problem_id int64 1 978 | question stringlengths 86 2.11k | source stringlengths 19 76 | solution stringlengths 94 14.7k | asymptote_code stringlengths 44 17.8k | solution_image_url stringclasses 16
values |
|---|---|---|---|---|---|
158 | Rectangle $ABCD$ and a semicircle with diameter $AB$ are coplanar and have nonoverlapping interiors. Let $\mathcal{R}$ denote the region enclosed by the semicircle and the rectangle. Line $\ell$ meets the semicircle, segment $AB$, and segment $CD$ at distinct points $N$, $U$, and $T$, respectively. Line $\ell$ divides ... | 2010 AIME I Problem 13 | Diagram
Solution 1
The center of the semicircle is also the midpoint of $AB$. Let this point be O. Let $h$ be the length of $AD$.
Rescale everything by 42, so $AU = 2, AN = 3, UB = 4$. Then $AB = 6$ so $OA = OB = 3$.
Since $ON$ is a radius of the semicircle, $ON = 3$. Thus $OAN$ is an equilateral triangle.
Let $X$,... | /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */ import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(500); pen zzttqq = rgb(0.6,0.2,0); pen xdxdff = rgb(0.4902,0.4902,1); /* segments and figures */ draw((0,-154.31785)--(0,0)); draw((0,0)--(252,0)); draw((0,0)--(126,0),zzttqq); draw((12... | [] |
159 | In $\triangle{ABC}$ with $AB = 12$, $BC = 13$, and $AC = 15$, let $M$ be a point on $\overline{AC}$ such that the incircles of $\triangle{ABM}$ and $\triangle{BCM}$ have equal radii. Then $\frac{AM}{CM} = \frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$. | 2010 AIME I Problem 15 | Solution 1
Let $AM = x$, then $CM = 15 - x$. Also let $BM = d$ Clearly, $\frac {[ABM]}{[CBM]} = \frac {x}{15 - x}$. We can also express each area by the rs formula. Then $\frac {[ABM]}{[CBM]} = \frac {p(ABM)}{p(CBM)} = \frac {12 + d + x}{28 + d - x}$. Equating and cross-multiplying yields $25x + 2dx = 15d + 180$ or $d ... | // Block 1
/* from geogebra: see azjps, userscripts.org/scripts/show/72997 */
import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(200);
/* segments and figures */
draw((0,0)--(15,0));
draw((15,0)--(6.66667,9.97775));
draw((6.66667,9.97775)--(0,0));
draw((7.33333,0)--(6.66667,9.97775));
draw(circle((4.66667,2... | [] |
160 | A point $P$ is chosen at random in the interior of a unit square $S$. Let $d(P)$ denote the distance from $P$ to the closest side of $S$. The probability that $\frac{1}{5}\le d(P)\le\frac{1}{3}$ is equal to $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2010 AIME II Problem 2 | Any point outside the square with side length $\frac{1}{3}$ that has the same center and orientation as the unit square and inside the square with side length $\frac{3}{5}$ that has the same center and orientation as the unit square has $\frac{1}{5}\le d(P)\le\frac{1}{3}$.
Since the area of the unit square is $1$, th... | // Block 1
unitsize(1mm);
defaultpen(linewidth(.8pt));
draw((0,0)--(0,30)--(30,30)--(30,0)--cycle);
draw((6,6)--(6,24)--(24,24)--(24,6)--cycle);
draw((10,10)--(10,20)--(20,20)--(20,10)--cycle);
fill((6,6)--(6,24)--(24,24)--(24,6)--cycle,gray);
fill((10,10)--(10,20)--(20,20)--(20,10)--cycle,white);
// Block 2
unitsize(... | [] |
161 | Triangle $ABC$ with right angle at $C$, $\angle BAC < 45^\circ$ and $AB = 4$. Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$. The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$, where $p$, $q$, $r$ are positive integers and $r$ is not divisible by the squa... | 2010 AIME II Problem 14 | Let $O$ be the circumcenter of $ABC$ and let the intersection of $CP$ with the circumcircle be $D$. It now follows that $\angle{DOA} = 2\angle ACP = \angle{APC} = \angle{DPB}$. Hence $ODP$ is isosceles and $OD = DP = 2$.
Denote $E$ the projection of $O$ onto $CD$. Now $CD = CP + DP = 3$. By the Pythagorean Theorem, $... | // Block 1
/* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */
import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(250); real lsf = 0.5; /* changes label-to-point distance */
pen xdxdff = rgb(0.49,0.49,1); pen qqwuqq = rgb(0,0.39,0); pen fftttt = rgb(1,0.2,0.2);
/* segments and figures */... | [] |
162 | 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... | 2011 AIME I Problem 2 | 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 similar. This i... | unitsize(0.5cm);
defaultpen(0.8);
pair A=(0,0), B=(12,0), C=(12,10), D=(0,10);
draw(A--B--C--D--cycle);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,NW);
pair E=(3,5), F=(9,5);
dot("$E$",E,N);
dot("$F$",F,N);
pair G = extension(A, D, E, F);
pair H = extension(B, C, E, F);
draw(G--H);
dot("$G... | [] |
163 | 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. Find ... | 2011 AIME I Problem 4 | 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, $AQ=AC=117$, a... | // Block 1
defaultpen(fontsize(10)+0.8); size(200);
pen p=fontsize(9)+linewidth(3);
pair A,B,C,D,K,L,M,N,P,Q;
A=origin; B=(125,0); C=IP(CR(A,117),CR(B,120)); L=extension(B,C,A,bisectorpoint(B,A,C)); K=extension(A,C,B,bisectorpoint(C,B,A)); M=foot(C,B,K); N=foot(C,A,L);
draw(A--B--C--A); draw(A--L^^B--K, gray+dashed+0.5... | [] |
164 | The probability that a set of three distinct vertices chosen at random from among the vertices of a regular n-gon determine an obtuse triangle is $\frac{93}{125}$ . Find the sum of all possible values of $n$. | 2011 AIME I Problem 10 | Let the regular $n$-gon be inscribed in a circle, allowing the use of the inscribed angle theorem.
We wish to determine the number of three distinct vertices such that they form an obtuse triangle.
WLOG fix vertex $A$
Let $B$ be $k$ edges away from $A$ clockwise
$C$ be $j$ edges away from $A$ in the other direction
... | // Block 1
/* Generated by Cloud's Excalidraw to Asymptote */
draw(shift(-950.5, -920.5) * rotate(0) * ellipse((0, 0), 295.5, 295.5), rgb(0.11, 0.11, 0.11)+linewidth(1)+solid);
draw((-1073, -654)--(-817, -654), rgb(0.87, 0.19, 0.19)+linewidth(2)+solid);
draw((-817, -658)--(-662, -855), rgb(0.87, 0.19, 0.19)+linewidth(2... | [] |
165 | A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$. The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\frac{r-\sqrt{s}}{t}$, where $r$, $s$, and $t$ are positive integ... | 2011 AIME I Problem 13 | First, shift the cube down by 11 so the vertices adjacent to $A$ are $-1$, $0$, and $1$ above the plane.
Consider the two points 1 above and 1 below the plane:
Now we rotate the cube so that all 3 vertices adjacent to $A$ are on the plane. We can calculate $A$ to be $\frac{10\sqrt3}{3}$ below the plane.
Notice that... | // Block 1
defaultpen(fontsize(10)+0.8); size(300);
pen p=fontsize(9)+linewidth(3);
// Define points
pair A=(-7,0), B=(0,0), C=(7,0);
pair D=(-7,-1), E=(7,1);
// Draw segments
draw(A--B--C, linewidth(1.2));
draw(A--D, linewidth(1.2));
draw(C--E, linewidth(1.2));
draw(D--E, linewidth(1.2));
// Label lengths
label("$7... | [] |
166 | On square $ABCD$, point $E$ lies on side $AD$ and point $F$ lies on side $BC$, so that $BE=EF=FD=30$. Find the area of the square $ABCD$. | 2011 AIME II Problem 2 | Drawing the square and examining the given lengths,
you find that the three segments cut the square into three equal horizontal sections. Therefore, ($x$ being the side length), $\sqrt{x^2+(x/3)^2}=30$, or $x^2+(x/3)^2=900$. Multiplying both sides by $9$ and simplifying, we find that $10x^2=8100$. Dividing by ten give... | // Block 1
size(2inch, 2inch);
currentpen = fontsize(8pt);
pair A = (0, 0); dot(A); label("$A$", A, plain.SW);
pair B = (3, 0); dot(B); label("$B$", B, plain.SE);
pair C = (3, 3); dot(C); label("$C$", C, plain.NE);
pair D = (0, 3); dot(D); label("$D$", D, plain.NW);
pair E = (0, 1); dot(E); label("$E$", E, plain.W);
pa... | [] |
167 | 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 positive ... | 2011 AIME II Problem 4 | 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}{PA} = \frac{1}{\frac{... | // Block 1
pointpen = black; pathpen = linewidth(0.7);
pair A = (0,0), C= (11,0), B=IP(CR(A,20),CR(C,18)), D = IP(B--C,CR(B,20/31*abs(B-C))), M = (A+D)/2, P = IP(M--2*M-B, A--C), D2 = IP(D--D+P-B, A--C);
D(MP("A",D(A))--MP("B",D(B),N)--MP("C",D(C))--cycle); D(A--MP("D",D(D),NE)--MP("D'",D(D2))); D(B--MP("P",D(P))); D... | [] |
168 | Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are re... | 2012 AIME I Problem 12 | We have $\angle BCE = \angle ECD = \angle DCA = \tfrac 13 \cdot 90^\circ = 30^\circ$. Drop the altitude from $D$ to $CB$ and call the foot $F$.
