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/2009_AIME_I_Problems/Problem_15 | null | 150 | In triangle $ABC$ $AB = 10$ $BC = 14$ , and $CA = 16$ . Let $D$ be a point in the interior of $\overline{BC}$ . Let points $I_B$ and $I_C$ denote the incenters of triangles $ABD$ and $ACD$ , respectively. The circumcircles of triangles $BI_BD$ and $CI_CD$ meet at distinct points $P$ and $D$ . The maximum possible area ... | [
"First, by the Law of Cosines , we have \\[\\cos BAC = \\frac {16^2 + 10^2 - 14^2}{2\\cdot 10 \\cdot 16} = \\frac {256+100-196}{320} = \\frac {1}{2},\\] so $\\angle BAC = 60^\\circ$\nLet $O_1$ and $O_2$ be the circumcenters of triangles $BI_BD$ and $CI_CD$ , respectively. We first compute \\[\\angle BO_1D = \\angl... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_19 | null | 21 | In triangle $ABC$ $AB = 13$ $BC = 14$ $AC = 15$ . Let $D$ denote the midpoint of $\overline{BC}$ and let $E$ denote the intersection of $\overline{BC}$ with the bisector of angle $BAC$ . Which of the following is closest to the area of the triangle $ADE$
$\text {(A)}\ 2 \qquad \text {(B)}\ 2.5 \qquad \text {(C)}\ 3 \qq... | [
"\nLet's find the area of $\\Delta ABC$ by Heron,\n$s=\\frac{a+b+c}{2}\\\\\\\\s=\\frac{14+15+13}{2}\\to\\boxed{21}$"
] |
https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_11 | null | 254 | In triangle $ABC$ $AB = AC = 100$ , and $BC = 56$ Circle $P$ has radius $16$ and is tangent to $\overline{AC}$ and $\overline{BC}$ . Circle $Q$ is externally tangent to $P$ and is tangent to $\overline{AB}$ and $\overline{BC}$ . No point of circle $Q$ lies outside of $\triangle ABC$ . The radius of circle $Q$ can be ex... | [
"Refer to the above diagram. Let the larger circle have center $O_1$ , the smaller have center $O_2$ , and the incenter be $I$ . We can easily calculate that the area of $\\triangle ABC = 2688$ , and $s = 128$ and $R = 21$ , where $R$ is the inradius.\nNow, Line $\\overline{AI}$ is the perpendicular bisector of $\\... |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_46 | E | 0 | In triangle $ABC$ $AB=12$ $AC=7$ , and $BC=10$ . If sides $AB$ and $AC$ are doubled while $BC$ remains the same, then:
$\textbf{(A)}\ \text{The area is doubled} \qquad\\ \textbf{(B)}\ \text{The altitude is doubled} \qquad\\ \textbf{(C)}\ \text{The area is four times the original area} \qquad\\ \textbf{(D)}\ \text{The m... | [
"If you double sides $AB$ and $AC$ , they become $24$ and $14$ respectively. If $BC$ remains $10$ , then this triangle has area $0$ because ${14} + {10} = {24}$ , so two sides overlap the third side. Therefore the answer is $\\boxed{0}$"
] |
https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_4 | null | 56 | In triangle $ABC$ $AB=125$ $AC=117$ and $BC=120$ . The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$ , and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$ . Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$ , respectively. Fin... | [
"Extend ${CM}$ and ${CN}$ such that they intersect line ${AB}$ at points $P$ and $Q$ , respectively. Since ${BM}$ is the angle bisector of angle $B$ and ${CM}$ is perpendicular to ${BM}$ $\\triangle BCP$ must be an isoceles triangle, so $BP=BC=120$ , and $M$ is the midpoint of ${CP}$ . For the same reason, $AQ=AC=... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_23 | B | 21 | In triangle $ABC$ $AB=13$ $BC=14$ , and $CA=15$ . Distinct points $D$ $E$ , and $F$ lie on segments $\overline{BC}$ $\overline{CA}$ , and $\overline{DE}$ , respectively, such that $\overline{AD}\perp\overline{BC}$ $\overline{DE}\perp\overline{AC}$ , and $\overline{AF}\perp\overline{BF}$ . The length of segment $\overli... | [
"Since $\\angle{AFB}=\\angle{ADB}=90^{\\circ}$ , quadrilateral $ABDF$ is cyclic. It follows that $\\angle{ADE}=\\angle{ABF}$ , so $\\triangle ABF \\sim \\triangle ADE$ are similar. In addition, $\\triangle ADE \\sim \\triangle ACD$ . We can easily find $AD=12$ $BD = 5$ , and $DC=9$ using Pythagorean triples.\nSo, t... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_19 | B | 21 | In triangle $ABC$ $AB=13$ $BC=14$ , and $CA=15$ . Distinct points $D$ $E$ , and $F$ lie on segments $\overline{BC}$ $\overline{CA}$ , and $\overline{DE}$ , respectively, such that $\overline{AD}\perp\overline{BC}$ $\overline{DE}\perp\overline{AC}$ , and $\overline{AF}\perp\overline{BF}$ . The length of segment $\overli... | [
"Since $\\angle{AFB}=\\angle{ADB}=90^{\\circ}$ , quadrilateral $ABDF$ is cyclic. It follows that $\\angle{ADE}=\\angle{ABF}$ , so $\\triangle ABF \\sim \\triangle ADE$ are similar. In addition, $\\triangle ADE \\sim \\triangle ACD$ . We can easily find $AD=12$ $BD = 5$ , and $DC=9$ using Pythagorean triples.\nSo, t... |
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_9 | null | 61 | In triangle $ABC$ $AB=13$ $BC=15$ and $CA=17$ . Point $D$ is on $\overline{AB}$ $E$ is on $\overline{BC}$ , and $F$ is on $\overline{CA}$ . Let $AD=p\cdot AB$ $BE=q\cdot BC$ , and $CF=r\cdot CA$ , where $p$ $q$ , and $r$ are positive and satisfy $p+q+r=2/3$ and $p^2+q^2+r^2=2/5$ . The ratio of the area of triangle $DEF... | [
"We let $[\\ldots]$ denote area; then the desired value is\nUsing the formula for the area of a triangle $\\frac{1}{2}ab\\sin C$ , we find that\nand similarly that $\\frac{[BDE]}{[ABC]} = q(1-p)$ and $\\frac{[CEF]}{[ABC]} = r(1-q)$ . Thus, we wish to find \\begin{align*}\\frac{[DEF]}{[ABC]} &= 1 - \\frac{[ADF]}{[AB... |
https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_4 | null | 51 | In triangle $ABC$ $AB=20$ and $AC=11$ . The angle bisector of $\angle A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of the intersection of $AC$ and $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime posit... | [
" Let $D'$ be on $\\overline{AC}$ such that $BP \\parallel DD'$ . It follows that $\\triangle BPC \\sim \\triangle DD'C$ , so \\[\\frac{PC}{D'C} = 1 + \\frac{BD}{DC} = 1 + \\frac{AB}{AC} = \\frac{31}{11}\\] by the Angle Bisector Theorem . Similarly, we see by the Midline Theorem that $AP = PD'$ . Thus, \\[\\frac{CP... |
https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_13 | null | 65 | In triangle $ABC$ $AB=\sqrt{30}$ $AC=\sqrt{6}$ , and $BC=\sqrt{15}$ . There is a point $D$ for which $\overline{AD}$ bisects $\overline{BC}$ , and $\angle ADB$ is a right angle. The ratio $\frac{[ADB]}{[ABC]}$ can be written in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+... | [
"Let $E$ be the midpoint of $\\overline{BC}$ . Since $BE = EC$ , then $\\triangle ABE$ and $\\triangle AEC$ share the same height and have equal bases, and thus have the same area. Similarly, $\\triangle BDE$ and $BAE$ share the same height, and have bases in the ratio $DE : AE$ , so $\\frac{[BDE]}{[BAE]} = \\frac{... |
https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_15 | null | 218 | In triangle $ABC$ $AC = 13$ $BC = 14$ , and $AB=15$ . Points $M$ and $D$ lie on $AC$ with $AM=MC$ and $\angle ABD = \angle DBC$ . Points $N$ and $E$ lie on $AB$ with $AN=NB$ and $\angle ACE = \angle ECB$ . Let $P$ be the point, other than $A$ , of intersection of the circumcircles of $\triangle AMN$ and $\triangle ADE... | [
"Define the function $f:\\mathbb{R}^{2}\\rightarrow\\mathbb{R}$ by \\[f(X)=\\text{Pow}_{(AMN)}(X)-\\text{Pow}_{(ADE)}(X)\\] for points $X$ in the plane. Then $f$ is linear, so $\\frac{BQ}{CQ}=\\frac{f(B)-f(Q)}{f(Q)-f(C)}$ . But $f(Q)=0$ since $Q$ lies on the radical axis of $(AMN)$ $(ADE)$ thus \\[\\frac{BQ}{CQ}=-\... |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_35 | B | 4 | In triangle $ABC$ $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is:
