link stringlengths 75 84 | letter stringclasses 5
values | answer float64 0 2,935,363,332B | problem stringlengths 14 5.33k | solution listlengths 1 13 |
|---|---|---|---|---|
https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_2 | null | 36 | In rectangle $ABCD$ $AB = 12$ and $BC = 10$ . Points $E$ and $F$ lie inside rectangle $ABCD$ so that $BE = 9$ $DF = 8$ $\overline{BE} \parallel \overline{DF}$ $\overline{EF} \parallel \overline{AB}$ , and line $BE$ intersects segment $\overline{AD}$ . The length $EF$ can be expressed in the form $m \sqrt{n} - p$ , wh... | [
"Let us call the point where $\\overline{EF}$ intersects $\\overline{AD}$ point $G$ , and the point where $\\overline{EF}$ intersects $\\overline{BC}$ point $H$ . Since angles $FHB$ and $EGA$ are both right angles, and angles $BEF$ and $DFE$ are congruent due to parallelism, right triangles $BHE$ and $DGF$ are simi... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_16 | E | 16 | In rectangle $ABCD$ $AB=1$ $BC=2$ , and points $E$ $F$ , and $G$ are midpoints of $\overline{BC}$ $\overline{CD}$ , and $\overline{AD}$ , respectively. Point $H$ is the midpoint of $\overline{GE}$ . What is the area of the shaded region?
[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (... | [
"Denote $D=(0,0)$ . Then $A= (0,2), F = \\left(\\frac12,0\\right), H = \\left(\\frac12,1\\right)$ . Let the intersection of $AF$ and $DH$ be $X$ , and the intersection of $BF$ and $CH$ be $Y$ . Then we want to find the coordinates of $X$ so we can find $XY$ . From our points, the slope of $AF$ is $\\bigg(\\dfrac{-2... |
https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_3 | null | 141 | In rectangle $ABCD$ $AB=100$ . Let $E$ be the midpoint of $\overline{AD}$ . Given that line $AC$ and line $BE$ are perpendicular, find the greatest integer less than $AD$ | [
"From the problem, $AB=100$ and triangle $FBA$ is a right triangle. As $ABCD$ is a rectangle, triangles $BCA$ , and $ABE$ are also right triangles. By $AA$ $\\triangle FBA \\sim \\triangle BCA$ , and $\\triangle FBA \\sim \\triangle ABE$ , so $\\triangle ABE \\sim \\triangle BCA$ . This gives $\\frac {AE}{AB}= \\fr... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_2 | A | 12 | In rectangle $ABCD$ $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$
[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label("$A$", (0,0), SW); label("$B$", (6, 0), SE); label("$C$", (6,8), NE); label("$D$", (0, 8), NW); label("$M$", (0, 4), W); lab... | [
"Using the triangle area formula for triangles: $A = \\frac{bh}{2},$ where $A$ is the area, $b$ is the base, and $h$ is the height. This equation gives us $A = \\frac{4 \\cdot 6}{2} = \\frac{24}{2} =\\boxed{12}$",
"A triangle with the same height and base as a rectangle is half of the rectangle's area. This mean... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_22 | E | 20 | In rectangle $ABCD$ $\overline{AB}=20$ and $\overline{BC}=10$ . Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$ . What is $\overline{AE}$
$\textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20$ | [
"Note that $\\tan 15^\\circ=2-\\sqrt{3}=\\frac{EC}{10} \\Rightarrow EC=20-10 \\sqrt 3$ . (It is important to memorize the sin, cos, and tan values of $15^\\circ$ and $75^\\circ$ .) Therefore, we have $DE=10\\sqrt 3$ . Since $ADE$ is a $30-60-90$ triangle, $AE=2 \\cdot AD=2 \\cdot 10=\\boxed{20}$",
"Let $F$ be a p... |
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_17 | E | 3 | In rectangle $ABCD$ , angle $C$ is trisected by $\overline{CF}$ and $\overline{CE}$ , where $E$ is on $\overline{AB}$ $F$ is on $\overline{AD}$ $BE=6$ and $AF=2$ . Which of the following is closest to the area of the rectangle $ABCD$ [asy] pair A=origin, B=(10,0), C=(10,7), D=(0,7), E=(5,0), F=(0,2); draw(A--B--C--D--c... | [
"Since $\\angle C = 90^\\circ$ , each of the three smaller angles is $30^\\circ$ , and $\\triangle BEC$ and $\\triangle CDF$ are both $30-60-90$ triangles.\n\nDefining the variables as illustrated above, we have $x = 6\\sqrt{3}$ from $\\triangle BEC$\nThen $x-2 = 6\\sqrt{3} - 2$ , and $y = \\sqrt{3} (6 \\sqrt{3} - ... |
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_17 | E | 150 | In rectangle $ABCD$ , angle $C$ is trisected by $\overline{CF}$ and $\overline{CE}$ , where $E$ is on $\overline{AB}$ $F$ is on $\overline{AD}$ $BE=6$ and $AF=2$ . Which of the following is closest to the area of the rectangle $ABCD$ [asy] pair A=origin, B=(10,0), C=(10,7), D=(0,7), E=(5,0), F=(0,2); draw(A--B--C--D--c... | [
"Use the process above, but use $\\sqrt{3} \\approx 1.73$ . You should get $[ABCD]=150.84$ , which then you select $\\boxed{150}$ . Notice that the actual area, when plugged into a calculator, yields about $151.0614872$"
] |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_22 | C | 3 | In rectangle $ABCD$ , points $F$ and $G$ lie on $AB$ so that $AF=FG=GB$ and $E$ is the midpoint of $\overline{DC}$ . Also, $\overline{AC}$ intersects $\overline{EF}$ at $H$ and $\overline{EG}$ at $J$ . The area of the rectangle $ABCD$ is $70$ . Find the area of triangle $EHJ$
$\text{(A) }\frac {5}{2} \qquad \text{(B) }... | [
"$[CEF] = \\frac{[ABCD]}{4} = \\frac{35}{2}$\n$\\triangle CEH \\sim \\triangle AFH$ $\\frac{HE}{HF} = \\frac{CE}{AF} = \\frac{3}{2}$ $\\frac{HE}{EF} = \\frac{3}{5}$\n$[CEH] = \\frac{HE}{EF} \\cdot [CEF] = \\frac{3}{5} \\cdot \\frac{35}{2} = \\frac{21}{2}$\n$\\triangle CEH \\sim \\triangle AFH$ $\\frac{AH}{HC} = \\f... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_22 | null | 3 | In rectangle $ABCD$ , points $F$ and $G$ lie on $AB$ so that $AF=FG=GB$ and $E$ is the midpoint of $\overline{DC}$ . Also, $\overline{AC}$ intersects $\overline{EF}$ at $H$ and $\overline{EG}$ at $J$ . The area of the rectangle $ABCD$ is $70$ . Find the area of triangle $EHJ$
$\text{(A) }\frac {5}{2} \qquad \text{(B) }... | [
" Note that the triangles $AFH$ and $CEH$ are similar, as they have the same angles. Hence $\\frac {AH}{HC} = \\frac{AF}{EC} = \\frac 23$\nAlso, triangles $AGJ$ and $CEJ$ are similar, hence $\\frac {AJ}{JC} = \\frac {AG}{EC} = \\frac 43$\nWe can now compute $[EHJ]$ as $[ACD]-[AHD]-[DEH]-[EJC]$ . We have:\nTherefore... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_20 | E | 40,400 | In rectangle $ABCD$ , we have $A=(6,-22)$ $B=(2006,178)$ $D=(8,y)$ , for some integer $y$ . What is the area of rectangle $ABCD$
$\mathrm{(A) \ } 4000\qquad \mathrm{(B) \ } 4040\qquad \mathrm{(C) \ } 4400\qquad \mathrm{(D) \ } 40,000\qquad \mathrm{(E) \ } 40,400$ | [
"This solution is the same as Solution 1 up to the point where we find that $y=-42$\nWe build right triangles so we can use the Pythagorean Theorem. The triangle with hypotenuse $AB$ has legs $200$ and $2000$ , while the triangle with hypotenuse $AD$ has legs $2$ and $20$ . Aha! The two triangles are similar by ... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_22 | B | 20 | In rectangle $ABCD$ , we have $AB=8$ $BC=9$ $H$ is on $BC$ with $BH=6$ $E$ is on $AD$ with $DE=4$ , line $EC$ intersects line $AH$ at $G$ , and $F$ is on line $AD$ with $GF \perp AF$ . Find the length of $GF$
[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair D=(0,0), Ep=(4,0), A=(9,0), B=(9,8), H=(3,... | [
"$\\angle GHC = \\angle AHB$ (Vertical angles are equal).\n$\\angle F = \\angle B$ (Both are 90 degrees).\n$\\angle BHA = \\angle HAD$ (Alt. Interior Angles are congruent).\nTherefore $\\triangle GFA$ and $\\triangle ABH$ are similar. $\\triangle GCH$ and $\\triangle GEA$ are also similar.\n$DA$ is 9, therefore $EA... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_19 | E | 20 | In rectangle $ABCD,$ $AB=6$ and $BC=3$ . Point $E$ between $B$ and $C$ , and point $F$ between $E$ and $C$ are such that $BE=EF=FC$ . Segments $\overline{AE}$ and $\overline{AF}$ intersect $\overline{BD}$ at $P$ and $Q$ , respectively. The ratio $BP:PQ:QD$ can be written as $r:s:t$ where the greatest common factor of $... | [
"\nUse similar triangles. Our goal is to put the ratio in terms of ${BD}$ . Since $\\triangle APD \\sim \\triangle EPB,$ $\\frac{DP}{PB}=\\frac{AD}{BE}=3.$ Therefore, $PB=\\frac{BD}{4}$ . Similarly, $\\frac{DQ}{QB}=\\frac{3}{2}$ . This means that ${DQ}=\\frac{3\\cdot BD}{5}$ . Therefore, $r:s:t=\\frac{1}{4}:\\frac{... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_17 | A | 12 | In rectangle $ADEH$ , points $B$ and $C$ trisect $\overline{AD}$ , and points $G$ and $F$ trisect $\overline{HE}$ . In addition, $AH=AC=2$ , and $AD=3$ . What is the area of quadrilateral $WXYZ$ shown in the figure?