Let $CD = 8a$. Using angle bisector theorem on $\triangle CDB$, we get $CB = 15a$. Now $CDF$ is a $30$-$60$-$90$ triangle, so $CF = 4a$, $FD = 4a\sqrt{3}$, and $FB = 11a$. ... | // Block 1
import cse5;size(200);
defaultpen(linewidth(0.4)+fontsize(8));
pair A,B,C,D,E0,F;
C=origin;
B=(10,0);
A=(0,5);
E0=extension(C,dir(30),A,B);
D=extension(C,dir(60),A,B);
F=foot(D,C,B);
draw(A--B--C--A, black+0.8);
draw(C--D--F^^C--E0, gray);
dot("$A$",A,N);
dot("$B$",B,SE);
dot("$C$",C,SW);
dot("$D$",D,NE);
... | [] |
169 | Three concentric circles have radii $3,$ $4,$ and $5.$ An equilateral triangle with one vertex on each circle has side length $s.$ The largest possible area of the triangle can be written as $a + \tfrac{b}{c} \sqrt{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers, $b$ and $c$ are relatively prime, and $d$ is not... | 2012 AIME I Problem 13 | Reinterpret the problem in the following manner. Equilateral triangle $ABC$ has a point $X$ on the interior such that $AX = 5,$ $BX = 4,$ and $CX = 3.$ A $60^\circ$ counter-clockwise rotation about vertex $A$ maps $X$ to $X'$ and $B$ to $C.$
Note that angle $XAX'$ is $60$ and $XA = X'A = 5$ which tells us that triang... | // Block 1
import cse5;
size(200);
defaultpen(linewidth(0.4)+fontsize(8));
pair O = origin;
pair A,B,C,Op,Bp,Cp;
path c3,c4,c5;
c3 = CR(O,3);
c4 = CR(O,4);
c5 = CR(O,5);
draw(c3^^c4^^c5, gray+0.25);
A = 5*dir(96.25);
Op = rotate(60,A)*O;
B = OP(CR(Op,4),c3);
Bp = IP(CR(Op,4),c3);
C = rotate(-60,A)*B;
Cp = rotate(-60,A... | [] |
169 | Three concentric circles have radii $3,$ $4,$ and $5.$ An equilateral triangle with one vertex on each circle has side length $s.$ The largest possible area of the triangle can be written as $a + \tfrac{b}{c} \sqrt{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers, $b$ and $c$ are relatively prime, and $d$ is not... | 2012 AIME I Problem 13 | Let's call the circle center $X$. It has a distance of 3, 4, 5 to an equilateral triangle $LMN$. Consider $X$’s pedal triangle $ABC$. Since $X$’s antipedal triangle is equilateral, $X$ must be the one of the isogonic centers of $\triangle{ABC}$. We’ll take the one inside $ABC$, i.e., the Fermat point, because it leads ... | // Block 1
import olympiad;
import cse5;
import graph;
dotfactor = 2;
unitsize(0.3inch);
pair B = (0,0), C= (5,0), A = (sqrt(9-2.4*2.4),2.4);
pair D = rotate(60,B)*A, E=rotate(60,A)*C, F=rotate(60,C)*B;
pair X = extension(A,F,D,C);
pair L = (-1.5,2), M = (6.2,3), N = rotate(-60,L)*M;
dot("$C$", C, dir(0)); dot("$... | [] |
169 | Three concentric circles have radii $3,$ $4,$ and $5.$ An equilateral triangle with one vertex on each circle has side length $s.$ The largest possible area of the triangle can be written as $a + \tfrac{b}{c} \sqrt{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers, $b$ and $c$ are relatively prime, and $d$ is not... | 2012 AIME I Problem 13 | We have $x=3$, $y=4$, and $z=5$. Because $AD=m$ is the median of $\triangle AOB$, by Stewart's Theorem we have \[m^2=\frac 12 (x^2+y^2)-\frac 14 \cdot z^2\quad \Rightarrow \quad m = \frac 52.\] Because $CD$ is the altitude of equilateral triangle $OBC$, we have $CD=\frac{\sqrt{3}}2\cdot z$. Then in $\triangle ADC$, we ... | // Block 1
import cse5; size(350); defaultpen(linewidth(0.6)+fontsize(12));
pair O=origin; pair Op,Bp,Cp;
path c3 = CR(O,3), c4 = CR(O,4), c5 = CR(O,5);
var theta=55.75;
pair A = 3*dir(theta), Ap = rotate(150,O)*A, F=IP(c4,O--2*Ap), C=rotate(60,A)*F, E=rotate(60,A)*C, B=IP(c5,O--E), D=foot(C,O,E);
filldraw(A--O-... | [] |
170 | Ana, Bob, and Cao bike at constant rates of $8.6$ meters per second, $6.2$ meters per second, and $5$ meters per second, respectively. They all begin biking at the same time from the northeast corner of a rectangular field whose longer side runs due west. Ana starts biking along the edge of the field, initially heading... | 2012 AIME II Problem 4 | Let $a,b,c$ be the labeled lengths as shown in the diagram. Also, assume WLOG the time taken is $1$ second.
Observe that $\dfrac{2a+b+c}{8.6}=1$ or $2a+b+c=8.6$, and $\dfrac{b+c}{6.2}=1$ or $b+c=6.2$. Subtracting the second equation from the first gives $2a=2.4$, or $a=1.2$.
Now, let us solve $b$ and $c$. Note that ... | // Block 1
draw((1.2,0)--(0,0)--(0,1.4)--(6,1.4)--(6,0)--(1.2,0)--(6,1.4));
label("$D$", (1.2,0),dir(-90));
dot((6,1.4));
dot((1.2,0));
label("$a$", (0.6,0),dir(-90));
label("$b$", (3.6,0),dir(-90));
label("$c$", (6,0.7),dir(0));
// Block 2
draw((1.2,0)--(0,0)--(0,1.4)--(6,1.4)--(6,0)--(1.2,0)--(6,1.4)); label("$D$", (... | [] |
171 | Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$, $BC=7$, and $AC=3$. The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $\overline{DE}$. Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$. Then $AF^2 = ... | 2012 AIME II Problem 15 | Use the angle bisector theorem to find $CD=\tfrac{21}{8}$, $BD=\tfrac{35}{8}$, and use Stewart's Theorem to find $AD=\tfrac{15}{8}$. Use Power of Point $D$ to find $DE=\tfrac{49}{8}$, and so $AE=8$. Use law of cosines to find $\angle CAD = \tfrac{\pi} {3}$, hence $\angle BAD = \tfrac{\pi}{3}$ as well, and $\triangle BC... | // Block 1
size(150);
defaultpen(fontsize(9pt));
picture pic;
pair A,B,C,D,E,F,W;
B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D... | [] |
171 | Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$, $BC=7$, and $AC=3$. The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $\overline{DE}$. Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$. Then $AF^2 = ... | 2012 AIME II Problem 15 | Let $a = BC$, $b = CA$, $c = AB$ for convenience. Let $M$ be the midpoint of segment $BC$. We claim that $\angle MAD=\angle DAF$.
$\textit{Proof}$. Since $AE$ is the angle bisector, it follows that $EB = EC$ and consequently $EM\perp BC$. Therefore, $M\in \gamma$. Now let $X = FD\cap \omega$. Since $\angle EFX=90^\ci... | // Block 1
size(175);
defaultpen(fontsize(10pt));
picture pic;
pair A,B,C,D,E,F,W;
B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(... | [] |
171 | Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$, $BC=7$, and $AC=3$. The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $\overline{DE}$. Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$. Then $AF^2 = ... | 2012 AIME II Problem 15 | Use the angle bisector theorem to find $CD=\tfrac{21}{8}$, $BD=\tfrac{35}{8}$, and use Stewart's Theorem to find $AD=\tfrac{15}{8}$. Use Power of Point $D$ to find $DE=\tfrac{49}{8}$, and so $AE=8$. Then use the Extended Law of Sine to find that the length of the circumradius of $\triangle ABC$ is $\tfrac{7\sqrt{3}}{3}... | // Block 1
size(175);
defaultpen(fontsize(9pt));
pair A,B,C,D,E,F,W;
B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D... | [] |
172 | A paper equilateral triangle $ABC$ has side length $12$. The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$. The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$, where $m$, $n$, and $p$ are positive ... | 2013 AIME I Problem 9 | Let $M$ and $N$ be the points on $\overline{AB}$ and $\overline{AC}$, respectively, where the paper is folded. Let $D$ be the point on $\overline{BC}$ where the folded $A$ touches it.
We have $AF=6\sqrt{3}$ and $FD=3$, so $AD=3\sqrt{13}$. Denote $\angle DAF = \theta$; we get $\cos\theta = 2\sqrt{3}/\sqrt{13}$.
In tr... | // Block 1
import cse5; size(8cm); pen tpen = defaultpen + 1.337;
real a = 39/5.0; real b = 39/7.0;
pair B = MP("B", (0,0), dir(200)); pair A = MP("A", 12*dir(60), dir(90)); pair C = MP("C", (12,0), dir(-20)); pair D = MP("D", (9,0), dir(-80)); pair Y = MP("Y", midpoint(A--D), dir(-50)); pair M = MP("M", extension(A,... | [] |
172 | A paper equilateral triangle $ABC$ has side length $12$. The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$. The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$, where $m$, $n$, and $p$ are positive ... | 2013 AIME I Problem 9 | We let the original position of $A$ be $A$, and the position of $A$ after folding be $D$. Also, we put the triangle on the coordinate plane such that $A=(0,0)$, $B=(-6,-6\sqrt3)$, $C=(6,-6\sqrt3)$, and $D=(3,-6\sqrt3)$.