$\textbf{(A)}\ 26\text{ in} \qquad \textbf{(B)}\ 4\text{ in} \qquad \textbf{(C)}\ 13\text{ in} \qquad \textbf{(D)}\ 8\text{ in} \qquad \textbf{(E)}\ \text{None of these}$ | [
"The inradius is equal to the area divided by semiperimeter. The area is $\\frac{(10)(24)}{2} = 120$ because it's a right triangle, as it's side length satisfies the Pythagorean Theorem. The semiperimeter is $30$ . Therefore the inradius is $\\boxed{4}$"
] |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_35 | null | 4 | In triangle $ABC$ $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is:
$\textbf{(A)}\ 26\text{ in} \qquad \textbf{(B)}\ 4\text{ in} \qquad \textbf{(C)}\ 13\text{ in} \qquad \textbf{(D)}\ 8\text{ in} \qquad \textbf{(E)}\ \text{None of these}$ | [
"Since this is a right triangle, we have \\[\\frac{a+b-c}{2}=\\boxed{4}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_8 | null | 318 | In triangle $ABC$ $BC = 23$ $CA = 27$ , and $AB = 30$ . Points $V$ and $W$ are on $\overline{AC}$ with $V$ on $\overline{AW}$ , points $X$ and $Y$ are on $\overline{BC}$ with $X$ on $\overline{CY}$ , and points $Z$ and $U$ are on $\overline{AB}$ with $Z$ on $\overline{BU}$ . In addition, the points are positioned so th... | [
"Note that triangles $\\triangle AUV, \\triangle BYZ$ and $\\triangle CWX$ all have the same height because when they are folded up to create the legs of the table, the top needs to be parallel to the floor. We want to find the maximum possible value of this height, given that no two of $\\overline{UV}, \\overline{... |
https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_7 | null | 110 | In triangle $ABC$ $\tan \angle CAB = 22/7$ , and the altitude from $A$ divides $BC$ into segments of length 3 and 17. What is the area of triangle $ABC$ | [
"Call $\\angle BAD$ $\\alpha$ and $\\angle CAD$ $\\beta$ . So, $\\tan \\alpha = \\frac {17}{h}$ and $\\tan \\beta = \\frac {3}{h}$ . Using the tangent addition formula $\\tan (\\alpha + \\beta) = \\dfrac {\\tan \\alpha + \\tan \\beta}{1 - \\tan \\alpha \\cdot \\tan \\beta}$ , we get $\\tan (\\alpha + \\beta) = \\df... |
https://artofproblemsolving.com/wiki/index.php/1991_USAMO_Problems/Problem_1 | null | 77 | In triangle $ABC$ , angle $A$ is twice angle $B$ , angle $C$ is obtuse , and the three side lengths $a, b, c$ are integers. Determine, with proof, the minimum possible perimeter | [
" (diagram by integralarefun)\nAfter drawing the triangle, also draw the angle bisector of $\\angle A$ , and let it intersect $\\overline{BC}$ at $D$ . Notice that $\\triangle ADC\\sim \\triangle BAC$ , and let $AD=x$ . Now from similarity, \\[x=\\frac{bc}{a}\\] However, from the angle bisector theorem, we have \\[... |
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_28 | B | 14 | In triangle $ABC$ , angle $C$ is a right angle and $CB > CA$ . Point $D$ is located on $\overline{BC}$ so that angle $CAD$ is twice angle $DAB$ . If $AC/AD = 2/3$ , then $CD/BD = m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
$\mathrm{(A) \ }10 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ ... | [
"Let $AC = 2$ $AD = 3$ $\\cos \\angle CAD = \\frac23$\nBy the pythagorean theorem $CD = \\sqrt{3^2-2^2} = \\sqrt{5}$\n$\\sin \\angle BDA = \\sin (180^{\\circ} - \\angle BDA) = \\sin \\angle CDA = \\cos \\angle (90^{\\circ} - CDA) = \\cos \\angle CAD = \\frac23$\n$\\sin \\angle BAD = \\sqrt{ \\frac{1-cos (2\\angle B... |
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_4 | null | 291 | In triangle $ABC$ , angles $A$ and $B$ measure $60$ degrees and $45$ degrees, respectively. The bisector of angle $A$ intersects $\overline{BC}$ at $T$ , and $AT=24$ . The area of triangle $ABC$ can be written in the form $a+b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the squ... | [
"After chasing angles, $\\angle ATC=75^{\\circ}$ and $\\angle TCA=75^{\\circ}$ , meaning $\\triangle TAC$ is an isosceles triangle and $AC=24$\nUsing law of sines on $\\triangle ABC$ , we can create the following equation:\n$\\frac{24}{\\sin(\\angle ABC)}$ $=$ $\\frac{BC}{\\sin(\\angle BAC)}$\n$\\angle ABC=45^{\\ci... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_16 | B | 13.5 | In triangle $ABC$ , medians $AD$ and $CE$ intersect at $P$ $PE=1.5$ $PD=2$ , and $DE=2.5$ . What is the area of $AEDC$
[asy] unitsize(0.2cm); pair A,B,C,D,E,P; A=(0,0); B=(80,0); C=(20,40); D=(50,20); E=(40,0); P=(33.3,13.3); draw(A--B); draw(B--C); draw(A--C); draw(C--E); draw(A--D); draw(D--E); dot(A); dot(B); dot(C... | [
"Let us use mass points:\nAssign $B$ mass $1$ . Thus, because $E$ is the midpoint of $AB$ $A$ also has a mass of $1$ . Similarly, $C$ has a mass of $1$ $D$ and $E$ each have a mass of $2$ because they are between $B$ and $C$ and $A$ and $B$ respectively. Note that the mass of $D$ is twice the mass of $A$ , so $A... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_16 | null | 13.5 | In triangle $ABC$ , medians $AD$ and $CE$ intersect at $P$ $PE=1.5$ $PD=2$ , and $DE=2.5$ . What is the area of $AEDC$
[asy] unitsize(0.2cm); pair A,B,C,D,E,P; A=(0,0); B=(80,0); C=(20,40); D=(50,20); E=(40,0); P=(33.3,13.3); draw(A--B); draw(B--C); draw(A--C); draw(C--E); draw(A--D); draw(D--E); dot(A); dot(B); dot(C... | [
"We know that $[AEDC]=\\frac{3}{4}[ABC]$ , and $[ABC]=3[PAC]$ using median properties. So Now we try to find $[PAC]$ . Since $\\triangle PAC\\sim \\triangle PDE$ , then the side lengths of $\\triangle PAC$ are twice as long as $\\triangle PDE$ since $D$ and $E$ are midpoints. Therefore, $\\frac{[PAC]}{[PDE]}=2^2=4$... |
https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_13 | null | 63 | In triangle $ABC$ the medians $\overline{AD}$ and $\overline{CE}$ have lengths $18$ and $27$ , respectively, and $AB=24$ . Extend $\overline{CE}$ to intersect the circumcircle of $ABC$ at $F$ . The area of triangle $AFB$ is $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of ... | [
"Applying Stewart's Theorem to medians $AD, CE$ , we have:\nSubstituting the first equation into the second and simplification yields $24^2 = 2\\left(3AC^2 + 2 \\cdot 24^2 - 4 \\cdot 18^2\\right)- 4 \\cdot 27^2$ $\\Longrightarrow AC = \\sqrt{2^5 \\cdot 3 + 2 \\cdot 3^5 + 2^4 \\cdot 3^3 - 2^7 \\cdot 3} = 3\\sqrt{70}... |
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_14 | null | 463 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | [
"By the Law of Sines and since $\\angle BAE = \\angle CAD, \\angle BAD = \\angle CAE$ , we have\n\\begin{align*} \\frac{CD \\cdot CE}{AC^2} &= \\frac{\\sin CAD}{\\sin ADC} \\cdot \\frac{\\sin CAE}{\\sin AEC} \\\\ &= \\frac{\\sin BAE \\sin BAD}{\\sin ADB \\sin AEB} \\\\ &= \\frac{\\sin BAE}{\\sin AEB} \\cdot \\fra... |
https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_6 | null | 112 | In triangle $ABC,$ $AB = 13,$ $BC = 14,$ $AC = 15,$ and point $G$ is the intersection of the medians. Points $A',$ $B',$ and $C',$ are the images of $A,$ $B,$ and $C,$ respectively, after a $180^\circ$ rotation about $G.$ What is the area of the union of the two regions enclosed by the triangles $ABC$ and $A'B'C'?$ | [
