
Y = midpoint(C--D);
Z = midpoint(E--F);
M = intersectionpoint(A--E,X--Z);
N = intersectionpoint(A--C,X... | [] |
828 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A... | 2018 AMC 10B Problem 24 | We could also subtract $\triangle APM,\triangle CQN,$ and $\triangle ERO$ from $\triangle ACE.$
Since $\angle BAF = 120^{\circ}$ and $\angle BAC = \angle FAE = 30^{\circ},$ we have $\angle CAE = \angle BAF-\angle BAC-\angle FAE=60^{\circ}.$
Since $AX=BX$ and $FZ=EZ,$ we have $AF \parallel XZ,$ from which $\angle AMX=... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
draw(polygon(6));
pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R;
A = dir(120);
B = dir(60);
C = dir(0);
D = dir(300);
E = dir(240);
F = dir(180);
X = midpoint(A--B);
Y = midpoint(C--D);
Z = midpoint(E--F);
M = intersectionpoint(A--E,X--Z);
N = intersectionpoint(A--C,X... | [] |
828 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A... | 2018 AMC 10B Problem 24 | We partition hexagon $ABCDEF$ into $48$ congruent $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangles, as shown below:
Let the brackets denote areas. Note that the desired region contains $15$ of the $48$ small triangles, so the answer is \[\frac{15}{48}[ABCDEF]=\frac{15}{48}\cdot\frac{3\sqrt3}{2}=\boxed{\textbf{(C)}... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R;
A = dir(120);
B = dir(60);
C = dir(0);
D = dir(300);
E = dir(240);
F = dir(180);
X = midpoint(A--B);
Y = midpoint(C--D);
Z = midpoint(E--F);
M = intersectionpoint(A--E,X--Z);
N = intersectionpoint(A--C,X--Y);
O = intersec... | [] |
828 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A... | 2018 AMC 10B Problem 24 | If we try to coordinate bash this problem, it's going to look very ugly with a lot of radicals. However, we can alter and skew the diagram in such a way that all ratios of lengths and areas stay the same while making it a lot easier to work with. Then, we can find the ratio of the area of the wanted region to the area ... | // Block 1
unitsize(1cm);
draw((0,0)--(4,0)--(6,3.464)--(2,3.464)--(0,0));
draw((2,0)--(1,1.732));
draw((5,1.732)--(4,3.464));
draw((1.5, 0.866)--(3, 3.464)--(4.5, 0.866)--cycle);
draw((2,0)--(2,3.464)--(5,1.732)--cycle);
// Block 2
unitsize(1cm);
fill((1,4)--(1,3.5)--(2,3)--cycle,red);
fill((1,1)--(1.5,1)--(1,2)--cycl... | [] |
828 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A... | 2018 AMC 10B Problem 24 | (~MRENTHUSIASM's diagram + modifications)
Lemma 1: $MP$ is midsegment of $\triangle ANE$.
As $ZX$ is a midsegment of trapezoid $FEBA$, we know that $FA||ZX||EB$. Furthermore, by subdividing trapezoid $FEBA$, $ZM,MP,PX$ are midsegments of $\triangle EFA, \triangle ANE, \triangle, ANB$ respectively.
Lemma 2: $[ACE] = \... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
draw(polygon(6));
pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R, G;
A = dir(120);
B = dir(60);
C = dir(0);
D = dir(300);
E = dir(240);
F = dir(180);
X = midpoint(A--B);
Y = midpoint(C--D);
Z = midpoint(E--F);
G = midpoint(A--D);
M = intersectionpoint(A--E,X--Z);
N =... | [] |
829 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$. How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$?
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | 2018 AMC 10B Problem 25 | This rewrites itself to $x^2=10,000\{x\}$ where $\lfloor x \rfloor + \{x\} = x$.
Graphing $y=10,000\{x\}$ and $y=x^2$ we see that the former is a set of line segments with slope $10,000$ from $0$ to $1$ with a hole at $x=1$, then $1$ to $2$ with a hole at $x=2$ etc.