Note that since $A$ is reflected over the fold line to $D$, the fold line is the perpendicular bi... | // Block 1
size(10cm);
pen tpen = defaultpen + 1.337;
real a = 39/5.0;
real b = 39/7.0;
pair B = MP("B", (0,0), dir(200));
pair A = (9,0);
pair C = MP("C", (12,0), dir(-20));
pair K = (6,10.392);
pair M = (a*B+(12-a)*K) / 12;
pair N = (b*C+(12-b)*K) / 12;
draw(B--M--N--C--cycle);
draw(M--A--N--cycle);
label("$D$", A, S... | [] |
172 | A paper equilateral triangle $ABC$ has side length $12$. The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$. The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$, where $m$, $n$, and $p$ are positive ... | 2013 AIME I Problem 9 | Thanks to Solution 1 for the diagram below:
We will use the notation already on the diagram, but our solution is slightly different.
We will only need M and N.
Let NC be length a, which implies NA be 12-a.
Also, AC = 3 because AB = 9
By the Law of Cosines on NCA,
$(12-a)^2=a^2+3^2-2(a)(3)(cos60)$
which sim... | // Block 1
import cse5; size(8cm); pen tpen = defaultpen + 1.337;
real a = 39/5.0; real b = 39/7.0;
pair B = MP("B", (0,0), dir(200)); pair A = MP("A", 12*dir(60), dir(90)); pair C = MP("C", (12,0), dir(-20)); pair D = MP("D", (9,0), dir(-80)); pair Y = MP("Y", midpoint(A--D), dir(-50)); pair M = MP("M", extension(A,... | [] |
172 | A paper equilateral triangle $ABC$ has side length $12$. The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$. The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$, where $m$, $n$, and $p$ are positive ... | 2013 AIME I Problem 9 | As shown in the diagram, let $MA'=x$. Since the triangle is equilateral, we have $BM=12-x$. Similarly, let $NA'=y$, so $NC=12-y$.
Because $\angle MBA=\angle NCA=60^\circ$, we may apply the Law of Cosines in the form $a^2+b^2-ab=c^2$. (Note that we could also find these side lengths through similar triangles as mention... | // Block 1
import cse5;
size(8cm);
pen tpen = defaultpen + 1.337;
real a = 39/5.0;
real b = 39/7.0;
pair B = (0,0);
pair Ap = (9,0);
pair C = (12,0);
pair K = (6, 10.392);
pair M = (a*B+(12-a)*K) / 12;
pair N = (b*C+(12-b)*K) / 12;
draw(B--M--N--C--cycle, tpen);
draw(M--Ap--N--cycle, tpen);
fill(M--Ap--N--cycle, mediu... | [] |
173 | In equilateral $\triangle ABC$ let points $D$ and $E$ trisect $\overline{BC}$. Then $\sin(\angle DAE)$ can be expressed in the form $\frac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is an integer that is not divisible by the square of any prime. Find $a+b+c$. | 2013 AIME II Problem 5 | Without loss of generality, assume the triangle sides have length 3. Then the trisected side is partitioned into segments of length 1, making your computation easier.
Let $M$ be the midpoint of $\overline{DE}$. Then $\Delta MCA$ is a 30-60-90 triangle with $MC = \dfrac{3}{2}$, $AC = 3$ and $AM = \dfrac{3\sqrt{3}}{2}$... | // Block 1
pair A = (1, sqrt(3)), B = (0, 0), C= (2, 0);
pair M = (1, 0);
pair D = (2/3, 0), E = (4/3, 0);
draw(A--B--C--cycle);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$M$", M, S);
label("$E$", E, S);
draw(A--D);
draw(A--M);
draw(A--E);
// Block 2
pair A = (1, sqrt(3)),... | [] |
174 | A hexagon that is inscribed in a circle has side lengths $22$, $22$, $20$, $22$, $22$, and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$, where $p$ and $q$ are positive integers. Find $p+q$. | 2013 AIME II Problem 8 | Solution 1
Let us call the hexagon $ABCDEF$, where $AB=CD=DE=AF=22$, and $BC=EF=20$.
We can just consider one half of the hexagon, $ABCD$, to make matters simpler.
Draw a line from the center of the circle, $O$, to the midpoint of $BC$, $X$. Now, draw a line from $O$ to the midpoint of $AB$, $Y$. Clearly, $\angle BXO... | // Block 1
import olympiad;
import math;
real a;
a=2*asin(11/(5+sqrt(267)));
pair A,B,C,D,E,F;
A=expi(pi); B=expi(pi-a); C=expi(a); D=expi(0); E=expi(-a); F=expi(pi+a);
draw(A--B--C--D--E--F--A--D);
dot(A); dot(B); dot(C); dot(D); dot(E); dot(F);
draw(unitcircle);
label("$A$",A,W);label("$B$",B,NW);label("$C$",C,NE);l... | [] |
174 | A hexagon that is inscribed in a circle has side lengths $22$, $22$, $20$, $22$, $22$, and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$, where $p$ and $q$ are positive integers. Find $p+q$. | 2013 AIME II Problem 8 | We know that $AD=x$ is a diameter, hence $ABD$ and $ACD$ are right triangles. Let $AB=BC=22$, and $CD=20.$ Hence, $ABD$ is a right triangle with legs $22,\sqrt{x^2-484},$ and hypotenuse, $x,$ and $ACD$ is a right triangle with legs $20, \sqrt{x^2-400},$ with hypotenuse $x$. By Ptolemy's we have \[22(x+20)=\sqrt{x^2-400... | // Block 1
import olympiad;
import math;
real a;
a=2*asin(11/(5+sqrt(267)));
pair A,B,C,D,E,F;
A=expi(pi); B=expi(pi-a); C=expi(a); D=expi(0); E=expi(-a); F=expi(pi+a);
draw(A--B--C--D--E--F--A--D);
draw(B--D);
draw(A--C);
dot(A); dot(B); dot(C); dot(D); dot(E); dot(F);
draw(unitcircle);
label("$A$",A,W);label("$B$",B... | [] |
175 | Given a circle of radius $\sqrt{13}$, let $A$ be a point at a distance $4 + \sqrt{13}$ from the center $O$ of the circle. Let $B$ be the point on the circle nearest to point $A$. A line passing through the point $A$ intersects the circle at points $K$ and $L$. The maximum possible area for $\triangle BKL$ can be writte... | 2013 AIME II Problem 10 | Now we put the figure in the Cartesian plane, let the center of the circle $O (0,0)$, then $B (\sqrt{13},0)$, and $A(4+\sqrt{13},0)$
The equation for Circle O is $x^2+y^2=13$, and let the slope of the line $AKL$ be $k$, then the equation for line $AKL$ is $y=k(x-4-\sqrt{13})$.
Then we get $(k^2+1)x^2-2k^2(4+\sqrt{13}... | // Block 1
import math;
import olympiad;
import graph;
pair A, B, K, L;
B = (sqrt(13), 0); A=(4+sqrt(13), 0);
dot(B);
dot(A);
draw(Circle((0,0), sqrt(13)));
label("$O$", (0,0), S);label("$B$", B, SW);label("$A$", A, S);
dot((0,0));
// Block 2
import math; import olympiad; import graph; pair A, B, K, L; B = (sqrt(13), ... | [] |
175 | Given a circle of radius $\sqrt{13}$, let $A$ be a point at a distance $4 + \sqrt{13}$ from the center $O$ of the circle. Let $B$ be the point on the circle nearest to point $A$. A line passing through the point $A$ intersects the circle at points $K$ and $L$. The maximum possible area for $\triangle BKL$ can be writte... | 2013 AIME II Problem 10 | Draw $OC$ perpendicular to $KL$ at $C$. Draw $BD$ perpendicular to $KL$ at $D$.
\[\frac{\triangle OKL}{\triangle BKL} = \frac{OC}{BD} = \frac{AO}{AB} = \frac{4+\sqrt{13}}{4}\]
Therefore, to maximize area of $\triangle BKL$, we need to maximize area of $\triangle OKL$.
\[\triangle OKL = \frac12 r^2 \sin{\angle KOL}\... | // Block 1
import math;
import olympiad;
import graph;
pair A, B, K, L;
B = (sqrt(13), 0); A=(4+sqrt(13), 0);
dot(B);
dot(A);
draw(Circle((0,0), sqrt(13)));
label("$O$", (0,0), S);label("$B$", B, SW);label("$A$", A, S);
dot((0,0));
// Block 2
import math; import olympiad; import graph; pair A, B, K, L; B = (sqrt(13), ... | [] |
176 | In $\triangle ABC$, $AC = BC$, and point $D$ is on $\overline{BC}$ so that $CD = 3\cdot BD$. Let $E$ be the midpoint of $\overline{AD}$. Given that $CE = \sqrt{7}$ and $BE = 3$, the area of $\triangle ABC$ can be expressed in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the ... | 2013 AIME II Problem 13 | Let $X$ be the foot of the altitude from $C$ with other points labelled as shown below.
Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. Th... | // Block 1
size(200);
pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7;
draw(A--B--C--cycle);draw(A--D^^B--L^^C--M);
label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE);label("$L$",L,NW);label("$M$",M,S);
pair X=foot(C,A,B), Y=foot(L,A,B);
pair EE=D/2;
label("$X$",X,S);label("$E$"... | [] |
176 | In $\triangle ABC$, $AC = BC$, and point $D$ is on $\overline{BC}$ so that $CD = 3\cdot BD$. Let $E$ be the midpoint of $\overline{AD}$. Given that $CE = \sqrt{7}$ and $BE = 3$, the area of $\triangle ABC$ can be expressed in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the ... | 2013 AIME II Problem 13 | Let $BD = x$. Then $CD = 3x$ and $AC = 4x$. Also, let $AE = ED = l$. Using Stewart's Theorem on $\bigtriangleup CEB$ gives us the equation $(x)(3x)(4x) + (4x)(l^2) = 27x + 7x$ or, after simplifying, $4l^2 = 34 - 12x^2$. We use Stewart's again on $\bigtriangleup CAD$: $(l)(l)(2l) + 7(2l) = (16x^2)(l) + (9x^2)(l)$, which... | // Block 1
size(200);
pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7;
draw(A--B--C--cycle);draw(A--D^^B--L^^C--M);
label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE);label("$L$",L,NW);label("$M$",M,S);
pair EE=D/2;
label("$\sqrt{7}$", C--EE, W); label("$x$", D--B, E); label("$3... | [] |
176 | In $\triangle ABC$, $AC = BC$, and point $D$ is on $\overline{BC}$ so that $CD = 3\cdot BD$. Let $E$ be the midpoint of $\overline{AD}$. Given that $CE = \sqrt{7}$ and $BE = 3$, the area of $\triangle ABC$ can be expressed in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the ... | 2013 AIME II Problem 13 | Let $AB = 2x$ and let $y = BD.$ Then $CD = 3y$ and $AC = 4y.$
By the Law of Cosines on triangle $ABC,$
\[\cos C = \frac{16y^2 + 16y^2 - 4x^2}{2 \cdot 4y \cdot 4y} = \frac{32y^2 - 4x^2}{32y^2} = \frac{8y^2 - x^2}{8y^2}.\]Then by the Law of Cosines on triangle $ACD,$
\begin{align*}
AD^2 &= 16y^2 + 9y^2 - 2 \cdot 4y \c... | // Block 1
unitsize(1.5 cm);
pair A, B, C, D, E;
A = (-sqrt(7),0);
B = (sqrt(7),0);
C = (0,3);
D = interp(B,C,1/4);
E = (A + D)/2;
draw(A--B--C--cycle);
draw(A--D);
draw(B--E--C);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("$D$", D, NE);
label("$E$", E, NW);
label("$2x$", (A + B)/2, S);
la... | [] |
177 | Let $w$ and $z$ be complex numbers such that $|w| = 1$ and $|z| = 10$. Let $\theta = \arg \left(\tfrac{w-z}{z}\right)$. The maximum possible value of $\tan^2 \theta$ can be written as $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. (Note that $\arg(w)$, for $w \neq 0$, denotes the ... | 2014 AIME I Problem 7 | Without the loss of generality one can let $z$ lie on the positive x axis and since $arg(\theta)$ is a measure of the angle if $z=10$ then $arg(\dfrac{w-z}{z})=arg(w-z)$ and we can see that the question is equivalent to having a triangle $OAB$ with sides $OA =10$ $AB=1$ and $OB=t$ and trying to maximize the angle $BOA$... | // Block 1
pair O = (0,0);
pair A = (100,0);
pair B = (80,30);
pair D = (sqrt(850),sqrt(850));
draw(A--B--O--cycle);
dotfactor = 3;
dot("$A$",A,dir(45));
dot("$B$",B,dir(45));
dot("$O$",O,dir(135));
dot("$ \theta$",O,(7,1.2));
label("$1$", ( A--B ));
label("$10$",(O--A));
label("$t$",(O--B));
// Block 2
pair O = (0,0)... | [] |
178 | A disk with radius $1$ is externally tangent to a disk with radius $5$. Let $A$ be the point where the disks are tangent, $C$ be the center of the smaller disk, and $E$ be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until t... | 2014 AIME I Problem 10 | Let $F$ be the new tangency point of the two disks. The smaller disk rolled along minor arc $\overset{\frown}{AF}$ on the larger disk.
Let $\alpha = \angle AEF$, in radians. The smaller disk must then have rolled along an arc of length $5\alpha$, since the larger disk has a radius of $5$. Since all of the points on m... | // Block 1
size(150);
pair a=(5,0),b=(2,3*sqrt(3)),c=(6,0),d=(3,3*sqrt(3)),e=(0,0);
draw(circle(e,5));
draw(circle(c,1));
draw(circle(d,1));
dot(a^^b^^c^^d^^e^^(5/2,5*sqrt(3)/2));
label("$A$",a,W,fontsize(9));
label("$B$",b,NW,fontsize(9));
label("$C$",c,E,fontsize(9));
label("$D$",d,E,fontsize(9));
label("$E$",e,SW,fo... | [] |
179 | On square $ABCD$, points $E,F,G$, and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$. Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$, and the areas of the quadrilaterals $AEPH, BFPE, CGP... | 2014 AIME I Problem 13 | Let $s$ be the side length of $ABCD$, let $Q$, and $R$ be the midpoints of $\overline{EG}$ and $\overline{FH}$, respectively, let $S$ be the foot of the perpendicular from $Q$ to $\overline{CD}$, let $T$ be the foot of the perpendicular from $R$ to $\overline{AD}$.
The fraction of the area of the square $ABCD$ which ... | size(150); defaultpen(fontsize(10pt)); pair A,B,C,D,E,F,Fp,G,Gp,H,O,I,J,R,S,T; A=dir(45*3); B=dir(-45*3); C=dir(-45); D=dir(45); O = origin; real theta=15; E=extension(A,B,O,dir(180+theta)); G=extension(C,D,O,dir(theta)); I=extension(A,D,O,dir(90+theta)); J=extension(B,C,O,dir(-90+theta)); H=(A+I)/2; F=H+(J-I); R=midp... | [] |
179 | On square $ABCD$, points $E,F,G$, and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$. Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$, and the areas of the quadrilaterals $AEPH, BFPE, CGP... | 2014 AIME I Problem 13 | Let $s$ be the side length of $ABCD$, let $[ABCD]=1360a$. Let $Q$ and $R$ be the midpoints of $\overline{EG}$ and $\overline{FH}$, respectively; because $269+411=275+405$, $Q$ is also the center of the square. Draw $\overline{IJ} \parallel \overline{HF}$ through $Q$, with $I$ on $\overline{AD}$, $J$ on $\overline{BC}$.... | // Block 1
size(150);
defaultpen(fontsize(9pt));
pair A,B,C,D,E,F,G,H,I,J,L,P,Q,R,S;
Q=MP("Q",origin,down); A=MP("A",(-1,1),dir(135)); B=MP("B",(-1,-1),dir(225)); C=MP("C",(1,-1),dir(-45)); D=MP("D",(1,1),dir(45)); real theta = 20; real shift=0.4; E=MP("E",extension(A,B,Q,dir(theta)),left); J=MP("J",extension(B,C,Q,dir... | [] |
179 | On square $ABCD$, points $E,F,G$, and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$. Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$, and the areas of the quadrilaterals $AEPH, BFPE, CGP... | 2014 AIME I Problem 13 | saw this at bottom of https://artofproblemsolving.com/community/c5h580667p3428922
Let $O$ be the center of the square. Then let $Q$ be on $HF$ susch that $OQ \perp CD$. Draw $H'F' \parallel HF$ through $O$. Let $s$ be the side length of the square, so that $s^2$ is the desired.
First, we note that as $w + z = x +... | // Block 1
size(200);
defaultpen(linewidth(0.8)+fontsize(10.6));
pair A = (0,sqrt(850));
pair B = (0,0);
pair C = (sqrt(850),0);
pair D = (sqrt(850),sqrt(850));
draw(A--B--C--D--cycle);
dotfactor = 3;
dot("$A$",A,dir(135));
dot("$B$",B,dir(215));
dot("$C$",C,dir(305));
dot("$D$",D,dir(45));
pair H = ((2sqrt(850)-sqrt(1... | [] |
180 | In $\triangle ABC$, $AB = 3$, $BC = 4$, and $CA = 5$. Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$, $\overline{BC}$ at $B$ and $D$, and $\overline{AC}$ at $F$ and $G$. Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4}$, length $DE=\frac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive inte... | 2014 AIME I Problem 15 | First we note that $\triangle DEF$ is an isosceles right triangle with hypotenuse $\overline{DE}$ the same as the diameter of $\omega$. We also note that $\triangle DGE \sim \triangle ABC$ since $\angle EGD$ is a right angle and the ratios of the sides are $3:4:5$.
From congruent arc intersections, we know that $\ang... | // Block 1
pair A = (0,3);
pair B = (0,0);
pair C = (4,0);
draw(A--B--C--cycle);
dotfactor = 3;
dot("$A$",A,dir(135));
dot("$B$",B,dir(215));
dot("$C$",C,dir(305));
pair D = (2.21, 0);
pair E = (0, 1.21);
pair F = (1.71, 1.71);
pair G = (2, 1.5);
dot("$D$",D,dir(270));
dot("$E$",E,dir(180));
dot("$F$",F,dir(90));
dot("... | [] |
181 | Arnold is studying the prevalence of three health risk factors, denoted by A, B, and C, within a population of men. For each of the three factors, the probability that a randomly selected man in the population has only this risk factor (and none of the others) is 0.1. For any two of the three factors, the probability t... | 2014 AIME II Problem 2 | We first assume a population of $100$ to facilitate solving. Then we simply organize the statistics given into a Venn diagram.