[asy] size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); pair A,B,C,D,E... | [
"By symmetry $WXYZ$ is a square.\n\nDraw $\\overline{BZ}$ $BZ = \\frac 12AH = 1$ , so $\\triangle BWZ$ is a $45-45-90 \\triangle$ . Hence $WZ = \\frac{1}{\\sqrt{2}}$ , and $[WXYZ] = \\left(\\frac{1}{\\sqrt{2}}\\right)^2 =\\boxed{12}$",
"\nDrawing lines as shown above and piecing together the triangles, we see tha... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_17 | B | 7 | In rectangle $PQRS$ $PQ=8$ and $QR=6$ . Points $A$ and $B$ lie on $\overline{PQ}$ , points $C$ and $D$ lie on $\overline{QR}$ , points $E$ and $F$ lie on $\overline{RS}$ , and points $G$ and $H$ lie on $\overline{SP}$ so that $AP=BQ<4$ and the convex octagon $ABCDEFGH$ is equilateral. The length of a side of this octag... | [
"Let $AP=BQ=x$ . Then $AB=8-2x$\nNow notice that since $CD=8-2x$ we have $QC=DR=x-1$\nThus by the Pythagorean Theorem we have $x^2+(x-1)^2=(8-2x)^2$ which becomes $2x^2-30x+63=0\\implies x=\\dfrac{15-3\\sqrt{11}}{2}$\nOur answer is $8-(15-3\\sqrt{11})=3\\sqrt{11}-7\\implies \\boxed{7}$ . (Mudkipswims42)",
"Denote... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_2 | D | 20 | In rhombus $ABCD$ , point $P$ lies on segment $\overline{AD}$ so that $\overline{BP}$ $\perp$ $\overline{AD}$ $AP = 3$ , and $PD = 2$ . What is the area of $ABCD$ ? (Note: The figure is not drawn to scale.)
[asy] import olympiad; size(180); real r = 3, s = 5, t = sqrt(r*r+s*s); defaultpen(linewidth(0.6) + fontsize(10))... | [
"\n\\[\\textbf{Figure redrawn to scale.}\\]\n$AD = AP + PD = 3 + 2 = 5$\n$ABCD$ is a rhombus, so $AB = AD = 5$\n$\\bigtriangleup APB$ is a $3-4-5$ right triangle, hence $BP = 4$\nThe area of the rhombus is base times height: $bh = (AD)(BP) = 5 \\cdot 4 = \\boxed{20}$",
"\nThe diagram is from as Solution 1, but a... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_2 | D | 20 | In rhombus $ABCD$ , point $P$ lies on segment $\overline{AD}$ so that $\overline{BP}$ $\perp$ $\overline{AD}$ $AP = 3$ , and $PD = 2$ . What is the area of $ABCD$ ? (Note: The figure is not drawn to scale.)
[asy] import olympiad; size(180); real r = 3, s = 5, t = sqrt(r*r+s*s); defaultpen(linewidth(0.6) + fontsize(10))... | [
"\n\\[\\textbf{Figure redrawn to scale.}\\]\n$AD = AP + PD = 3 + 2 = 5$\n$ABCD$ is a rhombus, so $AB = AD = 5$\n$\\bigtriangleup APB$ is a $3-4-5$ right triangle, hence $BP = 4$\nThe area of the rhombus is base times height: $bh = (AD)(BP) = 5 \\cdot 4 = \\boxed{20}$",
"\nThe diagram is from as Solution 1, but a... |
https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_12 | null | 11 | In right $\triangle ABC$ with hypotenuse $\overline{AB}$ $AC = 12$ $BC = 35$ , and $\overline{CD}$ is the altitude to $\overline{AB}$ . Let $\omega$ be the circle having $\overline{CD}$ as a diameter. Let $I$ be a point outside $\triangle ABC$ such that $\overline{AI}$ and $\overline{BI}$ are both tangent to circle $\o... | [
"First, note that $AB=37$ ; let the tangents from $I$ to $\\omega$ have length $x$ . Then the perimeter of $\\triangle ABI$ is equal to \\[2(x+AD+DB)=2(x+37).\\] It remains to compute $\\dfrac{2(x+37)}{37}=2+\\dfrac{2}{37}x$\nObserve $CD=\\dfrac{12\\cdot 35}{37}=\\dfrac{420}{37}$ , so the radius of $\\omega$ is $\\... |
https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_3 | D | 4 | In right $\triangle ABC$ with legs $5$ and $12$ , arcs of circles are drawn, one with center $A$ and radius $12$ , the other with center $B$ and radius $5$ . They intersect the hypotenuse in $M$ and $N$ . Then $MN$ has length
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(12,7), C=(12,0), M=12*dir(A--... | [
"Firstly, the Pythagorean theorem gives \\begin{align*}AB &=\\sqrt{AC^2+BC^2} \\\\ &= \\sqrt{12^2+5^2} \\\\ & =\\sqrt{144+25} \\\\ &=\\sqrt{169} \\\\ &= 13.\\end{align*} Also, $AM = AC = 12$ and $BN = BC = 5$ since they are both radii of the respective circles. Thus $MB = AB-AM = 13-12 = 1$ , and so $MN = BN-BM = 5... |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_9 | null | 737 | In right triangle $ABC$ with right angle $C$ $CA = 30$ and $CB = 16$ . Its legs $CA$ and $CB$ are extended beyond $A$ and $B$ Points $O_1$ and $O_2$ lie in the exterior of the triangle and are the centers of two circles with equal radii . The circle with center $O_1$ is tangent to the hypotenuse and to the extension ... | [
"A different approach is to plot the triangle on the Cartesian Plane with $C$ at $(0,0)$ $A$ at $(0,30)$ , and $B$ at $(16,0)$ . We wish to find the coordinates of $O_1$ and $O_2$ in terms of the radius, which will be expressed as $r$ in the rest of this solution. When we know the coordinates, we will set the dista... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_4 | E | 100 | In some countries, automobile fuel efficiency is measured in liters per $100$ kilometers while other countries use miles per gallon. Suppose that 1 kilometer equals $m$ miles, and $1$ gallon equals $l$ liters. Which of the following gives the fuel efficiency in liters per $100$ kilometers for a car that gets $x$ miles ... | [
"The formula for fuel efficiency is \\[\\frac{\\text{Distance}}{\\text{Gas Consumption}}.\\] Note that $1$ mile equals $\\frac 1m$ kilometers. We have \\[\\frac{x\\text{ miles}}{1\\text{ gallon}} = \\frac{\\frac{x}{m}\\text{ kilometers}}{l\\text{ liters}} = \\frac{1\\text{ kilometer}}{\\frac{lm}{x}\\text{ liters}} ... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_15 | D | 117 | In square $ABCD$ , points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$ , with $BR = 6$ and $PR = 7$ . What is the area of the square?