"Since a $13-14-15$ triangle is a $5-12-13$ triangle and a $9-12-15$ triangle \"glued\" together on the $12$ side, $[ABC]=\\frac{1}{2}\\cdot12\\cdot14=84$\nThere are six points of intersection between $\\Delta ABC$ and $\\Delta A'B'C'$ . Connect each of these points to $G$\n\nThere are $12$ smaller congruent triang... |
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_14 | null | 571 | In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent . Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$ | [
"Let $\\angle QPB=x^\\circ$ . Because $\\angle AQP$ is exterior to isosceles triangle $PQB$ its measure is $2x$ and $\\angle PAQ$ has the same measure. Because $\\angle BPC$ is exterior to $\\triangle BPA$ its measure is $3x$ . Let $\\angle PBC = y^\\circ$ . It follows that $\\angle ACB = x+y$ and that $4x+2y=180^\... |
https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_13 | null | 901 | In triangle $ABC,$ point $D$ is on $\overline{BC}$ with $CD = 2$ and $DB = 5,$ point $E$ is on $\overline{AC}$ with $CE = 1$ and $EA = 3,$ $AB = 8,$ and $\overline{AD}$ and $\overline{BE}$ intersect at $P.$ Points $Q$ and $R$ lie on $\overline{AB}$ so that $\overline{PQ}$ is parallel to $\overline{CA}$ and $\overline{P... | [
"First draw $\\overline{CP}$ and extend it so that it meets with $\\overline{AB}$ at point $X$\n\nWe have that $[ABC]=\\frac{1}{2}\\cdot AC \\cdot BC\\sin{C}=\\frac{1}{2}\\cdot 4\\cdot {7}\\sin{C}=14\\sin{C}$\nBy Ceva's, \\[3\\cdot{\\frac{2}{5}}\\cdot{\\frac{BX}{AX}}=1\\implies BX=\\frac{5\\cdot AX}{6}\\] That mean... |
https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_10 | null | 450 | In triangle $ABC,\,$ angle $C$ is a right angle and the altitude from $C\,$ meets $\overline{AB}\,$ at $D.\,$ The lengths of the sides of $\triangle ABC\,$ are integers, $BD=29^3,\,$ and $\cos B=m/n\,$ , where $m\,$ and $n\,$ are relatively prime positive integers. Find $m+n.\,$ | [
"Since $\\triangle ABC \\sim \\triangle CBD$ , we have $\\frac{BC}{AB} = \\frac{29^3}{BC} \\Longrightarrow BC^2 = 29^3 AB$ . It follows that $29^2 | BC$ and $29 | AB$ , so $BC$ and $AB$ are in the form $29^2 x$ and $29 x^2$ , respectively, where x is an integer.\nBy the Pythagorean Theorem , we find that $AC^2 + BC... |
https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_14 | null | 94 | In triangle $ABC^{}_{}$ $A'$ $B'$ , and $C'$ are on the sides $BC$ $AC^{}_{}$ , and $AB^{}_{}$ , respectively. Given that $AA'$ $BB'$ , and $CC'$ are concurrent at the point $O^{}_{}$ , and that $\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92$ , find $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}$ | [
"Let $K_A=[BOC], K_B=[COA],$ and $K_C=[AOB].$ Due to triangles $BOC$ and $ABC$ having the same base, \\[\\frac{AO}{OA'}+1=\\frac{AA'}{OA'}=\\frac{[ABC]}{[BOC]}=\\frac{K_A+K_B+K_C}{K_A}.\\] Therefore, we have \\[\\frac{AO}{OA'}=\\frac{K_B+K_C}{K_A}\\] \\[\\frac{BO}{OB'}=\\frac{K_A+K_C}{K_B}\\] \\[\\frac{CO}{OC'}=\\f... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_13 | C | 72 | In triangle $CAT$ , we have $\angle ACT =\angle ATC$ and $\angle CAT = 36^\circ$ . If $\overline{TR}$ bisects $\angle ATC$ , then $\angle CRT =$
[asy] pair A,C,T,R; C = (0,0); T = (2,0); A = (1,sqrt(5+sqrt(20))); R = (3/2 - sqrt(5)/2,1.175570); draw(C--A--T--cycle); draw(T--R); label("$A$",A,N); label("$T$",T,SE); labe... | [
"In $\\triangle ACT$ , the three angles sum to $180^\\circ$ , and $\\angle C = \\angle T$\n$\\angle CAT + \\angle ATC + \\angle ACT = 180$\n$36 + \\angle ATC + \\angle ATC = 180$\n$2 \\angle ATC = 144$\n$\\angle ATC = 72$\nSince $\\angle ATC$ is bisected by $\\overline{TR}$ $\\angle RTC = \\frac{72}{2} = 36$\nNow f... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24 | B | 30 | In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$
[asy] un... | [
"Draw $X$ on $\\overline{AF}$ such that $\\overline{XD}$ is parallel to $\\overline{BC}$\nTriangles $BEF$ and $EXD$ are similar, and since $BE = ED$ , they are also congruent, and so $XE=EF$ and $XD=BF$\n$AC:AD = 3$ implies $\\frac{AF}{AX} = 3 = \\frac{FC}{XD} = \\frac{FC}{BF}$ , so $BC=BF + 3BF = 4BF$ $BF=\\frac{B... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24 | null | 30 | In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$
[asy] un... | [
"We use the line-segment ratios to infer area ratios and height ratios.\nAreas:\n$AD:DC = 1:2 \\implies AD:AC = 1:3 \\implies [ABD] =\\frac{[ABC]}{3} = 120$\n$BE:BD = 1:2 \\text{ (midpoint)} \\implies [ABE] = \\frac{[ABD]}{2} = \\frac{120}{2} = 60$\nHeights:\nLet $h_A$ = height (of altitude) from $\\overline{BC}$... |
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_1 | null | 108 | Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three number... | [
"Let $r$ be the common ratio, where $r>1$ . We then have $ar-9-(a-5)=a(r-1)-4=ar^{2}-25-(ar-9)=ar(r-1)-16$ . We now have, letting, subtracting the 2 equations, $ar^{2}+-2ar+a=12$ , so we have $3ar=432,$ or $ar=144$ , which is how much Betty had. Now we have $144+\\dfrac{144}{r}+144r=444$ , or $144(r+\\dfrac{1}{r})=... |
https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_25 | C | 65 | Inscribed in a circle is a quadrilateral having sides of lengths $25,~39,~52$ , and $60$ taken consecutively. The diameter of this circle has length
$\textbf{(A) }62\qquad \textbf{(B) }63\qquad \textbf{(C) }65\qquad \textbf{(D) }66\qquad \textbf{(E) }69$ | [
"We note that $25^2+60^2=65^2$ and $39^2+52^2=65^2$ so our answer is $\\boxed{65}$"
] |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_16 | E | 58 | Integers $a, b, c,$ and $d$ , not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the probability that $ad-bc$ is even
$\mathrm{(A)}\ \frac 38\qquad \mathrm{(B)}\ \frac 7{16}\qquad \mathrm{(C)}\ \frac 12\qquad \mathrm{(D)}\ \frac 9{16}\qquad \mathrm{(E)}\ \frac 58$ | [
"The only time when $ad-bc$ is even is when $ad$ and $bc$ are of the same parity . The chance of $ad$ being odd is $\\frac 12 \\cdot \\frac 12 = \\frac 14$ , since the only way to have $ad$ be odd is to have both $a$ and $d$ be odd. As a result, $ad$ has a $\\frac 34$ probability of being even. $bc$ also has a $\\f... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_12 | E | 58 | Integers $a, b, c,$ and $d$ , not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the probability that $ad-bc$ is even
$\mathrm{(A)}\ \frac 38\qquad \mathrm{(B)}\ \frac 7{16}\qquad \mathrm{(C)}\ \frac 12\qquad \mathrm{(D)}\ \frac 9{16}\qquad \mathrm{(E)}\ \frac 58$ | [
"The only time when $ad-bc$ is even is when $ad$ and $bc$ are of the same parity . The chance of $ad$ being odd is $\\frac 12 \\cdot \\frac 12 = \\frac 14$ , since the only way to have $ad$ be odd is to have both $a$ and $d$ be odd. As a result, $ad$ has a $\\frac 34$ probability of being even. $bc$ also has a $\\f... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_10 | E | 26 | Integers $x$ and $y$ with $x>y>0$ satisfy $x+y+xy=80$ . What is $x$
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 18 \qquad\textbf{(E)}\ 26$ | [
"Use SFFT to get $(x+1)(y+1)=81$ . The terms $(x+1)$ and $(y+1)$ must be factors of $81$ , which include $1, 3, 9, 27, 81$ . Because $x > y$ $x+1$ is equal to $27$ or $81$ . But if $x+1=81$ , then $y=0$ and so $x=\\boxed{26}$"
] |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_3 | A | 8 | Isaac has written down one integer two times and another integer three times. The sum of the five numbers is $100$ , and one of the numbers is $28.$ What is the other number?