Let $x$ be the number of men with all three risk factors. Since "the probability that a randomly selected man has all three risk factors, given that he has A and B is $\frac{1}{3}$," we can... | // Block 1
pair A,B,C,D,E,F,G;
A=(0,55);
B=(60,55);
C=(60,0);
D=(0,0);
draw(A--B--C--D--A);
E=(30,35);
F=(20,20);
G=(40,20);
draw(circle(E,15));
draw(circle(F,15));
draw(circle(G,15));
draw("$A$",(30,52));
draw("$B$",(7,7));
draw("$C$",(53,7));
draw("100",(5,60));
draw("10",(30,40));
draw("10",(15,15));
draw("10",(45... | [] |
182 | A rectangle has sides of length $a$ and 36. A hinge is installed at each vertex of the rectangle, and at the midpoint of each side of length 36. The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with... | 2014 AIME II Problem 3 | When we squish the rectangle, the hexagon is composed of a rectangle and two isosceles triangles with side lengths 18, 18, and 24 as shown below.
By Heron's Formula, the area of each isosceles triangle is $\sqrt{(30)(12)(12)(6)}=\sqrt{180\times 12^2}=72\sqrt{5}$. So the area of both is $144\sqrt{5}$. From the rectan... | // Block 1
pair R,S,T,X,Y,Z;
dotfactor = 2;
unitsize(.1cm);
R = (12*2.236 +22,0);
S = (12*2.236 + 22 - 13.4164,12);
T = (12*2.236 + 22,24);
X = (12*4.472+ 22,24);
Y = (12*4.472+ 22 + 13.4164,12);
Z = (12*4.472+ 22,0);
draw(R--S--T--X--Y--Z--cycle);
draw(T--R,red);
draw(X--Z,red);
dot(" ",R,NW);
dot(" ",S,NW);
dot(" ",T... | [] |
183 | Circle $C$ with radius 2 has diameter $\overline{AB}$. Circle D is internally tangent to circle $C$ at $A$. Circle $E$ is internally tangent to circle $C$, externally tangent to circle $D$, and tangent to $\overline{AB}$. The radius of circle $D$ is three times the radius of circle $E$, and can be written in the form $... | 2014 AIME II Problem 8 | Using the diagram above, let the radius of $D$ be $3r$, and the radius of $E$ be $r$. Then, $EF=r$, and $CE=2-r$, so the Pythagorean theorem in $\triangle CEF$ gives $CF=\sqrt{4-4r}$. Also, $CD=CA-AD=2-3r$, so \[DF=DC+CF=2-3r+\sqrt{4-4r}.\] Noting that $DE=4r$, we can now use the Pythagorean theorem in $\triangle DEF$ ... | // Block 1
import graph; size(7.99cm);
real labelscalefactor = 0.5;
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
pen dotstyle = black;
real xmin = 4.087153740193288, xmax = 11.08175859031552, ymin = -4.938019122704778, ymax = 1.194137062512079;
draw(circle((7.780000000000009,-1.320000000000002), 2.0000... | [] |
184 | In $\triangle RED$, $\measuredangle DRE=75^{\circ}$ and $\measuredangle RED=45^{\circ}$. $RD=1$. Let $M$ be the midpoint of segment $\overline{RD}$. Point $C$ lies on side $\overline{ED}$ such that $\overline{RC}\perp\overline{EM}$. Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA=AR$. Then $AE=\fr... | 2014 AIME II Problem 11 | Let $P$ be the foot of the perpendicular from $A$ to $\overline{CR}$, so $\overline{AP}\parallel\overline{EM}$. Since triangle $ARC$ is isosceles, $P$ is the midpoint of $\overline{CR}$, and $\overline{PM}\parallel\overline{CD}$. Thus, $APME$ is a parallelogram and $AE = PM = \frac{CD}{2}$. We can then use coordinates.... | // Block 1
unitsize(8cm);
pair a, o, d, r, e, m, cm, c,p;
o =(0,0);
d = (0.5, 0);
r = (0,sqrt(3)/2);
e = (-sqrt(3)/2,0);
m = midpoint(d--r);
draw(e--m);
cm = foot(r, e, m);
draw(L(r, cm,1, 1));
c = IP(L(r, cm, 1, 1), e--d);
clip(r--d--e--cycle);
draw(r--d--e--cycle);
draw(rightanglemark(e, cm, c, 1.5));
a = -(4sqrt(3)... | [] |
184 | In $\triangle RED$, $\measuredangle DRE=75^{\circ}$ and $\measuredangle RED=45^{\circ}$. $RD=1$. Let $M$ be the midpoint of segment $\overline{RD}$. Point $C$ lies on side $\overline{ED}$ such that $\overline{RC}\perp\overline{EM}$. Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA=AR$. Then $AE=\fr... | 2014 AIME II Problem 11 | Let $P$ be the foot of the perpendicular from $A$ to $\overline{CR}$, so $\overline{AP}\parallel\overline{EM}$. Since $\triangle ARC$ is isosceles, $P$ is the midpoint of $\overline{CR}$, and by midpoint theorem $\overline{PM}\parallel\overline{CD}$. Thus, $APME$ is a parallelogram and therefore $AE = PM = \tfrac 12 C... | // Block 1
unitsize(8cm); pair a, d, r, e, m, cm, c,p;
d=origin; r=dir(60); e=extension(d,left,r,r+dir(75)*(d-r)); m = midpoint(d--r); cm = foot(r, e, m); c=extension(r,cm,d,e); p=midpoint(r--c); a=p+(e-m);
draw(e--m); draw(L(r, cm,1, 1)); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)... | [] |
185 | In $\triangle{ABC}, AB=10, \angle{A}=30^\circ$ , and $\angle{C=45^\circ}$. Let $H, D,$ and $M$ be points on the line $BC$ such that $AH\perp{BC}$, $\angle{BAD}=\angle{CAD}$, and $BM=CM$. Point $N$ is the midpoint of the segment $HM$, and point $P$ is on ray $AD$ such that $PN\perp{BC}$. Then $AP^2=\dfrac{m}{n}$, where ... | 2014 AIME II Problem 14 | Draw the $45-45-90 \triangle AHC$. Now, take the perpendicular bisector of $BC$ to intersect the circumcircle of $\triangle ABC$ and $AC$ at $F, L, G$ as shown, and denote $O$ to be the circumcenter of $\triangle ABC$. It is not difficult to see by angle chasing that $AHBGO$ is cyclic, namely with diameter $AB$. Then, ... | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(15cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black... | [] |
186 | 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 $\overline{AD}$ and $... | 2015 AIME I Problem 7 | 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 $FJGH=x$.
Now, 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 $BC$ in t... | // Block 1
pair A,B,C,D,E,F,G,H,J,K,L,M,N,P;
B=(0,0);
real m=7*sqrt(55)/5;
J=(m,0);
C=(7*m/2,0);
A=(0,7*m/2);
D=(7*m/2,7*m/2);
E=(A+D)/2;
H=(0,2m);
N=(0,2m+3*sqrt(55)/2);
G=foot(H,E,C);
F=foot(J,E,C);
draw(A--B--C--D--cycle);
draw(C--E);
draw(G--H--J--F);
pair X=foot(N,E,C);
M=extension(N,X,A,D);
K=foot(N,H,G);
L=foot(... | [] |
186 | 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 $\overline{AD}$ and $... | 2015 AIME I Problem 7 | We begin by denoting the length $ED$ $a$, giving us $DC = 2a$ and $EC = a\sqrt5$. Since angles $\angle DCE$ and $\angle FCJ$ are complementary, we have that $\triangle CDE \sim \triangle JFC$ (and similarly the rest of the triangles are $1-2-\sqrt5$ triangles). We let the sidelength of $FGHJ$ be $b$, giving us:
\[JC ... | // Block 1
pair A,B,C,D,E,F,G,H,J,K,L,M,N,P;
B=(0,0);
real m=7*sqrt(55)/5;
J=(m,0);
C=(7*m/2,0);
A=(0,7*m/2);
D=(7*m/2,7*m/2);
E=(A+D)/2;
H=(0,2m);
N=(0,2m+3*sqrt(55)/2);
G=foot(H,E,C);
F=foot(J,E,C);
draw(A--B--C--D--cycle);
draw(C--E);
draw(G--H--J--F);
pair X=foot(N,E,C);
M=extension(N,X,A,D);
K=foot(N,H,G);
L=foot(... | [] |
187 | In an isosceles trapezoid, the parallel bases have lengths $\log 3$ and $\log 192$, and the altitude to these bases has length $\log 16$. The perimeter of the trapezoid can be written in the form $\log 2^p 3^q$, where $p$ and $q$ are positive integers. Find $p + q$. | 2015 AIME II Problem 4 | Let $a=\log2$ and $b=\log3$ so that the base lengths are $\log3=b$ and $\log192=\log(3\cdot64)=\log3+\log\left(2^6\right)=6a+b$ and the altitudes are $\log16=4a$. Then we have the following picture:
Note that we have the two right triangles to the side; one of each of their bases is an altitude, which we know the l... | // Block 1
import graph;
unitsize(1cm);
draw((0,0)--(3,4)--(9,4)--(12,0)--cycle);
draw((3,4)--(3,0));draw((9,4)--(9,0));
label("$b$",(3,4)--(9,4),N);
label("$6a+b$",(0,0)--(12,0),S);
label("$b$",(0,0)--(12,0),N);
label("$4a$", (3,0)--(3,4),W);
label("$4a$", (9,0)--(9,4),E);
// Block 2
import graph; unitsize(1cm); dr... | [] |
188 | Triangle $ABC$ has side lengths $AB = 12$, $BC = 25$, and $CA = 17$. Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$, vertex $Q$ on $\overline{AC}$, and vertices $R$ and $S$ on $\overline{BC}$. In terms of the side length $PQ = \omega$, the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS) = \... | 2015 AIME II Problem 7 | Similar triangles can also solve the problem.