[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = o... | [
"Note that $\\triangle APB \\cong \\triangle BQC.$ Then, it follows that $\\overline{PB} \\cong \\overline{QC}.$ Thus, $QC = PB = PR + RB = 7 + 6 = 13.$ Define $x$ to be the length of side $CR,$ then $RQ = 13-x.$ Because $\\overline{BR}$ is the altitude of the triangle, we can use the property that $QR \\cdot RC = ... |
https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_9 | null | 20 | In tetrahedron $ABCD$ edge $AB$ has length 3 cm. The area of face $ABC$ is $15\mbox{cm}^2$ and the area of face $ABD$ is $12 \mbox { cm}^2$ . These two faces meet each other at a $30^\circ$ angle. Find the volume of the tetrahedron in $\mbox{cm}^3$ | [
"Position face $ABC$ on the bottom. Since $[\\triangle ABD] = 12 = \\frac{1}{2} \\cdot AB \\cdot h_{ABD}$ , we find that $h_{ABD} = 8$ . Because the problem does not specify, we may assume both $ABC$ and $ABD$ to be isosceles triangles. Thus, the height of $ABD$ forms a $30-60-90$ with the height of the tetrahedron... |
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_28 | C | 2 | In the $xy$ -plane, how many lines whose $x$ -intercept is a positive prime number and whose $y$ -intercept is a positive integer pass through the point $(4,3)$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$ | [
"The line with $x$ -intercept $a$ and $y$ -intercept $b$ is given by the equation $\\frac{x}{a} + \\frac{y}{b} = 1$ . We are told $(4,3)$ is on the line so\n\\[\\frac{4}{a} + \\frac{3}{b} = 1 \\implies ab - 4b - 3a = 0 \\implies (a-4)(b-3)=12\\]\nSince $a$ and $b$ are integers, this equation holds only if $(a-4)$ ... |
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_4 | A | 10 | In the $xy$ -plane, the segment with endpoints $(-5,0)$ and $(25,0)$ is the diameter of a circle. If the point $(x,15)$ is on the circle, then $x=$
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12.5 \qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 17.5 \qquad\textbf{(E)}\ 20$ | [
"We see that the center of this circle is at $\\left(\\frac{-5+25}{2},0\\right)=(10,0)$ . The radius is $\\frac{30}{2}=15$ . So the equation of this circle is \\[(x-10)^2+y^2=225.\\] Substituting $y=15$ yields $(x-10)^2=0$ so $x=\\boxed{10}$"
] |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_14 | B | 7 | In the BIG N, a middle school football conference, each team plays every other team exactly once. If a total of 21 conference games were played during the 2012 season, how many teams were members of the BIG N conference?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qqua... | [
"This problem is very similar to a handshake problem. We use the formula $\\frac{n(n-1)}{2}$ to usually find the number of games played (or handshakes). Now we have to use the formula in reverse.\nSo we have the equation $\\frac{n(n-1)}{2} = 21$ . Solving, we find that the number of teams in the BIG N conference is... |
https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_4 | null | 40 | In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$ . Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant. Let $P=(x,y)$ be the center of $\triangle ABC$ . Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$ , where $p$ and $r$ are relatively prime positive... | [
"The distance from point $A$ to point $B$ is $\\sqrt{13}$ . The vector that starts at point A and ends at point B is given by $B - A = (1, 2\\sqrt{3})$ . Since the center of an equilateral triangle, $P$ , is also the intersection of the perpendicular bisectors of the sides of the triangle, we need first find the eq... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_13 | B | 8 | In the United States, coins have the following thicknesses: penny, $1.55$ mm; nickel, $1.95$ mm; dime, $1.35$ mm; quarter, $1.75$ mm. If a stack of these coins is exactly $14$ mm high, how many coins are in the stack?
$\mathrm{(A) \ } 7 \qquad \mathrm{(B) \ } 8 \qquad \mathrm{(C) \ } 9 \qquad \mathrm{(D) \ } 10 \qquad ... | [
"All numbers in this solution will be in hundredths of a millimeter.\nThe thinnest coin is the dime, with thickness $135$ . A stack of $n$ dimes has height $135n$\nThe other three coin types have thicknesses $135+20$ $135+40$ , and $135+60$ . By replacing some of the dimes in our stack by other, thicker coins, we c... |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_5 | null | 750 | In the accompanying figure, the outer square $S$ has side length $40$ . A second square $S'$ of side length $15$ is constructed inside $S$ with the same center as $S$ and with sides parallel to those of $S$ . From each midpoint of a side of $S$ , segments are drawn to the two closest vertices of $S'$ . The result is a ... | [
"The volume of this pyramid can be found by the equation $V=\\frac{1}{3}bh$ , where $b$ is the base and $h$ is the height. The base is easy, since it is a square and has area $15^2=225$\nTo find the height of the pyramid, the height of the four triangles is needed, which will be called $h^\\prime$ . By drawing a li... |
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_20 | A | 1 | In the addition problem, each digit has been replaced by a letter. If different letters represent different digits then what is the value of $C$
[asy] unitsize(18); draw((-1,0)--(3,0)); draw((-3/4,1/2)--(-1/4,1/2)); draw((-1/2,1/4)--(-1/2,3/4)); label("$A$",(0.5,2.1),N); label("$B$",(1.5,2.1),N); label("$C$",(2.5,2.1)... | [
"From this we have \\[111A+11B+C=300.\\] Clearly, $A<3$ . Since $B,C\\leq 9$ \\[111A > 201 \\Rightarrow A\\geq 2.\\] Thus, $A=2$ and $11B+C=78$ . From here it becomes clear that $B=7$ and $C=1\\rightarrow \\boxed{1}$",
"Using logic, $a+b+c= 10$ , therefore $b+a+1$ (from the carry over) $= 10$ .\nSo $b+a=9$ $A+1... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_10 | C | 7 | In the addition shown below $A$ $B$ $C$ , and $D$ are distinct digits. How many different values are possible for $D$
\[\begin{array}[t]{r} ABBCB \\ + \ BCADA \\ \hline DBDDD \end{array}\]
$\textbf {(A) } 2 \qquad \textbf {(B) } 4 \qquad \textbf {(C) } 7 \qquad \textbf {(D) } 8 \qquad \textbf {(E) } 9$ | [
"Note from the addition of the last digits that $A+B=D\\text{ or }A+B=D+10$ . \nFrom the addition of the frontmost digits, $A+B$ cannot have a carry, since the answer is still a five-digit number. Also $A + B$ cant have a carry since then for the second column, $C + 1 + D$ cant equal $D$ . \nTherefore $A+B=D$\nUsin... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_8 | C | 7 | In the addition shown below $A$ $B$ $C$ , and $D$ are distinct digits. How many different values are possible for $D$
\[\begin{tabular}{cccccc}&A&B&B&C&B\\ +&B&C&A&D&A\\ \hline &D&B&D&D&D\end{tabular}\]
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$ | [
"From the first column, we see $A+B < 10$ because it yields a single digit answer. From the fourth column, we see that $C+D$ equals $D$ and therefore $C = 0$ . We know that $A+B = D$ . Therefore, the number of values $D$ can take is equal to the number of possible sums less than $10$ that can be formed by adding... |
https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_23 | A | 12 | In the adjoining figure the five circles are tangent to one another consecutively and to the lines $L_1$ and $L_2$ .
If the radius of the largest circle is $18$ and that of the smallest one is $8$ , then the radius of the middle circle is
[asy] size(250);defaultpen(linewidth(0.7)); real alpha=5.797939254, x=71.191836;... | [
"Pdfresizer.com-pdf-convert-q23.png\nConsider three consecutive circles, as shown in the diagram above; observe that their centres $P$ $Q$ , and $R$ are collinear by symmetry. Let $A$ $B$ , and $C$ be the points of tangency, and let $PS$ and $QT$ be segments parallel to the upper tangent (i.e. $L_1$ ), as also show... |
https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_24 | C | 16 | In the adjoining figure, circle $K$ has diameter $AB$ ; circle $L$ is tangent to circle $K$ and to $AB$ at the center of circle $K$ ; and circle $M$ tangent to circle $K$ , to circle $L$ and $AB$ . The ratio of the area of circle $K$ to the area of circle $M$ is [asy] /* Made by Klaus-Anton, Edited by MRENTHUSIASM */ s... | [
"Let $R$ and $r$ be the radius of $\\odot K$ and the radius of $\\odot M,$ respectively. It follows that the radius of $\\odot L$ is $\\frac{R}{2}.$\nSuppose $P$ is the foot of the perpendicular from $M$ to $\\overline{KL}.$ We construct the auxiliary lines, as shown below: In right $\\triangle KPM,$ we have $KP=r... |
https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_14 | A | 24 | In the adjoining figure, points $B$ and $C$ lie on line segment $AD$ , and $AB, BC$ , and $CD$ are diameters of circle $O, N$ , and $P$ , respectively. Circles $O, N$ , and $P$ all have radius $15$ and the line $AG$ is tangent to circle $P$ at $G$ . If $AG$ intersects circle $N$ at points $E$ and $F$ , then chord $EF$ ... | [
"Drop a perpendicular line from $N$ to $AG$ at point $H$ $AN=45$ , and since $\\triangle{AGP}$ is similar to $\\triangle{AHN}$ $NH=9$ $NE=NF=15$ so by the Pythagorean Theorem, $EH=HF=12$ . Thus $EF=\\boxed{24.}$"
] |
https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14 | null | 130 | In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$
[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(... | [
"Firstly, notice that if we reflect $R$ over $P$ , we get $Q$ . Since we know that $R$ is on circle $B$ and $Q$ is on circle $A$ , we can reflect circle $B$ over $P$ to get another circle (centered at a new point $C$ , and with radius $6$ ) that intersects circle $A$ at $Q$ . The rest is just finding lengths, as fo... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_15 | D | 24 | In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path only allows moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.