$\textbf{(A) }8\qquad\textbf{(B) }11\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }18$ | [
"Let the first number be $x$ and the second be $y$ . We have $2x+3y=100$ . We are given one of the numbers is $28$ . If $x$ were to be $28$ $y$ would not be an integer, thus $y=28$ $2x+3(28)=100$ , which gives $x=\\boxed{8}$"
] |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_3 | A | 8 | Isaac has written down one integer two times and another integer three times. The sum of the five numbers is 100, and one of the numbers is 28. What is the other number?
$\textbf{(A)}\; 8 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 14 \qquad\textbf{(D)}\; 15 \qquad\textbf{(E)}\; 18$ | [
"Let $a$ be the number written two times, and $b$ the number written three times. Then $2a + 3b = 100$ . Plugging in $a = 28$ doesn't yield an integer for $b$ , so it must be that $b = 28$ , and we get $2a + 84 = 100$ . Solving for $a$ , we obtain $a = \\boxed{8}$"
] |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_3 | B | 250 | Isabella had a week to read a book for a school assignment. She read an average of $36$ pages per day for the first three days and an average of $44$ pages per day for the next three days. She then finished the book by reading $10$ pages on the last day. How many pages were in the book?
$\textbf{(A) }240\qquad\textbf{(... | [
"Isabella read $3\\cdot 36+3\\cdot 44$ pages in the first 6 days. Although this can be calculated directly, it is simpler to calculate it as $3\\cdot (36+44)=3\\cdot 80$ , which gives that she read $240$ pages. On her last day, she read $10$ more pages for a total of $240+10=\\boxed{250}$ pages."
] |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_7 | B | 92 | Isabella must take four 100-point tests in her math class. Her goal is to achieve an average grade of 95 on the tests. Her first two test scores were 97 and 91. After seeing her score on the third test, she realized she can still reach her goal. What is the lowest possible score she could have made on the third test?
$... | [
"Isabella wants an average grade of $95$ on her 4 tests; this also means that she wants the sum of her test scores to be at least $95 \\times 4 = 380$ (if she goes over this number, she'll be over her goal!). She's already taken two tests, which sum to $97+91 = 188$ , which means she needs $192$ more points to achi... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_18 | B | 280 | Isabella uses one-foot cubical blocks to build a rectangular fort that is $12$ feet long, $10$ feet wide, and $5$ feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?
[asy]import three; currentprojection=orthographic(-8,15,15); triple A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P; A ... | [
"There are $10 \\cdot 12 = 120$ cubes on the base of the box. Then, for each of the 4 layers above the bottom (as since each cube is 1 foot by 1 foot by 1 foot and the box is 5 feet tall, there are 4 feet left), there are $9 + 11 + 9 + 11 = 40$ cubes. Hence, the answer is $120 + 4 \\cdot 40 = \\boxed{280}$",
"We ... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_1 | E | 876 | Isabella's house has $3$ bedrooms. Each bedroom is $12$ feet long, $10$ feet wide, and $8$ feet high. Isabella must paint the walls of all the bedrooms. Doorways and windows, which will not be painted, occupy $60$ square feet in each bedroom. How many square feet of walls must be painted?
$\mathrm{(A)}\ 678 \qquad \mat... | [
"There are four walls in each bedroom (she can't paint floors or ceilings). Therefore, we calculate the number of square feet of walls there is in one bedroom: \\[2\\cdot(12\\cdot8+10\\cdot8)-60=2\\cdot176-60=292\\] We have three bedrooms, so she must paint $292\\cdot3=\\boxed{876}$ square feet of walls."
] |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_1 | E | 876 | Isabella's house has $3$ bedrooms. Each bedroom is $12$ feet long, $10$ feet wide, and $8$ feet high. Isabella must paint the walls of all the bedrooms. Doorways and windows, which will not be painted, occupy $60$ square feet in each bedroom. How many square feet of walls must be painted?
$\mathrm{(A)}\ 678 \qquad \mat... | [
"There are four walls in each bedroom (she can't paint floors or ceilings). Therefore, we calculate the number of square feet of walls there is in one bedroom: \\[2\\cdot(12\\cdot8+10\\cdot8)-60=2\\cdot176-60=292\\] We have three bedrooms, so she must paint $292\\cdot3=\\boxed{876}$ square feet of walls."
] |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_23 | null | 127 | Isosceles $\triangle ABC$ has a right angle at $C$ . Point $P$ is inside $\triangle ABC$ , such that $PA=11$ $PB=7$ , and $PC=6$ . Legs $\overline{AC}$ and $\overline{BC}$ have length $s=\sqrt{a+b\sqrt{2}}$ , where $a$ and $b$ are positive integers. What is $a+b$
[asy] pathpen = linewidth(0.7); pen f = fontsize(10); ... | [
" Using the Law of Cosines on $\\triangle PBC$ , we have:\n\\begin{align*} PB^2&=BC^2+PC^2-2\\cdot BC\\cdot PC\\cdot \\cos(\\alpha) \\Rightarrow 49 = 36 + s^2 - 12s\\cos(\\alpha) \\Rightarrow \\cos(\\alpha) = \\dfrac{s^2-13}{12s}. \\end{align*}\nUsing the Law of Cosines on $\\triangle PAC$ , we have: \\begin{align*... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_24 | A | 14.6 | Isosceles $\triangle ABC$ has equal side lengths $AB$ and $BC$ . In the figure below, segments are drawn parallel to $\overline{AC}$ so that the shaded portions of $\triangle ABC$ have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of $h$ of $\triangle ABC$
... | [
"First, we notice that the smaller isosceles triangles are similar to the larger isosceles triangles. We can find that the area of the gray area in the first triangle is $[ABC]\\cdot\\left(1-\\left(\\tfrac{11}{h}\\right)^2\\right)$ . Similarly, we can find that the area of the gray part in the second triangle is $[... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_23 | null | 8 | Isosceles right triangle $ABC$ encloses a semicircle of area $2\pi$ . The circle has its center $O$ on hypotenuse $\overline{AB}$ and is tangent to sides $\overline{AC}$ and $\overline{BC}$ . What is the area of triangle $ABC$
[asy]pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2); draw(circle(o, 2)); clip(a--b--c--cycl... | [
"First, we notice half a square so first let's create a square. Once we have a square, we will have a full circle. This circle has a diameter of 4 which will be the side of the square. The area would be $4\\cdot 4 = 16.$ Divide 16 by 2 to get the original shape and you get $\\boxed{8}$"
] |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_15 | C | 26 | Isosceles triangle $ABC$ has $AB = AC = 3\sqrt6$ , and a circle with radius $5\sqrt2$ is tangent to line $AB$ at $B$ and to line $AC$ at $C$ . What is the area of the circle that passes through vertices $A$ $B$ , and $C?$
$\textbf{(A) }24\pi\qquad\textbf{(B) }25\pi\qquad\textbf{(C) }26\pi\qquad\textbf{(D) }27\pi\qquad\... | [
"Let $\\odot O_1$ be the circle with radius $5\\sqrt2$ that is tangent to $\\overleftrightarrow{AB}$ at $B$ and to $\\overleftrightarrow{AC}$ at $C.$ Note that $\\angle ABO_1 = \\angle ACO_1 = 90^\\circ.$ Since the opposite angles of quadrilateral $ABO_1C$ are supplementary, quadrilateral $ABO_1C$ is cyclic.\nLet $... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_20 | A | 3 | Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ $5$ , and $8$ , while those of $T'$ have lengths $a$ $a$ , and $b$ . Which of the following numbers is closest to $b$
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf... | [
"The area of $T$ is $\\dfrac{1}{2} \\cdot 8 \\cdot 3 = 12$ and the perimeter is 18.\nThe area of $T'$ is $\\dfrac{1}{2} b \\sqrt{a^2 - (\\dfrac{b}{2})^2}$ and the perimeter is $2a + b$\nThus $2a + b = 18$ , so $2a = 18 - b$\nThus $12 = \\dfrac{1}{2} b \\sqrt{a^2 - (\\dfrac{b}{2})^2}$ , so $48 = b \\sqrt{4a^2 - b^2}... |
https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_26 | C | 2 | It is desired to construct a right triangle in the coordinate plane so that its legs are parallel