First, solve for the area of the triangle. $[ABC] = 90$. This can be done by Heron's Formula or placing an $8-15-17$ right triangle on $AC$ and solving. (The $8$ side would be collinear with line $AB$)
After finding the area, solve for the altitude to $BC$. Let $E$ be the... | // Block 1
unitsize(20);
pair A,B,C,E,F,P,Q,R,S;
A=(48/5,36/5);
B=(0,0);
C=(25,0);
E=(48/5,0);
F=(48/5,18/5);
P=(24/5,18/5);
Q=(173/10,18/5);
S=(24/5,0);
R=(173/10,0);
draw(A--B--C--cycle);
draw(P--Q);
draw(Q--R);
draw(R--S);
draw(S--P);
draw(A--E,dashed);
label("$A$",A,N);
label("$B$",B,SW);
label("$C$",C,SE);
label("... | [] |
189 | A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$. | 2015 AIME II Problem 9 | We can use the same method as in Solution 2 to find the side length of the equilateral triangle, which is $4\sqrt3$. From here, its area is
\[\dfrac{\bigl(4\sqrt3\bigr)^2\sqrt3}4=12\sqrt3.\]
The leg of the isosceles right triangle is $\dfrac{4\sqrt3}{\sqrt2}=2\sqrt6$, and the horizontal distance from the vertex to the ... | // Block 1
import olympiad;
pair V, T, B;
V = (-4, 0);
B = origin;
T = (0, 2*sqrt(2));
draw(V--B--T--cycle);
draw(rightanglemark(V, B, T));
label("Vertex", V, W);
label("Tip", T, N);
label("Base", B, SE);
label("$4$", V--B, S);
label("$2\sqrt6$", V--T, NW);
// Block 2
import olympiad; pair V, T, B; V = (-4, 0); B ... | [] |
190 | The circumcircle of acute $\triangle ABC$ has center $O$. The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ at $P$ and $Q$, respectively. Also $AB=5$, $BC=4$, $BQ=4.5$, and $BP=\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2015 AIME II Problem 11 | Solution 1
Call $M$ and $N$ the feet of the altitudes from $O$ to $BC$ and $AB$, respectively. Let $OB = r$ . Notice that $\triangle{OMB} \sim \triangle{QOB}$ because both are right triangles, and $\angle{OBQ} \cong \angle{OBM}$. By $\frac{MB}{BO}=\frac{BO}{BQ}$, $MB = r\left(\frac{r}{4.5}\right) = \frac{r^2}{4.5}$. Ho... | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.4) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point ... | [] |
191 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$, respectively, and are externally tangent at point $A$. Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and in... | 2015 AIME II Problem 15 | Call $O_1$ and $O_2$ the centers of circles $\mathcal{P}$ and $\mathcal{Q}$, respectively, and extend $CB$ and $O_2O_1$ to meet at point $N$. Call $K$ and $L$ the feet of the altitudes from $B$ to $O_1N$ and $C$ to $O_2N$, respectively. Using the fact that $\triangle{O_1BN} \sim \triangle{O_2CN}$ and setting $NO_1 = k... | // Block 1
unitsize(35);
draw(Circle((-1,0),1));
draw(Circle((4,0),4));
pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y;
A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A);
label("$A$",A,NE);label("$O_1... | [] |
191 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$, respectively, and are externally tangent at point $A$. Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and in... | 2015 AIME II Problem 15 | Consider the homothety that takes triangle BDA onto CXY on the big circle, as plotted. Some hidden congruence angles are revealed which help reduce computation complexity. Just some angle chasing and straight forward trigs. Because $AE=XY$ and $AE \parallel XY$, $XYE$ is right angle.
First, $\frac{NO_1}{NO_1+5} = \f... | // Block 1
unitsize(35);
draw(Circle((-1,0),1));
draw(Circle((4,0),4));
pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y;
A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A);
label("$A$",A,NE);label("$O_1... | [] |
191 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$, respectively, and are externally tangent at point $A$. Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and in... | 2015 AIME II Problem 15 | Let the centers of circles $\mathcal{P}$ and $\mathcal{Q}$ be $P$ and $Q$, respectively. Let $F$ be the point on $\overline{QC}$ such that $\overline{PF}$ is parallel to $\overline{BC}$. We find that $BC=PF=\sqrt{(4+1)^{2}-(4-1)^2}=4$.
Now, we define $A$ to be $(0,0)$. It follows that:
\begin{align}
P &= \left(-\frac... | // Block 1
import graph;
size(300);
// This diagram was brought to you by ChatGPT!
// Define points
pair P = (0,0);
pair B = (0,-1);
pair C = (4,-1);
pair F = (4,0);
pair Q = (4,3);
pair A = (4/5, 3/5);
// Draw rectangle PBCF
draw(P--B--C--F--cycle);
// Draw right triangle PQF
draw(P--Q--F--P);
// Circles
draw(c... | [] |
192 | In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$. The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$. If $LI=2$ and $LD=3$, then $IC=\tfrac{p}{q}$, where $p$ and $q$ are relatively ... | 2016 AIME I Problem 6 | Suppose we label the angles as shown below.
As $\angle BCD$ and $\angle BAD$ intercept the same arc, we know that $\angle BAD=\gamma$. Similarly, $\angle ABD=\gamma$. Also, using $\triangle ICA$, we find $\angle CIA=180-\alpha-\gamma$. Therefore, $\angle AID=\alpha+\gamma=\angle DAI$, so $\triangle AID$ must be isosce... | // Block 1
size(150);
import olympiad;
real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6;
pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2));
pair C=intersectionpoints(circle(A,b),circle(B,a))[0];
pair I=incenter(A,B,C);
pair L=extension(C,D,A,B);
dot(I^^A^^B^^C^^D);
draw(C--D);
path midangle(pair d,pair e,pair f) {... | [] |
192 | In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$. The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$. If $LI=2$ and $LD=3$, then $IC=\tfrac{p}{q}$, where $p$ and $q$ are relatively ... | 2016 AIME I Problem 6 | Connect $D$ to $A$ and $D$ to $B$ to form quadrilateral $ACBD$. Since quadrilateral $ACBD$ is cyclic, we can apply Ptolemy's Theorem on the quadrilateral.
Denote the length of $BD$ and $AD$ as $z$ (they must be congruent, as $\angle ABD$ and $\angle DAB$ are both inscribed in arcs that have the same degree measure due... | // Block 1
size(150);
import olympiad;
real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6;
pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2));
pair C=intersectionpoints(circle(A,b),circle(B,a))[0];
pair I=incenter(A,B,C);
pair L=extension(C,D,A,B);
dot(I^^A^^B^^C^^D);
draw(C--D);
path midangle(pair d,pair e,pair f) {... | [] |
193 | Triangle $ABC$ has $AB=40,AC=31,$ and $\sin{A}=\frac{1}{5}$. This triangle is inscribed in rectangle $AQRS$ with $B$ on $\overline{QR}$ and $C$ on $\overline{RS}$. Find the maximum possible area of $AQRS$. | 2016 AIME I Problem 9 | We start by drawing a diagram;
We know that $\sin \alpha = \frac{1}{5}$. Since $\sin \alpha = \cos (90- \alpha)$,
\[\cos (90- \alpha) = \frac{1}{5} \implies \cos (\beta + \gamma) = \frac{1}{5}\]
Using our angle sum identities, we expand this to $\cos \beta \cdot \cos \gamma - \sin \beta \cdot \sin \gamma = \frac{... | // Block 1
size(400);
import olympiad;
import geometry;
pair A = (0, 20) ,B=(30,10) ,C=(15,0), Q=(30,20) ,R=(30,0), S=(0,0);
draw(A--B--C--cycle);
draw(A--Q);
draw(Q--R);
draw(R--S);
draw(S--A);
label("$A$", A, W);
label("$B$", B, E);
label("$C$", C, N);
label("$Q$", Q, E);
label("$R$", R, E);
label("$S$", S, W);
... | [] |
194 | Centered at each lattice point in the coordinate plane are a circle radius $\frac{1}{10}$ and a square with sides of length $\frac{1}{5}$ whose sides are parallel to the coordinate axes. The line segment from $(0,0)$ to $(1001, 429)$ intersects $m$ of the squares and $n$ of the circles. Find $m + n$. | 2016 AIME I Problem 14 | See if you can solve the problem with the following.
Solution to Solution 2
This is mostly a clarification to Solution 1, but let's take the diagram for the origin to $(7,3)$. We have the origin circle and square intersected, then two squares, then the circle and square at $(7,3)$. If we take the circle and square a... | // Block 1
size(12cm);draw((0,0)--(7,3));draw(box((0,0),(7,3)),dotted);
for(int i=0;i<8;++i)for(int j=0; j<4; ++j){dot((i,j),linewidth(1));draw(box((i-.1,j-.1),(i+.1,j+.1)),linewidth(.5));draw(circle((i,j),.1),linewidth(.5));}
// Block 2
size(12cm);draw((0,0)--(7,3));draw(box((0,0),(7,3)),dotted); for(int i=0;i<8;++i)f... | [] |
195 | Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$. Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$, respectively, with line $AB$ closer to point $X$ than to $Y$. Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at... | 2016 AIME I Problem 15 | First, we note that as $\triangle XDY$ and $\triangle XYC$ have bases along the same line, $\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{DY}{YC}$. We can also find the ratio of their areas using the circumradius area formula. If $R_1$ is the radius of $\omega_1$ and if $R_2$ is the radius of $\omega_2$, then
\[\frac{... | // Block 1
size(200);
import olympiad;
real R1=45,R2=67*R1/37;
real m1=sqrt(R1^2-23.5^2);
real m2=sqrt(R2^2-23.5^2);
pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5);
draw(circle(o1,R1));
draw(circle(o2,R2));
pair q=(-R1/(R2-R1)*o2.x,0);
pair a=tangent(q,o1,R1,2);
pair b=tangent(q,o2,R2,2);
pair d=intersectionpoints... | [] |
195 | Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$. Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$, respectively, with line $AB$ closer to point $X$ than to $Y$. Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at... | 2016 AIME I Problem 15 | First of all, since quadrilaterals $ADYX$ and $XYCB$ are cyclic, we can let $\angle DAX = \angle XYC = \theta$, and $\angle XYD = \angle CBX = 180 - \theta$, due to the properties of cyclic quadrilaterals. In addition, let $\angle BAX = x$ and $\angle ABX = y$. Thus, $\angle ADX = \angle AYX = x$ and $\angle XYB = \ang... | // Block 1
size(9cm);
import olympiad;
real R1=45,R2=67*R1/37;
real m1=sqrt(R1^2-23.5^2);
real m2=sqrt(R2^2-23.5^2);
pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5);
draw(circle(o1,R1));
draw(circle(o2,R2));
pair q=(-R1/(R2-R1)*o2.x,0);
pair a=tangent(q,o1,R1,2);
pair b=tangent(q,o2,R2,2);
pair d=intersectionpoints... | [] |
196 | Triangle $ABC_0$ has a right angle at $C_0$. Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$. Let $C_1$ be the foot of the altitude to $\overline{AB}$, and for $n \geq 2$, let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$. The sum $\... | 2016 AIME II Problem 5 | Let $a = BC_0$, $b = AC_0$, and $c = AB$.