[asy] fill((0.5, 4.5)-... | [
"Notice that the upper-most section contains a 3 by 3 square that looks like:\n\nIt has 6 paths in which you can spell out AMC8. You will find four identical copies of this square in the figure, so multiply ${6 \\cdot 4}$ to get $\\boxed{24}$ total paths.",
"There are three different kinds of paths that are on th... |
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_4 | null | 429 | In the array of 13 squares shown below, 8 squares are colored red, and the remaining 5 squares are colored blue. If one of all possible such colorings is chosen at random, the probability that the chosen colored array appears the same when rotated 90 degrees around the central square is $\frac{1}{n}$ , where is a posit... | [
"When the array appears the same after a 90-degree rotation, the top formation must look the same as the right formation, which looks the same as the bottom one, which looks the same as the right one. There are four of the same configuration. There are not enough red squares for these to be all red, nor are there e... |
https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_17 | D | 8 | In the base ten number system the number $526$ means $5 \times 10^2+2 \times 10 + 6$ .
In the Land of Mathesis, however, numbers are written in the base $r$ .
Jones purchases an automobile there for $440$ monetary units (abbreviated m.u).
He gives the salesman a $1000$ m.u bill, and receives, in change, $340$ m.u. T... | [
"If Jones received $340$ m.u. change after paying $1000$ m.u. for something that costs $440$ m.u., then \\[440_r + 340_r = 1000_r\\] This equation can be rewritten as \\[4r^2 + 4r + 3r^2 + 4r = r^3\\] Bring all of the terms to one side to get \\[r^3 - 7r^2 - 8r = 0\\] Factor to get \\[r(r-8)(r+1)=0\\] Since base nu... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_2 | B | 700 | In the country of East Westmore, statisticians estimate there is a baby born every $8$ hours and a death every day. To the nearest hundred, how many people are added to the population of East Westmore each year?
$\textbf{(A)}\hspace{.05in}600\qquad\textbf{(B)}\hspace{.05in}700\qquad\textbf{(C)}\hspace{.05in}800\qquad\t... | [
"There are $24\\text{ hours}\\div8\\text{ hours} = 3$ births and one death everyday in East Westmore. Therefore, the population increases by $3$ $1$ $2$ people everyday. Thus, there are $2 \\times 365 = 730$ people added to the population every year. Rounding, we find the answer is $\\boxed{700}$"
] |
https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_2 | null | 109 | In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$ , and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ [asy] ... | [
"Again, let the intersection of $AE$ and $BC$ be $G$ . By AA similarity, $\\triangle AFG \\sim \\triangle CDG$ with a $\\frac{7}{3}$ ratio. Define $x$ as $\\frac{[CDG]}{9}$ . Because of similar triangles, $[AFG] = 49x$ . Using $ABCD$ , the area of the parallelogram is $33-18x$ . Using $AECF$ , the area of the paral... |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_7 | null | 539 | In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$ . Points $F$ and $G$ lie on $\overline{CE}$ , and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$ , respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$ , and $M$ and $N$ lie on $\overline{AD}$ a... | [
"Let us find the proportion of the side length of $KLMN$ and $FJGH$ . Let the side length of $KLMN=y$ and the side length of $FJGH=x$\n\nNow, examine $BC$ . We know $BC=BJ+JC$ , and triangles $\\Delta BHJ$ and $\\Delta JFC$ are similar to $\\Delta EDC$ since they are $1-2-\\sqrt{5}$ triangles. Thus, we can rewrite ... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_15 | D | 1 | In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of $1$ square unit, then what is the area of the shaded region, in square units?
[asy] size(4cm); filldraw(scale(2)*unitcircle,gray,black); filldraw(shift(-1,0)*unitcircl... | [
"Let the radius of the large circle be $R$ . Then, the radius of the smaller circles are $\\frac R2$ . The areas of the circles are directly proportional to the square of the radii, so the ratio of the area of the small circle to the large one is $\\frac 14$ . This means the combined area of the 2 smaller circles i... |
https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_10 | null | 148 | In the diagram below, angle $ABC$ is a right angle. Point $D$ is on $\overline{BC}$ , and $\overline{AD}$ bisects angle $CAB$ . Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $AE=3$ and $AF=10$ . Given that $EB=9$ and $FC=27$ , find the integer closest to the area of quadrilateral... | [
"By the Pythagorean Theorem, $BC=35$ . Letting $BD=x$ we can use the Angle Bisector Theorem on triangle $ABC$ to get $x/12=(35-x)/37$ , and solving gives $BD=60/7$ and $DC=185/7$\nThe area of triangle $AGF$ is $10/3$ that of triangle $AEG$ , since they share a common side and angle, so the area of triangle $AGF$ is... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_5 | E | 5 | In the diagram, all angles are right angles and the lengths of the sides are given in centimeters. Note the diagram is not drawn to scale. What is , $X$ in centimeters?
[asy] pair A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R; A=(4,0); B=(7,0); C=(7,4); D=(8,4); E=(8,5); F=(10,5); G=(10,7); H=(7,7); I=(7,8); J=(5,8); K=(5,7); L=... | [
"2012amc85.png\n$1 + 1 + 1 + 2 + X = 1 + 2 + 1 + 6\\\\ 5 + X = 10\\\\ X = 5$\nThus, the answer is $\\boxed{5}$",
"Note that we only need to consider the value below the marked red line, so we have the equation: \\[X + 2 = 6 + 1\\] \\[X = 5\\]\nHence, the answer is $\\boxed{5}$"
] |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_17 | C | 25 | In the eight term sequence $A$ $B$ $C$ $D$ $E$ $F$ $G$ $H$ , the value of $C$ is $5$ and the sum of any three consecutive terms is $30$ . What is $A+H$
$\textbf{(A)}\ 17 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 43$ | [
"Let $A=x$ . Then from $A+B+C=30$ , we find that $B=25-x$ . From $B+C+D=30$ , we then get that $D=x$ . Continuing this pattern, we find $E=25-x$ $F=5$ $G=x$ , and finally $H=25-x$ . So $A+H=x+25-x=25 \\rightarrow \\boxed{25}$",
"Given that the sum of 3 consecutive terms is 30, we have $(A+B+C)+(C+D+E)+(F+G+H)=90$... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12A_Problems/Problem_8 | C | 25 | In the eight term sequence $A$ $B$ $C$ $D$ $E$ $F$ $G$ $H$ , the value of $C$ is $5$ and the sum of any three consecutive terms is $30$ . What is $A+H$
$\textbf{(A)}\ 17 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 43$ | [
"Let $A=x$ . Then from $A+B+C=30$ , we find that $B=25-x$ . From $B+C+D=30$ , we then get that $D=x$ . Continuing this pattern, we find $E=25-x$ $F=5$ $G=x$ , and finally $H=25-x$ . So $A+H=x+25-x=25 \\rightarrow \\boxed{25}$",
"Given that the sum of 3 consecutive terms is 30, we have $(A+B+C)+(C+D+E)+(F+G+H)=90$... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_11 | C | 13 | In the equation below, $A$ and $B$ are consecutive positive integers, and $A$ $B$ , and $A+B$ represent number bases: \[132_A+43_B=69_{A+B}.\] What is $A+B$
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 17$ | [
"Change the equation to base 10: \\[A^2 + 3A +2 + 4B +3= 6A + 6B + 9\\] \\[A^2 - 3A - 2B - 4=0\\]\nEither $B = A + 1$ or $B = A - 1$ , so either $A^2 - 5A - 6, B = A + 1$ or $A^2 - 5A - 2, B = A - 1$ . The second case has no integer roots, and the first can be re-expressed as $(A-6)(A+1) = 0, B = A + 1$ . Since A m... |
https://artofproblemsolving.com/wiki/index.php/1958_AHSME_Problems/Problem_49 | D | 66 | In the expansion of $(a + b)^n$ there are $n + 1$ dissimilar terms. The number of dissimilar terms in the expansion of $(a + b + c)^{10}$ is:
$\textbf{(A)}\ 11\qquad \textbf{(B)}\ 33\qquad \textbf{(C)}\ 55\qquad \textbf{(D)}\ 66\qquad \textbf{(E)}\ 132$ | [
"Expand the binomial $((a+b)+c)^n$ with the binomial theorem. We have:\n\\[\\sum\\limits_{k=0}^{10} \\binom{10}{k} (a+b)^k c^{10-k}\\]\nSo for each iteration of the summation operator, we add k+1 dissimilar terms. Therefore our answer is:\n\\[\\sum\\limits_{k=0}^{10} k+1 = \\frac{11(1+11)}{2} = 66 \\to \\boxed{66}\... |
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_3 | null | 667 | In the expansion of $(ax + b)^{2000},$ where $a$ and $b$ are relatively prime positive integers, the coefficients of $x^{2}$ and $x^{3}$ are equal. Find $a + b$ | [
"Using the binomial theorem $\\binom{2000}{1998} b^{1998}a^2 = \\binom{2000}{1997}b^{1997}a^3 \\Longrightarrow b=666a$\nSince $a$ and $b$ are positive relatively prime integers, $a=1$ and $b=666$ , and $a+b=\\boxed{667}$"
] |
https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_9 | C | 21 | In the expansion of $\left(a-\dfrac{1}{\sqrt{a}}\right)^7$ the coefficient of $a^{-\dfrac{1}{2}}$ is:
$\textbf{(A)}\ -7 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ -21 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 35$ | [
"By the Binomial Theorem , each term of the expansion is $\\binom{7}{n}(a)^{7-n}(\\frac{-1}{\\sqrt{a}})^n$\nWe want the exponent of $a$ to be $-\\frac{1}{2}$ , so \\[(7-n)-\\frac{1}{2}n=-\\frac{1}{2}\\] \\[-\\frac{3}{2}n = -\\frac{15}{2}\\] \\[n = 5\\]\nIf $n=5$ , then the corresponding term is \\[\\binom{7}{5}(a)^... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_19 | C | 224 | In the expansion of \[\left(1 + x + x^2 + \cdots + x^{27}\right)\left(1 + x + x^2 + \cdots + x^{14}\right)^2,\] what is the coefficient of $x^{28}$
$\mathrm{(A)}\ 195\qquad\mathrm{(B)}\ 196\qquad\mathrm{(C)}\ 224\qquad\mathrm{(D)}\ 378\qquad\mathrm{(E)}\ 405$ | [
"Let $A = \\left(1 + x + x^2 + \\cdots + x^{14}\\right)$ and $B = \\left(1 + x + x^2 + \\cdots + x^{27}\\right)$ . We are expanding $A \\cdot A \\cdot B$\nSince there are $15$ terms in $A$ , there are $15^2 = 225$ ways to choose one term from each $A$ . The product of the selected terms is $x^n$ for some integer $n... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_3 | B | 3 | In the expression $\left(\underline{\qquad}\times\underline{\qquad}\right)+\left(\underline{\qquad}\times\underline{\qquad}\right)$ each blank is to be filled in with one of the digits $1,2,3,$ or $4,$ with each digit being used once. How many different values can be obtained?