to the $x$ and $y$ axes and so that the medians to the midpoints of the legs lie on the lines $y = 3x + 1$ and $y = mx + 2$ . The number of different constants $m$ for which such a triangle exists is
$\textbf{(A)}\ 0\qqua... | [
"In any right triangle with legs parallel to the axes, one median to the midpoint of a leg has slope $4$ times that of the other. This can easily be shown with coordinates: any triangle of this sort may be labelled with right angle at $P(a,b)$ , other vertices $Q(a,b+2c)$ and $R(a-2d,b)$ , and thus midpoints $(a,b+... |
https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_3 | null | 111 | It is given that $\log_{6}a + \log_{6}b + \log_{6}c = 6,$ where $a,$ $b,$ and $c$ are positive integers that form an increasing geometric sequence and $b - a$ is the square of an integer. Find $a + b + c.$ | [
"$abc=6^6$ . Since they form an increasing geometric sequence, $b$ is the geometric mean of the product $abc$ $b=\\sqrt[3]{abc}=6^2=36$\nSince $b-a$ is the square of an integer, we can find a few values of $a$ that work: $11, 20, 27, 32,$ and $35$ . Out of these, the only value of $a$ that works is $a=27$ , from wh... |
https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_21 | E | 26 | It is given that one root of $2x^2 + rx + s = 0$ , with $r$ and $s$ real numbers, is $3+2i (i = \sqrt{-1})$ . The value of $s$ is:
$\textbf{(A)}\ \text{undetermined}\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ -13\qquad\textbf{(E)}\ 26$ | [
"If a quadratic with real coefficients has two non-real roots, the two roots must be complex conjugates of one another.\nThis means the other root of the given quadratic is $\\overline{3+2i}=3-2i$ . \nNow Vieta's formulas say that $s/2$ is equal to the product of the two roots, so $s = 2(3+2i)(3-2i) = \\boxed{26}$"... |
https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_7 | null | 112 | It is known that, for all positive integers $k$
Find the smallest positive integer $k$ such that $1^2+2^2+3^2+\ldots+k^2$ is a multiple of $200$ | [
"$\\frac{k(k+1)(2k+1)}{6}$ is a multiple of $200$ if $k(k+1)(2k+1)$ is a multiple of $1200 = 2^4 \\cdot 3 \\cdot 5^2$ . \nSo $16,3,25|k(k+1)(2k+1)$\nSince $2k+1$ is always odd, and only one of $k$ and $k+1$ is even, either $k, k+1 \\equiv 0 \\pmod{16}$\nThus, $k \\equiv 0, 15 \\pmod{16}$\nIf $k \\equiv 0 \\pmod{3}$... |
https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_6 | null | 142 | It is possible to place positive integers into the vacant twenty-one squares of the $5 \times 5$ square shown below so that the numbers in each row and column form arithmetic sequences. Find the number that must occupy the vacant square marked by the asterisk (*).
1988 AIME-6.png | [
"Let the coordinates of the square at the bottom left be $(0,0)$ , the square to the right $(1,0)$ , etc.\nLabel the leftmost column (from bottom to top) $0, a, 2a, 3a, 4a$ and the bottom-most row (from left to right) $0, b, 2b, 3b, 4b$ . Our method will be to use the given numbers to set up equations to solve for ... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_4 | A | 3 | It takes Anna $30$ minutes to walk uphill $1$ km from her home to school, but it takes her only $10$ minutes to walk from school to her home along the same route. What is her average speed, in km/hr, for the round trip?
$\mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 3.125\qquad \mathrm{(C) \ } 3.5\qquad \mathrm{(D) \ } 4\qqu... | [
"Since she walked $1$ km to school and $1$ km back home, her total distance is $1+1=2$ km.\nSince she spent $30$ minutes walking to school and $10$ minutes walking back home, her total time is $30+10=40$ minutes = $\\frac{40}{60}=\\frac{2}{3}$ hours.\nTherefore her average speed in km/hr is $\\frac{2}{\\frac{2}{3}}... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_4 | A | 3 | It takes Anna $30$ minutes to walk uphill $1$ km from her home to school, but it takes her only $10$ minutes to walk from school to her home along the same route. What is her average speed, in km/hr, for the round trip?
$\mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 3.125\qquad \mathrm{(C) \ } 3.5\qquad \mathrm{(D) \ } 4\qqu... | [
"Since she walked $1$ km to school and $1$ km back home, her total distance is $1+1=2$ km.\nSince she spent $30$ minutes walking to school and $10$ minutes walking back home, her total time is $30+10=40$ minutes = $\\frac{40}{60}=\\frac{2}{3}$ hours.\nTherefore her average speed in km/hr is $\\frac{2}{\\frac{2}{3}}... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_9 | B | 40 | It takes Clea 60 seconds to walk down an escalator when it is not moving, and 24 seconds when it is moving. How many seconds would it take Clea to ride the escalator down when she is not walking?
$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52$ | [
"She walks at a rate of $x$ units per second to travel a distance $y$ . As $vt=d$ , we find $60x=y$ and $24*(x+k)=y$ , where $k$ is the speed of the escalator. Setting the two equations equal to each other, $60x=24x+24k$ , which means that $k=1.5x$ . Now we divide $60$ by $1.5$ because you add the speed of the esca... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_13 | B | 40 | It takes Clea 60 seconds to walk down an escalator when it is not operating, and only 24 seconds to walk down the escalator when it is operating. How many seconds does it take Clea to ride down the operating escalator when she just stands on it?
$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf... | [
"Let $s$ be the speed of the escalator and $c$ be the speed of Clea. Using $d = v t$ , the first statement can be translated to the equation $d = 60c$ . The second statement can be translated to $d = 24(c+s)$ . Since the same distance is being covered in each scenario, we can set the two equations equal and solve f... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_3 | D | 9 | Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of $10$ miles per hour. Jack walks to the pool at a constant speed of $4$ miles per hour. How many minutes before Jack does Jill arrive?
$\textbf{(A) }5\qq... | [
"Using $d=rt$ , we can set up an equation for when Jill arrives at the swimming pool:\n$1=10t$\nSolving for $t$ , we get that Jill gets to the pool in $\\frac{1}{10}$ of an hour, which is $6$ minutes. Doing the same for Jack, we get that\nJack arrives at the pool in $\\frac{1}{4}$ of an hour, which in turn is $15$... |
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_21 | D | 71 | Jack had a bag of $128$ apples. He sold $25\%$ of them to Jill. Next he sold $25\%$ of those remaining to June. Of those apples still in his bag, he gave the shiniest one to his teacher. How many apples did Jack have then?
$\text{(A)}\ 7 \qquad \text{(B)}\ 63 \qquad \text{(C)}\ 65 \qquad \text{(D)}\ 71 \qquad \text... | [
"First he gives $128\\times .25 = 32$ apples to Jill, so he has $128-32=96$ apples left. Then he gives $96\\times .25 = 24$ apples to June, so he has $96-24=72$ left.\nFinally, he gives one to the teacher, leaving $71\\rightarrow \\boxed{71}$"
] |
https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_4 | null | 515 | Jackie and Phil have two fair coins and a third coin that comes up heads with probability $\frac47$ . Jackie flips the three coins, and then Phil flips the three coins. Let $\frac {m}{n}$ be the probability that Jackie gets the same number of heads as Phil, where $m$ and $n$ are relatively prime positive integers. Find... | [
"This can be solved quickly and easily with generating functions\nLet $x^n$ represent flipping $n$ heads.\nThe generating functions for these coins are $(1+x)$ $(1+x)$ ,and $(3+4x)$ in order.\nThe product is $3+10x+11x^2+4x^3$ . ( $ax^n$ means there are $a$ ways to get $n$ heads, eg there are $10$ ways to get $1$ h... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10B_Problems/Problem_4 | C | 1,625 | Jackson's paintbrush makes a narrow strip with a width of $6.5$ millimeters. Jackson has enough paint to make a strip $25$ meters long. How many square centimeters of paper could Jackson cover with paint?