Note that the total length of the red segments in the figure above is equal to the length of the blue segment times $\frac{a+c}{b}$.
The desired sum is equal to the total length of the infinite path $C_0 C_1 C_2 C_3 \cdots$, shown in red in the figure below.
Since each of the ... | // Block 1
size(10cm);
// Setup
pair A, B;
pair C0, C1, C2, C3, C4, C5, C6, C7, C8;
A = (5, 0);
B = (0, 3);
C0 = (0, 0);
C1 = foot(C0, A, B);
C2 = foot(C1, C0, B);
C3 = foot(C2, C1, B);
C4 = foot(C3, C2, B);
C5 = foot(C4, C3, B);
C6 = foot(C5, C4, B);
C7 = foot(C6, C5, B);
C8 = foot(C7, C6, B);
// Labels
label("$A$",... | [] |
197 | Squares $ABCD$ and $EFGH$ have a common center and $\overline{AB} || \overline{EF}$. The area of $ABCD$ is 2016, and the area of $EFGH$ is a smaller positive integer. Square $IJKL$ is constructed so that each of its vertices lies on a side of $ABCD$ and each vertex of $EFGH$ lies on a side of $IJKL$. Find the differenc... | 2016 AIME II Problem 7 | Letting $AI=a$ and $IB=b$, we have \[IJ^{2}=a^{2}+b^{2} \geq 1008\] by AM-GM inequality. Also, since $EFGH||ABCD$, the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and $2$ adjacent vertices of a smaller square are similar. Therefore, the areas form a geo... | // Block 1
pair A,B,C,D,E,F,G,H,I,J,K,L;
A=(0,0);
B=(2016,0);
C=(2016,2016);
D=(0,2016);
I=(1008,0);
J=(2016,1008);
K=(1008,2016);
L=(0,1008);
E=(504,504);
F=(1512,504);
G=(1512,1512);
H=(504,1512);
draw(A--B--C--D--A);
draw(I--J--K--L--I);
draw(E--F--G--H--E);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,NE);
l... | [] |
198 | Triangle $ABC$ is inscribed in circle $\omega$. Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$. Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$, then $ST=\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $... | 2016 AIME II Problem 10 | Let $\angle ACP=\alpha$, $\angle PCQ=\beta$, and $\angle QCB=\gamma$. Note that since $\triangle ACQ\sim\triangle TBQ$ we have $\tfrac{AC}{CQ}=\tfrac56$, so by the Ratio Lemma \[\dfrac{AP}{PQ}=\dfrac{AC}{CQ}\cdot\dfrac{\sin\alpha}{\sin\beta}\quad\implies\quad \dfrac{\sin\alpha}{\sin\beta}=\dfrac{24}{15}.\]Similarly, we... | // Block 1
import cse5;
pathpen = black; pointpen = black;
pointfontsize = 9;
size(8cm);
pair A = origin, B = (13,0), P = (4,0), Q = (7,0),
T = B + 5 dir(220), C = IP(circumcircle(A,B,T),Line(T,Q,-0.1,10)),
S = IP(circumcircle(A,B,C),Line(C,P,-0.1,10));
Drawing(A--B--C--cycle);
D(circumcircle(A,B,C),rgb(0,0.6,1));
... | [] |
199 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color. | 2016 AIME II Problem 12 | Let's number the regions $1,2,\dots 6$. Suppose we color regions $1,2,3$. Then, how many ways are there to color $4,5,6$?
Note: the numbers are numbered as shown:
$\textbf{Case 1:}$ The colors of $1,2,3$ are $BAB$, in that order.
Then the colors of $6,5,4$ can be $AXA$, $AXC$, $CXA$, $CXC$, or $CXD$ in that order,... | // Block 1
draw(Circle((0,0), 4));
draw(Circle((0,0), 3));
draw((0,4)--(0,3));
draw((0,-4)--(0,-3));
draw((-2.598, 1.5)--(-3.4641, 2));
draw((-2.598, -1.5)--(-3.4641, -2));
draw((2.598, -1.5)--(3.4641, -2));
draw((2.598, 1.5)--(3.4641, 2));
label("1",(-3.5,0));
label("2",(-1.6,3));
label("3",(1.6,3));
label("4",(3.5,0)... | [] |
200 | Beatrix is going to place six rooks on a $6 \times 6$ chessboard where both the rows and columns are labeled $1$ to $6$; the rooks are placed so that no two rooks are in the same row or the same column. The $value$ of a square is the sum of its row number and column number. The $score$ of an arrangement of rooks is the... | 2016 AIME II Problem 13 | If the score is $n+1$, then one of the rooks must appear in the $n$th antidiagonal, and this is the first antidiagonal in which a rook can appear. To demonstrate this, we draw the following diagram when $n=4$.
We first count the number of arrangements that avoid the squares above the $n$th diagonal, and then we subtr... | // Block 1
for (int i=0;i<7;++i) {draw((0,10*i)--(60,10*i));draw((10*i,0)--(10*i,60));}
path x=(1,1)--(9,9),y=(1,9)--(9,1);
for (int i=0;i<3;++i) {for (int j=3+i;j<6;++j) {draw(shift(10*i,10*j)*x,linewidth(1));draw(shift(10*i,10*j)*y,linewidth(1));}}
for (int i=0;i<4;++i) {filldraw((10*i,20+10*i)--(10*i+10,20+10*i)--(1... | [] |
201 | Equilateral $\triangle ABC$ has side length $600$. Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$, and $QA=QB=QC$, and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). Th... | 2016 AIME II Problem 14 | To make numbers more feasible, we'll scale everything down by a factor of $100$ so that $\overline{AB}=\overline{BC}=\overline{AC}=6$. We should also note that $P$ and $Q$ must lie on the line that is perpendicular to the plane of $ABC$ and also passes through the circumcenter of $ABC$ (due to $P$ and $Q$ being equidis... | pair C, D, I, P, Q, O; D=(0,0); C=(5.196152,0); P=(1.732051,7.37228); I=(1.732051,0); Q=(1.732051,-1.62772); O=(1.732051,2.87228); draw(C--Q--D--P--cycle); draw(C--D, dashed); draw(P--Q, dotted); draw(O--C, dotted); label("$C$", C, E); label("$D$", D, W); label("$I$", I, NW); label("$P$", P, N); label("$Q$", Q, S); l... | [] |
201 | Equilateral $\triangle ABC$ has side length $600$. Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$, and $QA=QB=QC$, and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). Th... | 2016 AIME II Problem 14 | Let $M$ be the midpoint of $\overline{AB}$ and $X$ the center of $\triangle ABC$. Then \[P, O, Q, M, X, C\] all lie in the same vertical plane. We can make the following observations:
The equilateral triangle has side length $600$, so $MC=300\sqrt{3}$ and $X$ divides $MC$ so that $MX=100\sqrt{3}$ and $XC=200\sqrt{3}$;... | // Block 1
unitsize(20);
pair P = (0, 12);
pair Q = (0, -3);
pair O = (P+Q)/2;
pair M = (-3, 0);
pair X = (0, 0);
pair C = (6, 0);
draw(P--O--Q);
draw(M--X--C);
draw(P--M--Q, blue);
draw(Q--C--P);
draw(circle((0, 4.5), 7.5));
label("$P$", P, N);
label("$Q$", Q, S);
label("$O$", O, E);
dot(O);
label("$M$", M, W);
label(... | [] |
202 | Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$. Let $O$ and $P$ be two points on the plane with $OP = 200$. Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\angle... | 2017 AIME I Problem 8 | Let $QR=x.$ Since we are given many angles in the problem, we can compute the lengths of some of the lines in terms of trigonometric functions: $OQ = 200 \cos a, PQ = 200 \sin a, PR = 200 \sin b, OR = 200 \cos b.$ Now observe that quadrilateral $OQRP$ is a cyclic quadrilateral. Thus, we are able to apply Ptolemy's Theo... | // Block 1
pair O, P, Q, R;
draw(circle(O, 10));
O = (10, 0);
P = (-10, 0);
Q = (10*cos(pi/3), 10*sin(pi/3));
R = (10*cos(5*pi/6), 10*sin(5*pi/6));
dot(Q);
dot(O);
dot(P);
dot(R);
draw(P--O--Q--P--R--O);
draw(Q--R, red);
label("$O$", O, 2*E);
label("$P$", P, 2*W);
label("$Q$", Q, NE);
label("$R$", R, NW);
label("$200$"... | [] |
203 | Let $z_1=18+83i,~z_2=18+39i,$ and $z_3=78+99i,$ where $i=\sqrt{-1}.$ Let $z$ be the unique complex number with the properties that $\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}$ is a real number and the imaginary part of $z$ is the greatest possible. Find the real part of $z$. | 2017 AIME I Problem 10 | This problem is pretty obvious how to bash, and indeed many of the solutions below explain how to do that. But there’s no fun in that, and let’s see if we can come up with a slicker solution that will be more enjoyable.