Here is a graph of $y=x^2$ and $y=16\{x\}$ for visua... | // Block 1
import graph;
size(400);
xaxis("$x$",Ticks(Label(fontsize(8pt)),new real[]{-5,-4,-3, -2, -1,0,1 2,3, 4,5}));
yaxis("$y$",Ticks(Label(fontsize(8pt)),new real[]{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18}));
real y(real x) {return x^2;}
draw(circle((-4,16), 0.1));
draw(circle((-3,16), 0.1));
draw(circle((-... | [] |
830 | A circle has a chord of length $10$, and the distance from the center of the circle to the chord is $5$. What is the area of the circle?
$\textbf{(A) }25\pi \qquad \textbf{(B) }50\pi \qquad \textbf{(C) }75\pi \qquad \textbf{(D) }100\pi \qquad \textbf{(E) }125\pi \qquad$ | 2018 AMC 12B Problem 4 | Let $O$ be the center of the circle, $\overline{AB}$ be the chord, and $M$ be the midpoint of $\overline{AB},$ as shown below.
Note that $\overline{OM}\perp\overline{AB}.$ Since $OM=AM=BM=5,$ we conclude that $\triangle OMA$ and $\triangle OMB$ are congruent isosceles right triangles. It follows that $r=5\sqrt2,$ so t... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
pair O, A, B, M;
O = (0,0);
A = (-5,5);
B = (5,5);
M = midpoint(A--B);
draw(Circle(O,5sqrt(2)));
dot("$O$", O, 1.5*S, linewidth(4.5));
dot("$A$", A, 1.5*NW, linewidth(4.5));
dot("$B$", B, 1.5*NE, linewidth(4.5));
dot("$M$", M, 1.5*N, linewidth(4.5));
draw(A--B^^M--O^^A... | [] |
831 | Line segment $\overline{AB}$ is a diameter of a circle with $AB = 24$. Point $C$, not equal to $A$ or $B$, lies on the circle. As point $C$ moves around the circle, the centroid (center of mass) of $\triangle ABC$ traces out a closed curve missing two points. To the nearest positive integer, what is the area of the reg... | 2018 AMC 12B Problem 8 | By the Inscribed Angle Theorem, $\triangle ABC$ is a right triangle with $\angle C=90^{\circ}.$ So, its circumcenter is the midpoint of $\overline{AB},$ and its median from $C$ is half as long as $\overline{AB}.$ For each $\triangle ABC,$ let $O$ and $G$ be its circumcenter and centroid, respectively. It follows that $... | /* Made by MRENTHUSIASM */ size(200); pair O, A, B, C1, C2, G1, G2, M1, M2; O = (0,0); A = (-12,0); B = (12,0); C1 = (36/5,48/5); C2 = (-96/17,-180/17); G1 = O + 1/3 * C1; G2 = O + 1/3 * C2; M1 = (4,0); M2 = (-4,0); draw(Circle(O,12)); draw(Circle(O,4),red); dot("$O$", O, (3/5,-4/5), linewidth(4.5)); dot("$A$", A, W,... | [] |
832 | A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the side... | 2018 AMC 12B Problem 11 | Consider one-quarter of the image (the wrapping paper is divided up into $4$ congruent squares). The length of each dotted line is $h$. The area of the rectangle that is $w$ by $h$ is $wh$. The combined figure of the two triangles with base $h$ is a square with $h$ as its diagonal. Using the Pythagorean Theorem, each s... | /* Edited by MRENTHUSIASM */ size(180pt); defaultpen(fontsize(10pt)); fill(((0,3)--(-3,3)--(-3,0)--(0,0)--cycle),lightgrey); dot((-3,3)); label("$A$",(-3,3),NW); draw((-3,0)--(-3,3)--(0,3),linewidth(.5)); draw((0,2)--(-1,3)--(-3,1)--(-2,0),dashed+linewidth(.5)); draw((0,2)--(-2,0),linewidth(.5)); draw((0,3)--(0,0),line... | [] |
832 | A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the side... | 2018 AMC 12B Problem 11 | The sheet of paper is made out of the surface area of the box plus the sum of the four yellow triangles, as shown below.