$\textbf{(A) }2 \qquad \textbf{(B) }3\qqua... | [
"We have $\\binom{4}{2}$ ways to choose the pairs, and we have $2!$ ways for the values to be rearranged, hence $\\frac{6}{2}=\\boxed{3}$",
"We have four available numbers $(1, 2, 3, 4)$ . Because different permutations do not matter because they are all addition and multiplication, if we put $1$ on the first spa... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_5 | D | 9 | In the expression $c\cdot a^b-d$ , the values of $a$ $b$ $c$ , and $d$ are $0$ $1$ $2$ , and $3$ , although not necessarily in that order. What is the maximum possible value of the result?
$\mathrm{(A)\ }5\qquad\mathrm{(B)\ }6\qquad\mathrm{(C)\ }8\qquad\mathrm{(D)\ }9\qquad\mathrm{(E)\ }10$ | [
"If $a=0$ or $c=0$ , the expression evaluates to $-d<0$ If $b=0$ , the expression evaluates to $c-d\\leq 2$ Case $d=0$ remains.\nIn that case, we want to maximize $c\\cdot a^b$ where $\\{a,b,c\\}=\\{1,2,3\\}$ . Trying out the six possibilities we get that the greatest is $(a,b,c)=(3,2,1)$ , where $c\\cdot a^b=1\\cd... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_2 | D | 9 | In the expression $c\cdot a^b-d$ , the values of $a$ $b$ $c$ , and $d$ are $0$ $1$ $2$ , and $3$ , although not necessarily in that order. What is the maximum possible value of the result?
$\mathrm{(A)\ }5\qquad\mathrm{(B)\ }6\qquad\mathrm{(C)\ }8\qquad\mathrm{(D)\ }9\qquad\mathrm{(E)\ }10$ | [
"If $a=0$ or $c=0$ , the expression evaluates to $-d<0$ If $b=0$ , the expression evaluates to $c-d\\leq 2$ Case $d=0$ remains.\nIn that case, we want to maximize $c\\cdot a^b$ where $\\{a,b,c\\}=\\{1,2,3\\}$ . Trying out the six possibilities we get that the greatest is $(a,b,c)=(3,2,1)$ , where $c\\cdot a^b=1\\cd... |
https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_13 | E | 2,700 | In the fall of 1996, a total of 800 students participated in an annual school clean-up day. The organizers of the event expect that in each of the years 1997, 1998, and 1999, participation will increase by 50% over the previous year. The number of participants the organizers will expect in the fall of 1999 is
$\text{... | [
"If the participation increases by $50\\%$ , then it is the same as saying participation is multipled by a factor of $100\\% + 50\\% = 1 + 0.5 = 1.5$\nIn 1997, participation will be $800 \\cdot 1.5 = 1200$\nIn 1998, participation will be $1200 \\cdot 1.5 = 1800$\nIn 1999, participation will be $1800 \\cdot 1.5 = 27... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_13 | D | 12 | 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?
[asy] size(100); pair A,... | [
"Looking at the answer choices, we see that the possibilities are indeed countable. Thus, we will utilize that approach in the form of two separate cases, as rotation and reflection take care of numerous possibilities. First, consider the case that the green disk is in a corner. This yields $6$ possible arrangement... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_18 | D | 12 | 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?
[asy] size(110); pair A,... | [
"First we figure out the number of ways to put the $3$ blue disks. Denote the spots to put the disks as $1-6$ from left to right, top to bottom. The cases to put the blue disks are $(1,2,3),(1,2,4),(1,2,5),(1,2,6),(2,3,5),(1,4,6)$ . For each of those cases we can easily figure out the number of ways for each case, ... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_7 | D | 19 | In the figure below, $N$ congruent semicircles lie on the diameter of a large semicircle, with their diameters covering the diameter of the large semicircle with no overlap. Let $A$ be the combined area of the small semicircles and $B$ be the area of the region inside the large semicircle but outside the semicircles. T... | [
"Use the answer choices and calculate them. The one that works is $\\bold{\\boxed{19}$",
"Let the number of semicircles be $n$ and let the radius of each semicircle be $r$ . To find the total area of all of the small semicircles, we have $n \\cdot \\frac{\\pi \\cdot r^2}{2}$\nNext, we have to find the area of the... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_7 | null | 19 | In the figure below, $N$ congruent semicircles lie on the diameter of a large semicircle, with their diameters covering the diameter of the large semicircle with no overlap. Let $A$ be the combined area of the small semicircles and $B$ be the area of the region inside the large semicircle but outside the semicircles. T... | [
"Each small semicircle is $\\frac{1}{N^2}$ of the large semicircle. Since $N$ small semicircles make $\\frac{1}{19}$ of the large one, $\\frac{N}{N^2} = \\frac1{19}$ . Solving this, we get $\\boxed{19}$ . ..."