$\textbf{(A) } 162,500 \qquad\textbf{(B) } 162.5 \qquad\textbf{(C) }1,625 \qquad\textbf{(D) }1,625,000 \qquad\textb... | [
"$6.5$ millimeters is equal to $0.65$ centimeters. $25$ meters is $2500$ centimeters. The answer is $0.65 \\times 2500$ , so the answer is $\\boxed{1,625}$",
"$6.5$ millimeters can be represented as $65 \\times 10^{-2}$ centimeters. $25$ meters is $25 \\times 10^{2}$ centimeters. Multiplying out these results in ... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_4 | C | 1,625 | Jackson's paintbrush makes a narrow strip with a width of $6.5$ millimeters. Jackson has enough paint to make a strip $25$ meters long. How many square centimeters of paper could Jackson cover with paint?
$\textbf{(A) } 162,500 \qquad\textbf{(B) } 162.5 \qquad\textbf{(C) }1,625 \qquad\textbf{(D) }1,625,000 \qquad\textb... | [
"$6.5$ millimeters is equal to $0.65$ centimeters. $25$ meters is $2500$ centimeters. The answer is $0.65 \\times 2500$ , so the answer is $\\boxed{1,625}$",
"$6.5$ millimeters can be represented as $65 \\times 10^{-2}$ centimeters. $25$ meters is $25 \\times 10^{2}$ centimeters. Multiplying out these results in ... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_13 | B | 9 | Jamal has a drawer containing $6$ green socks, $18$ purple socks, and $12$ orange socks. After adding more purple socks, Jamal noticed that there is now a $60\%$ chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?
$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) ... | [
"After Jamal adds $x$ purple socks, he has $(18+x)$ purple socks and $6+18+12+x=(36+x)$ total socks. This means the probability of drawing a purple sock is $\\frac{18+x}{36+x}$ , so we obtain \\[\\frac{18+x}{36+x}=\\frac{3}{5}\\] Since $\\frac{18+9}{36+9}=\\frac{27}{45}=\\frac{3}{5}$ , the answer is $\\boxed{9}$",
... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_11 | B | 13 | Jamal wants to save 30 files onto disks, each with 1.44 MB space. 3 of the files take up 0.8 MB, 12 of the files take up 0.7 MB, and the rest take up 0.4 MB. It is not possible to split a file onto 2 different disks. What is the smallest number of disks needed to store all 30 files?
$\textbf{(A)}\ 12 \qquad \textbf{(B)... | [
"$0.8$ MB file can either be on its own disk, or share it with a $0.4$ MB. Clearly it is better to pick the second possibility. Thus we will have $3$ disks, each with one $0.8$ MB file and one $0.4$ MB file.\nWe are left with $12$ files of $0.7$ MB each, and $12$ files of $0.4$ MB each. Their total size is $12\\cdo... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_11 | null | 13 | Jamal wants to save 30 files onto disks, each with 1.44 MB space. 3 of the files take up 0.8 MB, 12 of the files take up 0.7 MB, and the rest take up 0.4 MB. It is not possible to split a file onto 2 different disks. What is the smallest number of disks needed to store all 30 files?
$\textbf{(A)}\ 12 \qquad \textbf{(B)... | [
"Similarly to Solution 1, we see that there must be $3$ disks to account for the $0.8$ MB file. Secondly, since there are $[30-(3+12)]-3 = 12$ files(for both 0.4 MB and 0.7 MB) left, it is easy to see that the optimal way to place the files would be $3$ files per disks for the 0.4 MB files and hence would require 4... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_9 | B | 13 | Jamal wants to save 30 files onto disks, each with 1.44 MB space. 3 of the files take up 0.8 MB, 12 of the files take up 0.7 MB, and the rest take up 0.4 MB. It is not possible to split a file onto 2 different disks. What is the smallest number of disks needed to store all 30 files?
$\textbf{(A)}\ 12 \qquad \textbf{(B)... | [
"$0.8$ MB file can either be on its own disk, or share it with a $0.4$ MB. Clearly it is better to pick the second possibility. Thus we will have $3$ disks, each with one $0.8$ MB file and one $0.4$ MB file.\nWe are left with $12$ files of $0.7$ MB each, and $12$ files of $0.4$ MB each. Their total size is $12\\cdo... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_9 | null | 13 | Jamal wants to save 30 files onto disks, each with 1.44 MB space. 3 of the files take up 0.8 MB, 12 of the files take up 0.7 MB, and the rest take up 0.4 MB. It is not possible to split a file onto 2 different disks. What is the smallest number of disks needed to store all 30 files?
$\textbf{(A)}\ 12 \qquad \textbf{(B)... | [
"Similarly to Solution 1, we see that there must be $3$ disks to account for the $0.8$ MB file. Secondly, since there are $[30-(3+12)]-3 = 12$ files(for both 0.4 MB and 0.7 MB) left, it is easy to see that the optimal way to place the files would be $3$ files per disks for the 0.4 MB files and hence would require 4... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_13 | C | 4 | Jamar bought some pencils costing more than a penny each at the school bookstore and paid $\textdollar 1.43$ . Sharona bought some of the same pencils and paid $\textdollar 1.87$ . How many more pencils did Sharona buy than Jamar?
$\textbf{(A)}\hspace{.05in}2\qquad\textbf{(B)}\hspace{.05in}3\qquad\textbf{(C)}\hspace{.0... | [
"We assume that the price of the pencils remains constant. Convert $\\textdollar 1.43$ and $\\textdollar 1.87$ to cents. Since the price of the pencils is more than one penny, we can find the price of one pencil (in cents) by taking the greatest common divisor of $143$ and $187$ , which is $11$ . Therefore, Jamar ... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_1 | E | 26 | Jamie counted the number of edges of a cube, Jimmy counted the numbers of corners, and Judy counted the number of faces. They then added the three numbers. What was the resulting sum?
$\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 16 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 22 \qquad\mathrm{(E)}\ 26$ | [
"On a cube, there are $12$ edges, $8$ corners, and $6$ faces. Adding them up gets $12+8+6= \\boxed{26}$"
] |
https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_3 | null | 25 | Jane is 25 years old. Dick is older than Jane. In $n$ years, where $n$ is a positive integer, Dick's age and Jane's age will both be two-digit numbers and will have the property that Jane's age is obtained by interchanging the digits of Dick's age. Let $d$ be Dick's present age. How many ordered pairs of positive integ... | [
"Let Jane's age $n$ years from now be $10a+b$ , and let Dick's age be $10b+a$ . If $10b+a>10a+b$ , then $b>a$ . The possible pairs of $a,b$ are:\nThat makes 36. But $10a+b>25$ , so we subtract all the extraneous pairs: $(1,2), (1,3), (2,3), (1,4), (2,4), (1,5), (2,5), (1,6), (1,7), (1,8),$ and $(1,9)$ $36-11=\\boxe... |
https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_1 | null | 85 | Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C contains one liter of a solution that is $k\%$ acid. From jar C, $\frac{m}{n}$ liters of the solution is added to jar A, and the remainder of the solution in jar C is added to jar B. At the en... | [
"Jar A contains $\\frac{11}{5}$ liters of water, and $\\frac{9}{5}$ liters of acid; jar B contains $\\frac{13}{5}$ liters of water and $\\frac{12}{5}$ liters of acid.\nThe gap between the amount of water and acid in the first jar, $\\frac{2}{5}$ , is double that of the gap in the second jar, $\\frac{1}{5}$ . Theref... |
https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_24 | B | 5 | Jean has made a piece of stained glass art in the shape of two mountains, as shown in the figure below. One mountain peak is $8$ feet high while the other peak is $12$ feet high. Each peak forms a $90^\circ$ angle, and the straight sides form a $45^\circ$ angle with the ground. The artwork has an area of $183$ square f... | [
"Extend the \"inner part\" of the mountain so that the image is two right triangles that overlap in a third right triangle as shown. The side length of the largest right triangle is $12\\sqrt{2},$ which means its area is $144.$ Similarly, the area of the second largest right triangle is $64$ (the side length is $8... |
https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_4 | null | 116 | Jen enters a lottery by picking $4$ distinct numbers from $S=\{1,2,3,\cdots,9,10\}.$ $4$ numbers are randomly chosen from $S.$ She wins a prize if at least two of her numbers were $2$ of the randomly chosen numbers, and wins the grand prize if all four of her numbers were the randomly chosen numbers. The probability of... | [
"This is a conditional probability problem. Bayes' Theorem states that \\[P(A|B)=\\dfrac{P(B|A)\\cdot P(A)}{P(B)}\\]\nLet us calculate the probability of winning a prize. We do this through casework: how many of Jen's drawn numbers match the lottery's drawn numbers?\nTo win a prize, Jen must draw at least $2$ numbe... |