The surface area is $2w^2 + 2wh + 2wh$ which equals $2w^2 + 4wh$.The four triangles each have a height and a base of $h$, so they each have an area of $\frac{h^2}{2}$. There are four of them, so ... | /* Edited by MRENTHUSIASM */ size(180pt); defaultpen(fontsize(10pt)); fill(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),lightgrey); fill((2,0)--(3,1)--(3,-1)--cycle,yellow); fill((-2,0)--(-3,1)--(-3,-1)--cycle,yellow); fill((0,2)--(1,3)--(-1,3)--cycle,yellow); fill((0,-2)--(1,-3)--(-1,-3)--cycle,yellow); draw(((3,3)--(-3,3)... | [] |
833 | Square $ABCD$ has side length $30$. Point $P$ lies inside the square so that $AP = 12$ and $BP = 26$. The centroids of $\triangle{ABP}$, $\triangle{BCP}$, $\triangle{CDP}$, and $\triangle{DAP}$ are the vertices of a convex quadrilateral. What is the area of that quadrilateral?
$\textbf{(A) }100\sqrt{2}\qquad\textbf... | 2018 AMC 12B Problem 13 | As shown below, let $M_1,M_2,M_3,M_4$ be the midpoints of $\overline{AB},\overline{BC},\overline{CD},\overline{DA},$ respectively, and $G_1,G_2,G_3,G_4$ be the centroids of $\triangle{ABP},\triangle{BCP},\triangle{CDP},\triangle{DAP},$ respectively.
By SAS, we conclude that $\triangle G_1G_2P\sim\triangle M_1M_2P, \tr... | // Block 1
/* Made by MRENTHUSIASM */
unitsize(210);
pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3);
pair M1 = midpoint(A--B);
pair M2 = midpoint(B--C);
pair M3 = midpoint(C--D);
pair M4 = midpoint(D--A);
pair G1 = centroid(A,B,P);
pair G2 = centroid(B,C,P);
pair G3 = centroid(C,D,P);
pair G4 = cen... | [] |
834 | The solutions to the equation $(z+6)^8=81$ are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled $A,B,$ and $C$. What is the least possible area of $\triangle ABC?$
$\textbf{(A) } \frac{1}{6}\sqrt{6} \qquad \textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2} \qquad \textbf{... | 2018 AMC 12B Problem 16 | Recall that translations preserve the shapes and the sizes for all objects. We translate the solutions to the given equation $6$ units right, so they become the solutions to the equation $z^8=81.$
We rewrite $z$ to the polar form \[z=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta,\] where $r$ is the magnitude of ... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
int xMin = -2;
int xMax = 2;
int yMin = -2;
int yMax = 2;
int numRays = 24;
//Draws a polar grid that goes out to a number of circles
//equal to big, with numRays specifying the number of rays:
void polarGrid(int big, int numRays)
{
for (int i = 1; i < big+1; ++i... | [] |
835 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A... | 2018 AMC 12B Problem 20 | The desired area (hexagon $MPNQOR$) consists of an equilateral triangle ($\triangle MNO$) and three right triangles ($\triangle MPN,\triangle NQO,$ and $\triangle ORM$).
Notice that $\overline {AD}$ (not shown) and $\overline {BC}$ are parallel. $\overline {XY}$ divides transversals $\overline {AB}$ and $\overline {CD... | /* Made by MRENTHUSIASM */ size(200); draw(polygon(6)); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); M = intersectionpoint(A--E,X--Z); N = intersectionpoint(A--C,X--Y); O = i... | [] |
835 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A... | 2018 AMC 12B Problem 20 | Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of $3$ isosceles trapezoids (namely $AXZF,CYXB,$ and $EZYD$) and $3$ right triangles (namely $\triangle XPN,\triangle YQO,$ and $\triangle ZRM$).