] |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_24 | C | 7.6 | In the figure, $ABCD$ is a rectangle and $EFGH$ is a parallelogram. Using the measurements given in the figure, what is the length $d$ of the segment that is perpendicular to $\overline{HE}$ and $\overline{FG}$
[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair D=(0,0), C=(10,0), B=(10,8), A=(0,8); ... | [
"The area of the parallelogram can be found in two ways. The first is by taking rectangle $ABCD$ and subtracting the areas of the triangles cut out to create parallelogram $EFGH$ . This is \\[(4+6)(5+3) - 2 \\cdot \\frac12 \\cdot 6 \\cdot 5 - 2 \\cdot \\frac12 \\cdot 3 \\cdot 4 = 80 - 30 - 12 = 38\\] The second way... |
https://artofproblemsolving.com/wiki/index.php/1989_AHSME_Problems/Problem_4 | null | 4 | In the figure, $ABCD$ is an isosceles trapezoid with side lengths $AD=BC=5$ $AB=4$ , and $DC=10$ . The point $C$ is on $\overline{DF}$ and $B$ is the midpoint of hypotenuse $\overline{DE}$ in right triangle $DEF$ . Then $CF=$
[asy] defaultpen(fontsize(10)); pair D=origin, A=(3,4), B=(7,4), C=(10,0), E=(14,8), F=(14,0);... | [
"Drop perpendiculars from $A$ and $B$ ; then the triangle $DBY$ is similar to $DEF$ but with corresponding sides of half the length. $XY=AB=4$ and $DX=YC=3$ , hence $DY=7\\implies DF=14\\implies CF=\\boxed{4.0}$"
] |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_8 | C | 36 | In the figure, the length of side $AB$ of square $ABCD$ is $\sqrt{50}$ and $BE=1$ . What is the area of the inner square $EFGH$
AMC102005Aq.png
$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 42$ | [
"We see that side $BE$ , which we know is $1$ , is also the shorter leg of one of the four right triangles (which are congruent, I won’t prove this). So, $AH = 1$ . Then $HB = HE + BE = HE + 1$ , and $HE$ is one of the sides of the square whose area we want to find. So:\n\\[1^2 + (HE+1)^2=\\sqrt{50}^2\\]\n\\[1 + (H... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_13 | null | 12 | In the five-sided star shown, the letters $A$ $B$ $C$ $D$ and $E$ are replaced by the
numbers 3, 5, 6, 7 and 9, although not necessarily in that order. The sums of the
numbers at the ends of the line segments $\overline{AB}$ $\overline{BC}$ $\overline{CD}$ $\overline{DE}$ , and $\overline{EA}$ form an
arithmetic sequen... | [
"Let the terms in the arithmetic sequence be $a$ $a + d$ $a + 2d$ $a + 3d$ , and $a + 4d$ . We seek the middle term $a + 2d$\nThese five terms are $A + B$ $B + C$ $C + D$ $D + E$ , and $E + A$ , in some order. The numbers $A$ $B$ $C$ $D$ , and $E$ are equal to 3, 5, 6, 7, and 9, in some order, so \\[A + B + C + D +... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_17 | D | 12 | In the five-sided star shown, the letters $A, B, C, D,$ and $E$ are replaced by the numbers $3, 5, 6, 7,$ and $9$ , although not necessarily in this order. The sums of the numbers at the ends of the line segments $AB$ $BC$ $CD$ $DE$ , and $EA$ form an arithmetic sequence, although not necessarily in that order. What is... | [
"Each corner $(A,B,C,D,E)$ goes to two sides/numbers. ( $A$ goes to $AE$ and $AB$ $D$ goes to $DC$ and $DE$ ). The sum of every term is equal to $2(3+5+6+7+9)=60$\nSince the middle term in an arithmetic sequence is the average of all the terms in the sequence, the middle number is $\\frac{60}{5}=\\boxed{12}$"
] |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_17 | null | 12 | In the five-sided star shown, the letters $A, B, C, D,$ and $E$ are replaced by the numbers $3, 5, 6, 7,$ and $9$ , although not necessarily in this order. The sums of the numbers at the ends of the line segments $AB$ $BC$ $CD$ $DE$ , and $EA$ form an arithmetic sequence, although not necessarily in that order. What is... | [
"We know that the smallest number in this sequence must be $3 + 5 = 8$ , and the biggest number must be $7 + 9 = 16$ . Since there are $5$ terms in this sequence, we know that $8 + 4d = 16$ , or that $d = 2$ . Thus, we know that the middle term must be $8 + 2 \\cdot 2 = \\boxed{12}.$ ~yk2007 (Daniel K.)"
] |
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_31 | C | 21 | In the following equation, each of the letters represents uniquely a different digit in base ten:
\[(YE) \cdot (ME) = TTT\]
The sum $E+M+T+Y$ equals
$\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 24$ | [
"The right side of the equation can be rewritten as $111T = 37 \\cdot 3T$ . With trial and error and prime factorization as a guide, we can test different digits of $T$ to see if we can find two two-digit numbers that have the same units digit and multiply to $111T$\nThe only possibility that works is $37 \\cdot 2... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_16 | C | 142 | In the following list of numbers, the integer $n$ appears $n$ times in the list for $1 \leq n \leq 200$ \[1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots, 200, 200, \ldots , 200\] What is the median of the numbers in this list?
$\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\t... | [
"The $x$ th number of this sequence is $\\left\\lceil\\frac{-1\\pm\\sqrt{1+8x}}{2}\\right\\rceil$ via the quadratic formula. We can see that if we halve $x$ we end up getting $\\left\\lceil\\frac{-1\\pm\\sqrt{1+4x}}{2}\\right\\rceil$ . This is approximately the number divided by $\\sqrt{2}$ $\\frac{200}{\\sqrt{2}} ... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_16 | C | 142 | In the following list of numbers, the integer $n$ appears $n$ times in the list for $1 \leq n \leq 200$ \[1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots, 200, 200, \ldots , 200\] What is the median of the numbers in this list?
$\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\t... | [
"The $x$ th number of this sequence is $\\left\\lceil\\frac{-1\\pm\\sqrt{1+8x}}{2}\\right\\rceil$ via the quadratic formula. We can see that if we halve $x$ we end up getting $\\left\\lceil\\frac{-1\\pm\\sqrt{1+4x}}{2}\\right\\rceil$ . This is approximately the number divided by $\\sqrt{2}$ $\\frac{200}{\\sqrt{2}} ... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_17 | C | 130 | In the given circle, the diameter $\overline{EB}$ is parallel to $\overline{DC}$ , and $\overline{AB}$ is parallel to $\overline{ED}$ . The angles $AEB$ and $ABE$ are in the ratio $4 : 5$ . What is the degree measure of angle $BCD$
$\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 125 \qquad\textbf{(C)}\ 130 \qquad\textbf{(D)}\ 1... | [
"We can let $\\angle AEB$ be $4x$ and $\\angle ABE$ be $5x$ because they are in the ratio $4 : 5$ . When an inscribed angle contains the diameter , the inscribed angle is a right angle . Therefore by triangle sum theorem, $4x+5x+90=180 \\longrightarrow x=10$ and $\\angle ABE = 50$\n$\\angle ABE = \\angle BED$ becau... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_21 | C | 12 | In the given figure, hexagon $ABCDEF$ is equiangular, $ABJI$ and $FEHG$ are squares with areas $18$ and $32$ respectively, $\triangle JBK$ is equilateral and $FE=BC$ . What is the area of $\triangle KBC$
[asy] draw((-4,6*sqrt(2))--(4,6*sqrt(2))); draw((-4,-6*sqrt(2))--(4,-6*sqrt(2))); draw((-8,0)--(-4,6*sqrt(2))); draw... | [
"Clearly, since $\\overline{FE}$ is a side of a square with area $32$ $\\overline{FE} = \\sqrt{32} = 4 \\sqrt{2}$ . Now, since $\\overline{FE} = \\overline{BC}$ , we have $\\overline{BC} = 4 \\sqrt{2}$\nNow, $\\overline{JB}$ is a side of a square with area $18$ , so $\\overline{JB} = \\sqrt{18} = 3 \\sqrt{2}$ . Sin... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_22 | D | 46 | In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by $v$ $w$ $x$ $y$ , and $z$ . Find $y + z$
[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5... | [
"We know that $y+z=2v$ , so we could find one variable rather than two.\n$v+24+w=43+v$\n$24+w=43$\n$w=19$\n\n$44+x=24+x+z \\implies z=20$\n\nThe sum per row is $25+21+20=66$\nThus $66-18-25=66-43=v=23$\nSince we needed $2v$ and we know $v=23$ $23 \\times 2 = \\boxed{46}$",
"$v+24+w=43+v$\n$24+w=43$\n$w=19$\n\n$44... |
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_13 | null | 731 | In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at $50$ miles per hour along the highways and at $14$ miles per hour across the prairie. Consider the set of points that can be reached by the firetruck within six minutes. The area of... | [
"Let the intersection of the highways be at the origin $O$ , and let the highways be the x and y axes. We consider the case where the truck moves in the positive x direction.\nAfter going $x$ miles, $t=\\frac{d}{r}=\\frac{x}{50}$ hours has passed. If the truck leaves the highway it can travel for at most $t=\\frac{... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_18 | B | 24 | In the non-convex quadrilateral $ABCD$ shown below, $\angle BCD$ is a right angle, $AB=12$ $BC=4$ $CD=3$ , and $AD=13$ . What is the area of quadrilateral $ABCD$
[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label("$B$", (0, 0), SW); label("$A$", (12, 0), ESE); label("$C$", (2.4, 3.6), SE); label("$D$", (0, 5), N)... | [
"We first connect point $B$ with point $D$\n\nWe can see that $\\triangle BCD$ is a 3-4-5 right triangle. We can also see that $\\triangle BDA$ is a right triangle, by the 5-12-13 Pythagorean triple. With these lengths, we can solve the problem. The area of $\\triangle BDA$ is $\\frac{5\\cdot 12}{2}$ , and the area... |
https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_25 | C | 2 | In the non-decreasing sequence of odd integers $\{a_1,a_2,a_3,\ldots \}=\{1,3,3,3,5,5,5,5,5,\ldots \}$ each odd positive integer $k$ appears $k$ times. It is a fact that there are integers $b, c$ , and $d$ such that for all positive integers $n$ $a_n=b\lfloor \sqrt{n+c} \rfloor +d$ ,
where $\lfloor x \rfloor$ denotes ... | [
"Solution by e_power_pi_times_i\nBecause the set consists of odd numbers, and since $\\lfloor{}\\sqrt{n+c}\\rfloor{}$ is an integer and can be odd or even, $b = 2$ and $|a| = 1$ . However, given that $\\lfloor{}\\sqrt{n+c}\\rfloor{}$ can be $0$ $a = 1$ . Then, $a_1 = 1 = 2\\lfloor{}\\sqrt{1+c}\\rfloor{}+1$ , and $\... |
https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_6 | C | 100,000 | In the number $74982.1035$ the value of the place occupied by the digit 9 is how many times as great as the value of the place occupied by the digit 3?