https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_2 | null | 29 | Jenn randomly chooses a number $J$ from $1, 2, 3,\ldots, 19, 20$ . Bela then randomly chooses a number $B$ from $1, 2, 3,\ldots, 19, 20$ distinct from $J$ . The value of $B - J$ is at least $2$ with a probability that can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. F... | [
"$B-J \\ne 0$ because $B \\ne J$ , so the probability that $B-J < 0$ is $\\frac{1}{2}$ by symmetry.\nThe probability that $B-J = 1$ is $\\frac{19}{20 \\times 19} = \\frac{1}{20}$ because there are 19 pairs: $(B,J) = (2,1), \\ldots, (20,19)$\nThe probability that $B-J \\ge 2$ is $1-\\frac{1}{2}-\\frac{1}{20} = \\fra... |
https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_13 | null | 163 | Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and K... | [
"Consider the unit cicle of radius 50. Assume that they start at points $(-50,100)$ and $(-50,-100).$ Then at time $t$ , they end up at points $(-50+t,100)$ and $(-50+3t,-100).$ The equation of the line connecting these points and the equation of the circle are \\begin{align}y&=-\\frac{100}{t}x+200-\\frac{5000}{t}\... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_17 | D | 9 | Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad ... | [
"For starters, we identify d as distance and v as velocity (speed)\nWriting the equation gives us: $\\frac{d}{v}=\\frac{1}{3}$ and $\\frac{d}{v+18}=\\frac{1}{5}$\nThis gives $d=\\frac{1}{5}v+3.6=\\frac{1}{3}v$ , which gives $v=27$ , which then gives $d=\\boxed{9}$",
"$d = rt$ $d=\\frac{1}{3} \\times r = \\frac{1}... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_7 | A | 30 | Jerry and Silvia wanted to go from the southwest corner of a square field to the northeast corner. Jerry walked due east and then due north to reach the goal, but Silvia headed northeast and reached the goal walking in a straight line. Which of the following is closest to how much shorter Silvia's trip was, compared to... | [
"Let $j$ represent how far Jerry walked, and $s$ represent how far Silvia walked. Since the field is a square, and Jerry walked two sides of it, while Silvia walked the diagonal, we can simply define the side of the square field to be one, and find the distances they walked. Since Jerry walked two sides, $j = 2$ Si... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_4 | A | 30 | Jerry and Silvia wanted to go from the southwest corner of a square field to the northeast corner. Jerry walked due east and then due north to reach the goal, but Silvia headed northeast and reached the goal walking in a straight line. Which of the following is closest to how much shorter Silvia's trip was, compared to... | [
"Let $j$ represent how far Jerry walked, and $s$ represent how far Sylvia walked. Since the field is a square, and Jerry walked two sides of it, while Silvia walked the diagonal, we can simply define the side of the square field to be one, and find the distances they walked. Since Jerry walked two sides, $j = 2$ Si... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_21 | C | 151 | Jerry cuts a wedge from a $6$ -cm cylinder of bologna as shown by the dashed curve. Which answer choice is closest to the volume of his wedge in cubic centimeters?
[asy] defaultpen(linewidth(0.65)); real d=90-63.43494882; draw(ellipse((origin), 2, 4)); fill((0,4)--(0,-4)--(-8,-4)--(-8,4)--cycle, white); draw(ellipse((-... | [
"The slice is cutting the cylinder into two equal wedges with equal volume. The cylinder's volume is $\\pi r^2 h = \\pi (4^2)(6) = 96\\pi$ . The volume of the wedge is half this which is $48\\pi \\approx \\boxed{151}$"
] |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_19 | B | 151 | Jerry starts at $0$ on the real number line. He tosses a fair coin $8$ times. When he gets heads, he moves $1$ unit in the positive direction; when he gets tails, he moves $1$ unit in the negative direction. The probability that he reaches $4$ at some time during this process $\frac{a}{b},$ where $a$ and $b$ are relati... | [
"For $6$ to $8$ heads, we are guaranteed to hit $4$ heads, so the sum here is $\\binom{8}{2}+\\binom{8}{1}+\\binom{8}{0}=28+8+1=37$\nFor $4$ heads, you have to hit the $4$ heads at the start so there's only one way, $1$\nFor $5$ heads, we either start off with $4$ heads, which gives us $_4\\text{C}_1=4$ ways to arr... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_25 | B | 3 | Jim starts with a positive integer $n$ and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with $n = 55$ , then his sequence contains $5$ numbers:
\[\begin{arra... | [
"We can find the answer by working backwards. We begin with $1-1^2=0$ on the bottom row, then the $1$ goes to the right of the equal's sign in the row above. We find the smallest value $x$ for which $x-1^2=1$ and $x>1^2$ , which is $x=2$\nWe repeat the same procedure except with $x-1^2=1$ for the next row and $x-1^... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_13 | E | 8 | Jo and Blair take turns counting from $1$ to one more than the last number said by the other person. Jo starts by saying $``1"$ , so Blair follows by saying $``1, 2"$ . Jo then says $``1, 2, 3"$ , and so on. What is the $53^{\text{rd}}$ number said?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad ... | [
"We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. Thus, after nine \"turns\", $1+2+3+4+5+6+7+8+9=45$ numbers have been said. In the tenth turn, the eighth number will be the 53rd number said, becau... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_7 | E | 8 | Jo and Blair take turns counting from $1$ to one more than the last number said by the other person. Jo starts by saying $``1"$ , so Blair follows by saying $``1, 2"$ . Jo then says $``1, 2, 3"$ , and so on. What is the $53^{\text{rd}}$ number said?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad ... | [
"We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. Thus, after nine \"turns\", $1+2+3+4+5+6+7+8+9=45$ numbers have been said. In the tenth turn, the eighth number will be the 53rd number said, becau... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_10 | D | 8 | Joe had walked half way from home to school when he realized he was late. He ran the rest of the way to school. He ran 3 times as fast as he walked. Joe took 6 minutes to walk half way to school. How many minutes did it take Joe to get from home to school?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 7.3\qquad\textbf{(C)}\ 7.7\... | [
"We can use the equation $d=rt$ where $d$ is the distance, $r$ is the rate, and $t$ is the time. The distances he ran and walked are equal, so $r_rt_r=r_wt_w$ , where $r_r$ is the rate at which he ran, $t_r$ is the time for which he ran, $r_w$ is the rate at which he walked, and $t_w$ is the time for which he walke... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_8 | C | 2 | Joe has a collection of $23$ coins, consisting of $5$ -cent coins, $10$ -cent coins, and $25$ -cent coins. He has $3$ more $10$ -cent coins than $5$ -cent coins, and the total value of his collection is $320$ cents. How many more $25$ -cent coins does Joe have than $5$ -cent coins?
$\textbf{(A) } 0 \qquad \t... | [
"Let $x$ be the number of $5$ -cent coins that Joe has. Therefore, he must have $(x+3) \\ 10$ -cent coins and $(23-(x+3)-x) \\ 25$ -cent coins. Since the total value of his collection is $320$ cents, we can write \\begin{align*} 5x + 10(x+3) + 25(23-(x+3)-x) &= 320 \\\\ 5x + 10x + 30 + 500 - 50x &= 320 \\\\ 35x &= ... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_19 | E | 11 | Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age i... | [
"Suppose that Chloe is $c$ years old today, so Joey is $c+1$ years old today. After $n$ years, Chloe and Zoe will be $n+c$ and $n+1$ years old, respectively. We are given that \\[\\frac{n+c}{n+1}=1+\\frac{c-1}{n+1}\\] is an integer for $9$ nonnegative integers $n.$ It follows that $c-1$ has $9$ positive divisors. T... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_14 | E | 11 | Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age i... | [
"Suppose that Chloe is $c$ years old today, so Joey is $c+1$ years old today. After $n$ years, Chloe and Zoe will be $n+c$ and $n+1$ years old, respectively. We are given that \\[\\frac{n+c}{n+1}=1+\\frac{c-1}{n+1}\\] is an integer for $9$ nonnegative integers $n.$ It follows that $c-1$ has $9$ positive divisors. T... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_6 | D | 11 | Joey and his five brothers are ages $3$ $5$ $7$ $9$ $11$ , and $13$ . One afternoon two of his brothers whose ages sum to $16$ went to the movies, two brothers younger than $10$ went to play baseball, and Joey and the $5$ -year-old stayed home. How old is Joey?