Finding the trapezoid's area, we know that one base of each tr... | /* Made by MRENTHUSIASM */ size(200); draw(polygon(6)); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); M = intersectionpoint(A--E,X--Z); N = intersectionpoint(A--C,X--Y); O = i... | [] |
835 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A... | 2018 AMC 12B Problem 20 | We could also subtract $\triangle APM,\triangle CQN,$ and $\triangle ERO$ from $\triangle ACE.$
Since $\angle BAF = 120^{\circ}$ and $\angle BAC = \angle FAE = 30^{\circ},$ we have $\angle CAE = \angle BAF-\angle BAC-\angle FAE=60^{\circ}.$
Since $AX=BX$ and $FZ=EZ,$ we have $AF \parallel XZ,$ from which $\angle AMX=... | /* Made by MRENTHUSIASM */ size(200); draw(polygon(6)); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); M = intersectionpoint(A--E,X--Z); N = intersectionpoint(A--C,X--Y); O = i... | [] |
835 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A... | 2018 AMC 12B Problem 20 | We partition hexagon $ABCDEF$ into $48$ congruent $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangles, as shown below:
Let the brackets denote areas. Note that the desired region contains $15$ of the $48$ small triangles, so the answer is \[\frac{15}{48}[ABCDEF]=\frac{15}{48}\cdot\frac{3\sqrt3}{2}=\boxed{\textbf{(C)}... | /* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); M = intersectionpoint(A--E,X--Z); N = intersectionpoint(A--C,X--Y); O = intersectionpoint(C... | [] |
835 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A... | 2018 AMC 12B Problem 20 | If we try to coordinate bash this problem, it's going to look very ugly with a lot of radicals. However, we can alter and skew the diagram in such a way that all ratios of lengths and areas stay the same while making it a lot easier to work with. Then, we can find the ratio of the area of the wanted region to the area ... | // Block 1
unitsize(1cm); draw((0,0)--(4,0)--(6,3.464)--(2,3.464)--(0,0)); draw((2,0)--(1,1.732)); draw((5,1.732)--(4,3.464)); draw((1.5, 0.866)--(3, 3.464)--(4.5, 0.866)--cycle); draw((2,0)--(2,3.464)--(5,1.732)--cycle);
// Block 2
unitsize(1cm); fill((1,4)--(1,3.5)--(2,3)--cycle,red); fill((1,1)--(1.5,1)--(1,2)--cycl... | [] |
835 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A... | 2018 AMC 12B Problem 20 | (~MRENTHUSIASM's diagram + modifications)
Lemma 1: $MP$ is midsegment of $\triangle ANE$.
As $ZX$ is a midsegment of trapezoid $FEBA$, we know that $FA||ZX||EB$. Furthermore, by subdividing trapezoid $FEBA$, $ZM,MP,PX$ are midsegments of $\triangle EFA, \triangle ANE, \triangle, ANB$ respectively.
Lemma 2: $[ACE] = \... | /* Made by MRENTHUSIASM */ size(200); draw(polygon(6)); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R, G; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); G = midpoint(A--D); M = intersectionpoint(A--E,X--Z); N = intersecti... | [] |
836 | Ajay is standing at point $A$ near Pontianak, Indonesia, $0^\circ$ latitude and $110^\circ \text{ E}$ longitude. Billy is standing at point $B$ near Big Baldy Mountain, Idaho, USA, $45^\circ \text{ N}$ latitude and $115^\circ \text{ W}$ longitude. Assume that Earth is a perfect sphere with center $C.$ What is the degre... | 2018 AMC 12B Problem 23 | This solution refers to the Diagram section.