$\text{(A)}\ 1,000 \qquad \text{(B)}\ 10,000 \qquad \text{(C)}\ 100,000 \qquad \text{(D)}\ 1,000,000 \qquad \text{(E)}\ 10,000,000$ | [
"The digit $9$ is $5$ places to the left of the digit $3$ . Thus, it has a place value that is $10^5 = 100,000$ times greater. $\\boxed{100,000}$",
"The digit $9$ is in the $100$ s place. The digit $3$ is in the $\\frac{1}{1000}$ ths place. The ratio of these two numbers is $\\frac{100}{\\frac{1}{1000}} = 100\\... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_9 | B | 4 | In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150)... | [
"Looking, we see that the area of $[\\triangle EBA]$ is 16 and the area of $[\\triangle ABC]$ is 12. Set the area of $[\\triangle ADB]$ to be x. We want to find $[\\triangle ADE]$ $[\\triangle CDB]$ . So, that would be $[\\triangle EBA]-[\\triangle ADB]=16-x$ and $[\\triangle ABC]-[\\triangle ADB]=12-x$ . Therefore... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_8 | B | 4 | In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150)... | [
"Looking, we see that the area of $[\\triangle EBA]$ is 16 and the area of $[\\triangle ABC]$ is 12. Set the area of $[\\triangle ADB]$ to be x. We want to find $[\\triangle ADE]$ $[\\triangle CDB]$ . So, that would be $[\\triangle EBA]-[\\triangle ADB]=16-x$ and $[\\triangle ABC]-[\\triangle ADB]=12-x$ . Therefore... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_6 | D | 7 | In the plane figure shown below, $3$ of the unit squares have been shaded. What is the least number of additional unit squares that must be shaded so that the resulting figure has two lines of symmetry?
[asy] import olympiad; unitsize(25); filldraw((1,3)--(1,4)--(2,4)--(2,3)--cycle, gray(0.7)); filldraw((2,1)--(2,2)--(... | [
"The two lines of symmetry must be horizontally and vertically through the middle. We can then fill the boxes in like so:\n\nwhere the light gray boxes are the ones we have filled. Counting these, we get $\\boxed{7}$ total boxes.",
"We label the three shaded unit squares $A,B,$ and $C,$ then construct the two lin... |
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_8 | C | 100 | In the polygon shown, each side is perpendicular to its adjacent sides, and all 28 of the sides are congruent. The perimeter of the polygon is $56$ . The area of the region bounded by the polygon is [asy] draw((0,0)--(1,0)--(1,-1)--(2,-1)--(2,-2)--(3,-2)--(3,-3)--(4,-3)--(4,-2)--(5,-2)--(5,-1)--(6,-1)--(6,0)--(7,0)--(7... | [
"Since the perimeter is $56$ and all of the sides are congruent, the length of each side is $2$ . We break the figure into squares as shown below.\n\nWe see that there are a total of $2(1+3+5)+7=25$ squares with side length $2$ . Therefore, the total area is $4\\cdot 25=\\boxed{100.}$"
] |
https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_1 | null | 86 | In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$ , the product of $2$ of its roots is $- 32$ . Find $k$ | [
"Using Vieta's formulas, we have:\n\\begin{align*}a+b+c+d &= 18,\\\\ ab+ac+ad+bc+bd+cd &= k,\\\\ abc+abd+acd+bcd &=-200,\\\\ abcd &=-1984.\\\\ \\end{align*}\nFrom the last of these equations, we see that $cd = \\frac{abcd}{ab} = \\frac{-1984}{-32} = 62$ . Thus, the second equation becomes $-32+ac+ad+bc+bd+62=k$ , a... |
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_8 | E | 8 | In the product shown, $\text{B}$ is a digit. The value of $\text{B}$ is
\[\begin{array}{rr} &\text{B}2 \\ \times& 7\text{B} \\ \hline &6396 \\ \end{array}\]
$\text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$ | [
"Note that in any multiplication problem, the only 2 digits that will influence the last digit of the number will be the last digits of each number being multiplied.\nSo, $2 \\times \\text{B}$ has a units digit of $6$ , so $\\text{B}$ is either $3$ or $8$ . If $\\text{B}=3$ , then the product is $32\\times 73$ , wh... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_10 | E | 2 | In the rectangular parallelepiped shown, $AB$ $3$ $BC$ $1$ , and $CG$ $2$ . Point $M$ is the midpoint of $\overline{FG}$ . What is the volume of the rectangular pyramid with base $BCHE$ and apex $M$
[asy] size(250); defaultpen(fontsize(10pt)); pair A =origin; pair B = (4.75,0); pair E1=(0,3); pair F = (4.75,3); pair G ... | [
"Consider the cross-sectional plane and label its area $b$ . Note that the volume of the triangular prism that encloses the pyramid is $\\frac{bh}{2}=3$ , and we want the rectangular pyramid that shares the base and height with the triangular prism. The volume of the pyramid is $\\frac{bh}{3}$ , so the answer is $\... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_10 | null | 2 | In the rectangular parallelepiped shown, $AB$ $3$ $BC$ $1$ , and $CG$ $2$ . Point $M$ is the midpoint of $\overline{FG}$ . What is the volume of the rectangular pyramid with base $BCHE$ and apex $M$
[asy] size(250); defaultpen(fontsize(10pt)); pair A =origin; pair B = (4.75,0); pair E1=(0,3); pair F = (4.75,3); pair G ... | [
"If you look carefully, you will see that on the either side of the pyramid in question, there are two congruent tetrahedra. The volume of one is $\\frac{1}{3}Bh$ , with its base being half of one of the rectangular prism's faces and its height being half of one of the edges, so its volume is $\\frac{1}{3} (3 \\tim... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_19 | null | 0 | In the sequence $2001$ $2002$ $2003$ $\ldots$ , each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is $2001 + 2002 - 2003 = 2000$ . What is the $2004^\textrm{th}$ term in this sequence?
$\mathrm{(A) \ } -2004 \qquad \mat... | [
"We already know that $a_1=2001$ $a_2=2002$ $a_3=2003$ , and $a_4=2000$ . Let's compute the next few terms to get the idea how the sequence behaves. We get $a_5 = 2002+2003-2000 = 2005$ $a_6=2003+2000-2005=1998$ $a_7=2000+2005-1998=2007$ , and so on.\nWe can now discover the following pattern: $a_{2k+1} = 2001+2k$ ... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_12 | null | 0 | In the sequence $2001$ $2002$ $2003$ $\ldots$ , each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is $2001 + 2002 - 2003 = 2000$ . What is the $2004^\textrm{th}$ term in this sequence?