$\textbf{(A)}\ 3 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C... | [
"Because the $5$ -year-old stayed home, we know that the $11$ -year-old did not go to the movies, as the $5$ -year-old did not and $11+5=16$ . Also, the $11$ -year-old could not have gone to play baseball, as he is older than $10$ . Thus, the $11$ -year-old must have stayed home, so Joey is $\\boxed{11}$",
"The... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_18 | D | 56 | Johann has $64$ fair coins. He flips all the coins. Any coin that lands on tails is tossed again. Coins that land on tails on the second toss are tossed a third time. What is the expected number of coins that are now heads?
$\textbf{(A) } 32 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 48 \qquad\textbf{(D) } 56 \qquad\te... | [
"We can simplify the problem first, then apply reasoning to the original problem. Let's say that there are $8$ coins. Shaded coins flip heads, and blank coins flip tails. So, after the first flip;\n\nThen, after the second (new heads in blue);\n\nAnd after the third (new head in green);\n\nSo in total, $7$ of the $... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_18 | null | 56 | Johann has $64$ fair coins. He flips all the coins. Any coin that lands on tails is tossed again. Coins that land on tails on the second toss are tossed a third time. What is the expected number of coins that are now heads?
$\textbf{(A) } 32 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 48 \qquad\textbf{(D) } 56 \qquad\te... | [
"(Similar to solution 2)\nNotice how:\n$E(\\text{Heads on 1st flip, 2nd flip, 3rd flip}) = E(\\text{Heads on 1st flip}) + E(\\text{Heads on 2nd flip}) + E(\\text{Heads on 3rd flip})$\nThe expected number of heads for the first flip is simply $64 \\cdot \\frac{1}{2}$ , since each coin has a 1 in 2 chance of being he... |
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_4 | null | 49 | Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east at $20$ miles per hour, and Steve rides west at $20$ miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds each pass the two... | [
"For the purposes of this problem, we will use miles and minutes as our units; thus, the bikers travel at speeds of $\\dfrac{1}{3}$ mi/min.\nLet $d$ be the length of the trains, $r_1$ be the speed of train 1 (the faster train), and $r_2$ be the speed of train 2.\nConsider the problem from the bikers' moving frame o... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_2 | E | 95 | Josanna's test scores to date are $90, 80, 70, 60,$ and $85$ . Her goal is to raise here test average at least $3$ points with her next test. What is the minimum test score she would need to accomplish this goal?
$\textbf{(A)}\ 80 \qquad\textbf{(B)}\ 82 \qquad\textbf{(C)}\ 85 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ ... | [
"The average of her current scores is $77$ . To raise it $3$ points, she needs an average of $80$ , and so after her $6$ tests, a sum of $480$ . Her current sum is $385$ , so she needs a $480 - 385 = \\boxed{95}$"
] |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_2 | E | 95 | Josanna's test scores to date are $90, 80, 70, 60,$ and $85.$ Her goal is to raise her test average at least $3$ points with her next test. What is the minimum test score she would need to accomplish this goal?
$\textbf{(A)}\ 80 \qquad \textbf{(B)}\ 82 \qquad \textbf{(C)}\ 85 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}... | [
"Take the average of her current test scores, which is \\[\\frac{90+80+70+60+85}{5} = \\frac{385}{5} = 77\\]\nThis means that she wants her test average after the sixth test to be $80.$ Let $x$ be the score that Josanna receives on her sixth test. Thus, our equation is\n\\[\\frac{90+80+70+60+85+x}{6} = 80\\]\n\\[38... |
https://artofproblemsolving.com/wiki/index.php/1995_AJHSME_Problems/Problem_2 | C | 14 | Jose is $4$ years younger than Zack. Zack is $3$ years older than Inez. Inez is $15$ years old. How old is Jose?
$\text{(A)}\ 8 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 14 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 22$ | [
"Zack is $3$ years older than Inez, who is $15$ . Therefore, Zack is $15 + 3 = 18$ years old.\nJose is $4$ years younger than Zack, who is $18$ . Therefore, Jose is $18 - 4 = 14$ years old, and the answer is $\\boxed{14}$"
] |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_7 | B | 5 | Josh and Mike live $13$ miles apart. Yesterday Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?
$... | [
"Let $m$ be the distance in miles that Mike rode.\nSince Josh rode for twice the length of time as Mike and at four-fifths of Mike's rate, he rode $2\\cdot\\frac{4}{5}\\cdot m = \\frac{8}{5}m$ miles.\nSince their combined distance was $13$ miles,\n$\\frac{8}{5}m + m = 13$\n$\\frac{13}{5}m = 13$\n$m = \\boxed{5}$"
] |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_7 | D | 64 | Josh writes the numbers $1,2,3,\dots,99,100$ . He marks out $1$ , skips the next number $(2)$ , marks out $3$ , and continues skipping and marking out the next number to the end of the list. Then he goes back to the start of his list, marks out the first remaining number $(2)$ , skips the next number $(4)$ , marks out ... | [
"Following the pattern, you are crossing out...\nTime 1: Every non-multiple of $2$\nTime 2: Every non-multiple of $4$\nTime 3: Every non-multiple of $8$\nFollowing this pattern, you are left with every multiple of $64$ which is only $\\boxed{64}$"
] |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_6 | B | 17 | Joy has $30$ thin rods, one each of every integer length from $1 \text{ cm}$ through $30 \text{ cm}$ . She places the rods with lengths $3 \text{ cm}$ $7 \text{ cm}$ , and $15 \text{cm}$ on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How m... | [
"The quadrilateral cannot be a straight line. Thus, the fourth side must be longer than $15 - (3 + 7) = 5$ and shorter than $15 + 3 + 7 = 25$ . This means Joy can use the $19$ possible integer rod lengths that fall into $[6, 24]$ . However, she has already used the rods of length $7$ cm and $15$ cm so the answer is... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_10 | B | 17 | Joy has $30$ thin rods, one each of every integer length from $1$ cm through $30$ cm. She places the rods with lengths $3$ cm, $7$ cm, and $15$ cm on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose... | [
"The triangle inequality generalizes to all polygons, so $x < 3+7+15$ and $15<x+3+7$ yields $5<x<25$ . Now, we know that there are $19$ numbers between $5$ and $25$ exclusive, but we must subtract $2$ to account for the 2 lengths already used that are between those numbers, which gives $19-2=\\boxed{17}$"
] |
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_16 | E | 8 | Joyce made $12$ of her first $30$ shots in the first three games of this basketball game, so her seasonal shooting average was $40\%$ . In her next game, she took $10$ shots and raised her seasonal shooting average to $50\%$ . How many of these $10$ shots did she make?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text... | [
"After the fourth game, she took $40$ shots, $50\\%$ of which she made, so she made $40\\times .5=20$ shots. Twelve of them were made in the first three games, so in the last game she made $20-12=8$ shots.\n$\\boxed{8}$"
] |
https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_4 | E | 5,650 | Julie is preparing a speech for her class. Her speech must last between one-half hour and three-quarters of an hour. The ideal rate of speech is 150 words per minute. If Julie speaks at the ideal rate, which of the following number of words would be an appropriate length for her speech?
$\text{(A)}\ 2250 \qquad \tex... | [
"Since there are $60$ minutes in an hour, we need a number of words between $(\\frac{1}{2}\\times 60)\\times150=4500$ and $(\\frac{3}{4}\\times60)\\times150=6750$ . The only such choice is $\\boxed{5650}$"
] |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_2 | C | 2.5 | Karl bought five folders from Pay-A-Lot at a cost of $\textdollar 2.50$ each.
Pay-A-Lot had a 20%-off sale the following day. How much could
Karl have saved on the purchase by waiting a day?
$\textbf{(A)}\ \textdollar 1.00 \qquad\textbf{(B)}\ \textdollar 2.00 \qquad\textbf{(C)}\ \textdollar 2.50\qquad\textbf{(D)}\ \te... | [
"Karl paid $5 \\cdot 2.50 = \\textdollar 12.50$ $20 \\%$ of this cost that he saved is $12.50 \\cdot .2 = \\boxed{2.50}$",
"Each folder can also be $5/2$ dollars, and $20\\%$ can be shown as $(1/5)$ . We can multiply $(5/2) \\cdot (1/5) = (1/2)$ $(1/2)$ is also $50$ cents or the amount of money that is saved afte... |
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