Let $D$ be the orthogonal projection of $B$ onto the equator. Note that $\angle BDA = \angle BDC = 90^\circ$ and $\angle BCD = 45^\circ.$ Recall that $115^\circ \text{ W}$ longitude is the same as $245^\circ \text{ E}$ longitude, so $\angle ACD=135^\circ.$
We obtain the fo... | // Block 1
/* Made by MRENTHUSIASM */
size(250);
import graph3;
import solids;
currentprojection=orthographic((0.2,-0.5,0.2));
triple A, B, C, D;
A = (1,0,0);
B = (-1/2,1/2,sqrt(2)/2);
C = (0,0,0);
D = (-1/2,1/2,0);
draw(unitsphere,white,light=White);
draw(surface(A--B--C--cycle),yellow);
dot(A^^B^^C^^D,linewidth(4.5)... | [] |
836 | Ajay is standing at point $A$ near Pontianak, Indonesia, $0^\circ$ latitude and $110^\circ \text{ E}$ longitude. Billy is standing at point $B$ near Big Baldy Mountain, Idaho, USA, $45^\circ \text{ N}$ latitude and $115^\circ \text{ W}$ longitude. Assume that Earth is a perfect sphere with center $C.$ What is the degre... | 2018 AMC 12B Problem 23 | This solution refers to the Diagram section.
Let $D$ be the orthogonal projection of $B$ onto the equator. Note that $\angle BDA = \angle BDC = 90^\circ$ and $\angle BCD = 45^\circ.$ Recall that $115^\circ \text{ W}$ longitude is the same as $245^\circ \text{ E}$ longitude, so $\angle ACD=135^\circ.$
Without loss of ... | // Block 1
/* Made by MRENTHUSIASM */
size(300);
import graph3;
import solids;
currentprojection=orthographic((0.2,-0.5,0.2));
triple A, B, C, D;
A = (1,0,0);
B = (-1/2,1/2,sqrt(2)/2);
C = (0,0,0);
D = (-1/2,1/2,0);
draw(unitsphere,white,light=White);
dot(A^^B^^C^^D,linewidth(4.5));
draw(Circle(C,1,(0,0,1))^^B--C--D--... | [] |
837 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$. How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$?
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | 2018 AMC 12B Problem 24 | This rewrites itself to $x^2=10,000\{x\}$ where $\lfloor x \rfloor + \{x\} = x$.
Graphing $y=10,000\{x\}$ and $y=x^2$ we see that the former is a set of line segments with slope $10,000$ from $0$ to $1$ with a hole at $x=1$, then $1$ to $2$ with a hole at $x=2$ etc.
Here is a graph of $y=x^2$ and $y=16\{x\}$ for visua... | import graph; size(400); xaxis("$x$",Ticks(Label(fontsize(8pt)),new real[]{-5,-4,-3, -2, -1,0,1 2,3, 4,5})); yaxis("$y$",Ticks(Label(fontsize(8pt)),new real[]{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18})); real y(real x) {return x^2;} draw(circle((-4,16), 0.1)); draw(circle((-3,16), 0.1)); draw(circle((-2,16), 0.1)... | [] |
838 | Circles $\omega_1$, $\omega_2$, and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$, $P_2$, and $P_3$ lie on $\omega_1$, $\omega_2$, and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\om... | 2018 AMC 12B Problem 25 | First, note that because the $\angle P_1=\angle P_2=\angle P_3=\pi/3$, the arcs inside the shaded equilateral triangle are each $2\pi/3$. Also, the distances between the centers of any two of the $3$ given circles are each $8$.