$\mathrm{(A) \ } -2004 \qquad \mat... | [
"We already know that $a_1=2001$ $a_2=2002$ $a_3=2003$ , and $a_4=2000$ . Let's compute the next few terms to get the idea how the sequence behaves. We get $a_5 = 2002+2003-2000 = 2005$ $a_6=2003+2000-2005=1998$ $a_7=2000+2005-1998=2007$ , and so on.\nWe can now discover the following pattern: $a_{2k+1} = 2001+2k$ ... |
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_6 | A | 3 | In the sequence \[..., a, b, c, d, 0, 1, 1, 2, 3, 5, 8,...\] each term is the sum of the two terms to its left. Find $a$
$\textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ 3$ | [
"We work backwards to find $a$\n\\[d+0=1\\implies d=1\\] \\[c+1=0\\implies c=-1\\] \\[b+(-1)=1\\implies b=2\\] \\[a+2=-1\\implies a=\\boxed{3.}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_15 | A | 5 | In the sequence of numbers $1, 3, 2, \ldots$ each term after the first two is equal to the term preceding it minus the term preceding that. The sum of the first one hundred terms of the sequence is
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ -1$ | [
"First, write a few terms of the sequence: $1, 3, 2, -1, -3, -2, 1, 3, 2, \\ldots$ Notice how the pattern repeats every six terms and every six terms have a sum of 0. Then, find that the $16*6=96$ th term is $-2$ and the sum of the all those previous terms is $0$ . Then, write the 97th to the 100th terms down: $1, ... |
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_11 | D | 29 | In the sixth, seventh, eighth, and ninth basketball games of the season, a player scored $23$ $14$ $11$ , and $20$ points, respectively. Her points-per-game average was higher after nine games than it was after the first five games. If her average after ten games was greater than $18$ , what is the least number of poin... | [
"The sum of the scores for games $6$ through $9$ is $68$ . The average in these four games is $\\frac{68}{4} = 17$\nThe total points in all ten games is greater than $10\\cdot 18 = 180$ . Thus, it must be at least $181$\nThere are at least $181 - 68 = 113$ points in the other six games: games $1-5$ and game $10$... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_16 | D | 11 | In the state of Coinland, coins have values $6,10,$ and $15$ cents. Suppose $x$ is the value in cents of the most expensive item in Coinland that cannot be purchased using these coins with exact change. What is the sum of the digits of $x?$
$\textbf{(A) }8\qquad\textbf{(B) }10\qquad\textbf{(C) }7\qquad\textbf{(D) }11\q... | [
"This problem asks to find largest $x$ that cannot be written as \\[ 6 a + 10 b + 15 c = x, \\hspace{1cm} (1) \\] where $a, b, c \\in \\Bbb Z_+$\nDenote by $r \\in \\left\\{ 0, 1 \\right\\}$ the remainder of $x$ divided by 2.\nModulo 2 on Equation (1), we get\nBy using modulus $m \\in \\left\\{ 2, 3, 5 \\right\\}$ ... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_1 | E | 671 | In the year $2001$ , the United States will host the International Mathematical Olympiad . Let $I,M,$ and $O$ be distinct positive integers such that the product $I \cdot M \cdot O = 2001$ . What is the largest possible value of the sum $I + M + O$
$\textbf{(A)}\ 23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\ 99 \qq... | [
"First, we need to recognize that a number is going to be lowest only if, of the $3$ factors , two of them are small. If we want to make sure that this is correct, we could test with a smaller number, like $30$ . It becomes much more clear that this is true, and in this situation, the value of $I + M + O$ would be ... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_1 | E | 671 | In the year $2001$ , the United States will host the International Mathematical Olympiad . Let $I,M,$ and $O$ be distinct positive integers such that the product $I \cdot M \cdot O = 2001$ . What is the largest possible value of the sum $I + M + O$
$\textbf{(A)}\ 23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\ 99 \qq... | [
"First, we need to recognize that a number is going to be lowest only if, of the $3$ factors , two of them are small. If we want to make sure that this is correct, we could test with a smaller number, like $30$ . It becomes much more clear that this is true, and in this situation, the value of $I + M + O$ would be ... |
https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_29 | C | 17,865 | In their base $10$ representations, the integer $a$ consists of a sequence of $1985$ eights and the integer $b$ consists of a sequence of $1985$ fives. What is the sum of the digits of the base $10$ representation of the integer $9ab$
$\mathrm{(A)\ } 15880 \qquad \mathrm{(B) \ }17856 \qquad \mathrm{(C) \ } 17865 \qqua... | [
"By the formula for the sum of a geometric series, \\begin{align*}a &= 8 \\cdot 10^0 + 8 \\cdot 10^1 + \\dotsb + 8 \\cdot 10^{1984} \\\\ &= \\frac{8\\left(10^{1985}-1\\right)}{10-1} \\\\ &= \\frac{8\\left(10^{1985}-1\\right)}{9},\\end{align*} and similarly \\[b = \\frac{5\\left(10^{1985}-1\\right)}{9},\\] so \\begi... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_14 | D | 3 | In this addition problem, each letter stands for a different digit.
$\setlength{\tabcolsep}{0.5mm}\begin{array}{cccc}&T & W & O\\ +&T & W & O\\ \hline F& O & U & R\end{array}$
If T = 7 and the letter O represents an even number, what is the only possible value for W?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(... | [
"Since both T's are 7, then O has to equal 4, because 7 + 7 = 14. Then, F has to equal 1. To get R, we do 4 + 4 (since O = 4) to get R = 8. The value for W then has to be a number less than 5, otherwise it will change the value of O, and can't be a number that has already been used, like 4 or 1. The only other poss... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_24 | B | 12.25 | In trapezoid $ABCD$ $\overline{AB}$ and $\overline{CD}$ are perpendicular to $\overline{AD}$ , with $AB+CD=BC$ $AB<CD$ , and $AD=7$ . What is $AB\cdot CD$
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 12.25 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 12.75 \qquad \textbf{(E)}\ 13$ | [
"\nIf $AB=x$ and $CD=y$ , then $BC=x+y$ . By the Pythagorean theorem , we have $(x+y)^2=(y-x)^2+49.$ Solving the equation, we get $4xy=49 \\implies xy = \\boxed{12.25}$",
"Simpler is just drawing the trapezoid and then using what is given to solve.\nDraw a line parallel to $\\overline{AD}$ that connects the longe... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_8 | B | 4.5 | In trapezoid $ABCD$ $\overline{AD}$ is perpendicular to $\overline{DC}$ $AD = AB = 3$ , and $DC = 6$ . In addition, $E$ is on $\overline{DC}$ , and $\overline{BE}$ is parallel to $\overline{AD}$ . Find the area of $\triangle BEC$ [asy] defaultpen(linewidth(0.7)); pair A=(0,3), B=(3,3), C=(6,0), D=origin, E=(3,0); draw(... | [
"Clearly, $ABED$ is a square with side-length $3.$ By segment subtraction, we have $EC = DC - DE = 6 - 3 = 3.$\nThe area of $\\triangle BEC$ is \\[\\frac12\\cdot EC\\cdot BE = \\frac12\\cdot3\\cdot3 = \\boxed{4.5}.\\] ~Aplus95 (Solution)",
"Clearly, $ABED$ is a square with side-length $3.$\nLet the brackets denot... |
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_8 | null | 110 | In trapezoid $ABCD$ , leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD}$ , and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001}$ , find $BC^2$ | [
"Let $x = BC$ be the height of the trapezoid, and let $y = CD$ . Since $AC \\perp BD$ , it follows that $\\triangle BAC \\sim \\triangle CBD$ , so $\\frac{x}{\\sqrt{11}} = \\frac{y}{x} \\Longrightarrow x^2 = y\\sqrt{11}$\nLet $E$ be the foot of the altitude from $A$ to $\\overline{CD}$ . Then $AE = x$ , and $ADE$ i... |
https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_14 | D | 34 | In trapezoid $ABCD$ , the sides $AB$ and $CD$ are equal. The perimeter of $ABCD$ is
[asy] draw((0,0)--(4,3)--(12,3)--(16,0)--cycle); draw((4,3)--(4,0),dashed); draw((3.2,0)--(3.2,.8)--(4,.8)); label("$A$",(0,0),SW); label("$B$",(4,3),NW); label("$C$",(12,3),NE); label("$D$",(16,0),SE); label("$8$",(8,3),N); label("$16... | [
"\nThere is a rectangle present, with both horizontal bases being $8$ units in length. The excess units on the bottom base must then be $16-8=8$ . The fact that $AB$ and $CD$ are equal in length indicate, by the Pythagorean Theorem, that these excess lengths are equal. There are two with a total length of $8$ units... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_23 | C | 5 | In trapezoid $ABCD$ we have $\overline{AB}$ parallel to $\overline{DC}$ $E$ as the midpoint of $\overline{BC}$ , and $F$ as the midpoint of $\overline{DA}$ . The area of $ABEF$ is twice the area of $FECD$ . What is $AB/DC$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 6 \qquad \tex... | [
"Since the heights of both trapezoids are equal, and the area of $ABEF$ is twice the area of $FECD$\n$\\frac{AB+EF}{2}=2\\left(\\frac{DC+EF}{2}\\right)$\n$\\frac{AB+EF}{2}=DC+EF$ , so\n$AB+EF=2DC+2EF$\n$EF$ is exactly halfway between $AB$ and $DC$ , so $EF=\\frac{AB+DC}{2}$\n$AB+\\frac{AB+DC}{2}=2DC+AB+DC$ , so\n$\... |
https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_5 | null | 504 | In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$ , let $BC = 1000$ and $AD = 2008$ . Let $\angle A = 37^\circ$ $\angle D = 53^\circ$ , and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively. Find the length $MN$ | [
"Extend $\\overline{AB}$ and $\\overline{CD}$ to meet at a point $E$ . Then $\\angle AED = 180 - 53 - 37 = 90^{\\circ}$\nAs $\\angle AED = 90^{\\circ}$ , note that the midpoint of $\\overline{AD}$ $N$ , is the center of the circumcircle of $\\triangle AED$ . We can do the same with the circumcircle about $\\triangl... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_25 | C | 210 | In trapezoid $ABCD$ with bases $AB$ and $CD$ , we have $AB = 52$ $BC = 12$ $CD = 39$ , and $DA = 5$ . The area of $ABCD$ is
[asy] pair A,B,C,D; A=(0,0); B=(52,0); C=(38,20); D=(5,20); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--D--cycle); label("$A$",A,S); label("$B$",B,S); label("$C$",C,N); label("$D$",D,N); label("... | [
"It shouldn't be hard to use trigonometry to bash this and find the height, but there is a much easier way. Extend $\\overline{AD}$ and $\\overline{BC}$ to meet at point $E$\nSince $\\overline{AB} || \\overline{CD}$ we have $\\triangle AEB \\sim \\triangle DEC$ , with the ratio of proportionality being $\\frac {39}... |
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