Draw the circle $\Gamma$ concentric with $\omega_1$ with radius $2$. Because the arc of $\om... | // Block 1
unitsize(12);
pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A;
real theta = 41.5;
pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1;
filldraw(P1--P2--P3--cycle, gray(0.9));
draw(Circle(A, 4));
draw(Circle(B, 4));
draw(Circle(C, 4));
dot(P1);
dot(P2);
dot(P3... | [] |
838 | Circles $\omega_1$, $\omega_2$, and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$, $P_2$, and $P_3$ lie on $\omega_1$, $\omega_2$, and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\om... | 2018 AMC 12B Problem 25 | Suppose $P_1P_2 = x$, noticed that $OMP_1H$ is rectangle.
\[OO_1 = \frac{O_1O_2}{\sqrt{3}} = \frac83\sqrt{3}\]
\[OH = P_1M = \frac{1}{2}P_1P_2 = \frac12 x\]
\[O_1H = O_1P_1 - HP_1 = O_1P_1 - OM = O_1P_1 - \frac{P_1P_2}{2\sqrt{3}} = 4 - \frac{\sqrt{3}}{6}x\]
$\triangle O_1HO$ is right triangle, we can use Pythagorean th... | // Block 1
import geometry;
unitsize(12);
pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A;
real theta = 41.5;
pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1;
filldraw(P1--P2--P3--cycle, gray(0.9));
draw(Circle(A, 4));
draw(Circle(B, 4));
draw(Circle(C, 4));
dot(P1);... | [] |
839 | A rectangular floor that is $10$ feet wide and $17$ feet long is tiled with $170$ one-foot square tiles. A bug walks from one corner to the opposite corner in a straight line. Including the first and the last tile, how many tiles does the bug visit?
$\textbf{(A) } 17 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 26 \qqua... | 2019 AMC 10A Problem 10 | We can also draw a diagram or scale model of the entire rectangular floor (optionally with grid paper and/or a ruler so it will be to scale), then simply count the number of tiles the path crosses. To make this slightly easier, we can divide the full grid into $4$ sections, and just draw one of these $5$ feet by $8.5$ ... | // Block 1
unitsize(20);
for(int i =0; i<= 7; ++i) {
for(int j =0; j<= 4; ++j) {
draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle);
}
for(int k =0; k<= 4; ++k) {
draw((8,k)--(8.5,k)--(8.5,k+1)--(8,k+1)--cycle);
}
}
draw((0,5)--(8.5,0)--cycle);
// Block 2
unitsize(20); for(int i =0; i<= 7; ++i) { for(int j =0; j<= 4; +... | [] |
840 | Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$. Construct the circle with diameter $\overline{BC}$, and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$, respectively. Let $F$ be the intersection of the diagonals... | 2019 AMC 10A Problem 13 | Drawing it out, we see $\angle BDC$ and $\angle BEC$ are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find $\angle ABC=70^{\circ}$. We can find $\angle ECB=20^{\circ}$ and $\angle DBC=50^{\circ}$ by the triangle angle sum on $\triangle ECB$ and $\triangle DBC$... | unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label("$A$",(1,0),SE);label("$C$",(0,2.75),N);label("$B$",(-1,0),SW);label("$E$",(0,0),S);label("$D$",(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64)); | [] |
841 | For a set of four distinct lines in a plane, there are exactly $N$ distinct points that lie on two or more of the lines. What is the sum of all possible values of $N$?
$\textbf{(A) } 14 \qquad \textbf{(B) } 16 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 21$ | 2019 AMC 10A Problem 14 | It is possible to obtain $0$, $1$, $3$, $4$, $5$, and $6$ points of intersection, as demonstrated in the following figures:
It is clear that the maximum number of possible intersections is ${4 \choose 2} = 6$, since each pair of lines can intersect at most once. In addition, by looking at the answer choices, we kn... | // Block 1
unitsize(2cm);
real d = 2.5;
draw((-1,.6)--(1,.6),Arrows);
draw((-1,.2)--(1,.2),Arrows);
draw((-1,-.2)--(1,-.2),Arrows);
draw((-1,-.6)--(1,-.6),Arrows);
draw((-1+d,0)--(1+d,0),Arrows);
draw((0+d,1)--(0+d,-1),Arrows);
draw(dir(45)+(d,0)--dir(45+180)+(d,0),Arrows);
draw(dir(135)+(d,0)--dir(135+180)+(d,0),Arro... | [] |
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