Instead of thinking of complex numbers as purely a real plus a constant times $i$, let’s graph them... | // Block 1
size(1200,300);
real xMin = 0;
real xMax = 100;
real yMin = 0;
real yMax = 120;
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("$R$",(xMax,0),(2,0));
label("$Im$",(0,yMax),(0,2));
pair A,B,C,D;
A = (18,83); B = (18,39); C = (78,... | [] |
204 | A triangle has vertices $A(0,0)$, $B(12,0)$, and $C(8,10)$. The probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to either vertex $A$ or vertex $C$ can be written as $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. | 2017 AIME II Problem 3 | The set of all points closer to point $B$ than to point $A$ lie to the right of the perpendicular bisector of $AB$ (line $PZ$ in the diagram), and the set of all points closer to point $B$ than to point $C$ lie below the perpendicular bisector of $BC$ (line $PX$ in the diagram). Therefore, the set of points inside the ... | // Block 1
pair A,B,C,D,X,Z,P;
A=(0,0); B=(12,0); C=(8,10); X=(10,5); Z=(6,0); P=(6,3.4);
fill(B--X--P--Z--cycle,lightgray);
draw(A--B--C--cycle);
dot(A);
label("$A$",A,SW);
dot(B);
label("$B$",B,SE);
dot(C);
label("$C$",C,N);
draw(X--P,dashed);
draw(Z--P,dashed);
dot(X);
label("$X$",X,NE);
dot(Z);
label("$Z$",Z,S);
do... | [] |
204 | A triangle has vertices $A(0,0)$, $B(12,0)$, and $C(8,10)$. The probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to either vertex $A$ or vertex $C$ can be written as $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. | 2017 AIME II Problem 3 | Since we know the coordinates of all three vertices of the triangle, we can find the side lengths: $AB=12$, $AC=2\sqrt{41}$, and $BC=2\sqrt{29}$. We notice that the point where the three distances are the same is the circumcenter - so we use one of the triangle area formulas to find the circumradius, since we know what... | // Block 1
draw((0,0)--(12,0)--(8,10)--(0,0));
draw((6,0)--(6,3.4)--(10,5));
draw((6,3.4)--(4,5));
label("$A$", (0,0), SW);
label("$B$", (12,0), SE);
label("$C$", (8, 10), N);
label("$P$", (6, 3.4), NNE);
label("$R$", (10, 5), NE);
label("$S$", (6, 0), S);
label("$T$", (4, 5), NW);
// Block 2
draw((0,0)--(12,0)--(8,10)... | [] |
205 | Find the number of integer values of $k$ in the closed interval $[-500,500]$ for which the equation $\log(kx)=2\log(x+2)$ has exactly one real solution. | 2017 AIME II Problem 7 | Note the equation $\log(kx)=2\log(x+2)$ is valid for $kx>0$ and $x>-2$.
Now, we simplify the given expression:
$\log(kx)=2\log(x+2)=\log((x+2)^2) \implies kx=(x+2)^2$. The last equation is derived by taking away the outside logs from the previous equation.
Because $(x+2)^2$ is always non-negative, $kx$ must also be... | // Block 1
Label f;
f.p=fontsize(5);
xaxis(-3,3,Ticks(f,1.0));
yaxis(-3,26,Ticks(f,1.0));
real f(real x){return (x+2)^2;}
real g(real x){return x*-1;}
real h(real x){return x*-2;}
real i(real x){return x*-3;}
real j(real x){return x*8;}
draw(graph(f,-2,3),green);
draw(graph(g,-2,2),red);
draw(graph(h,-2,1),red);
draw... | [] |
206 | Rectangle $ABCD$ has side lengths $AB=84$ and $AD=42$. Point $M$ is the midpoint of $\overline{AD}$, point $N$ is the trisection point of $\overline{AB}$ closer to $A$, and point $O$ is the intersection of $\overline{CM}$ and $\overline{DN}$. Point $P$ lies on the quadrilateral $BCON$, and $\overline{BP}$ bisects the a... | 2017 AIME II Problem 10 | Impose a coordinate system on the diagram where point $D$ is the origin. Therefore $A=(0,42)$, $B=(84,42)$, $C=(84,0)$, and $D=(0,0)$. Because $M$ is a midpoint and $N$ is a trisection point, $M=(0,21)$ and $N=(28,42)$. The equation for line $DN$ is $y=\frac{3}{2}x$ and the equation for line $CM$ is $\frac{1}{84}x+\fra... | // Block 1
pair A,B,C,D,M,n,O,P;
A=(0,42);B=(84,42);C=(84,0);D=(0,0);M=(0,21);n=(28,42);O=(12,18);P=(32,13);
fill(C--D--P--cycle,blue);
draw(A--B--C--D--cycle);
draw(C--M);
draw(D--n);
draw(B--P);
draw(D--P);
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
label("$M$",M,W);
label("$N$",n,N);... | [] |
207 | Circle $C_0$ has radius $1$, and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r<1$ and is internally tangent to $C_0$ at point $A_0$. Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\circ}$ counterclockwise from $A_0$ on $C_1$. Circle $C_2$ has radius $r^2$ and is internally tangent... | 2017 AIME II Problem 12 | Using the invariance again as in Solution 3, assume $B$ is $d$ away from the origin. The locus of possible points is a circle with radius $d$. Consider the following diagram.
Let the distance from $B$ to $(1,0)$ be $x$. As $B$ is invariant, $x = r(BB' + x) \implies x = r\frac{d\sqrt{2}}{1-r}$. Then by Power of a Poin... | // Block 1
size(7cm);
draw(circle((0,0), 49/61));
draw((0,0)--(0.790110185, 0.144853534));
draw((0,0)--(-0.144853534, 0.790110185));
draw((-0.144853534, 0.790110185)--(1,0));
draw((0,0)--(1,0));
draw(rightanglemark((-0.144853534, 0.790110185), (0,0), (0.790110185, 0.144853534), 3));
label("$O$", (0,0), SW);
label("$(1... | [] |
208 | A $10\times10\times10$ grid of points consists of all points in space of the form $(i,j,k)$, where $i$, $j$, and $k$ are integers between $1$ and $10$, inclusive. Find the number of different lines that contain exactly $8$ of these points. | 2017 AIME II Problem 14 | A line through the point $(a,b,c)$ must contain $(a \pm 1, b \pm 1, c \pm 1)$ on it as well, as otherwise, the line would not pass through more than 5 points. This corresponds to the 4 diagonals of the cube. With this information, we proceed with casework:
$\textbf{Case 1}:$ The lines are not parallel to the faces
We... | // Block 1
import graph;
size(8cm);
// This diagram was brought to you by ChatGPT!!!
// Settings for dots
real dotSize = 5pt;
// Ring 1: 1 Point (Black)
dot((0,0), black + dotSize);
// Ring 2: 6 Points (Blue)
// Angles: 30, 90, 150, 210, 270, 330
for(int i=0; i<6; ++i) {
dot(dir(30 + 60*i), blue + dotSize);
}
//... | [] |
209 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$. Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$. Then $AD$ can be expressed in the form $\dfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find ... | 2018 AIME I Problem 4 | We draw the altitude from $B$ to $\overline{AC}$ to get point $F$. We notice that the triangle's height from $A$ to $\overline{BC}$ is 8 because it is a $3-4-5$ Right Triangle. To find the length of $\overline{BF}$, we let $h$ represent $\overline{BF}$ and set up an equation by finding two ways to express the area. The... | // Block 1
import olympiad;
import cse5;
unitsize(10mm);
pathpen=black;
dotfactor=3;
pair B = (0,0), A = (6,8), C = (12,0), D = intersectionpoints(circle(A,250/39),A--B)[0], E = intersectionpoints(circle(D,250/39),A--C)[0], F=intersectionpoints(circle(B,9.6),A--C)[0], G=A/2+E/2;
pair[] dotted = {A,B,C,D,E,F,G};
D(A--... | [] |
210 | Suppose that $x$, $y$, and $z$ are complex numbers such that $xy = -80 - 320i$, $yz = 60$, and $zx = -96 + 24i$, where $i$ $=$ $\sqrt{-1}$. Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$. Find $a^2 + b^2$. | 2018 AIME II Problem 5 | First we evaluate the magnitudes. $|xy|=80\sqrt{17}$, $|yz|=60$, and $|zx|=24\sqrt{17}$. Therefore, $|x^2y^2z^2|=17\cdot80\cdot60\cdot24$, or $|xyz|=240\sqrt{34}$. Divide to find that $|z|=3\sqrt{2}$, $|x|=4\sqrt{34}$, and $|y|=10\sqrt{2}$.
This allows us to see that the argument of $y$ is $\frac{\pi}{4}$, and the arg... | // Block 1
draw((0,0)--(4,0));
dot((4,0),red);
draw((0,0)--(-4,0));
draw((0,0)--(0,-4));
draw((0,0)--(-4,1));
dot((-4,1),red);
draw((0,0)--(-1,-4));
dot((-1,-4),red);
draw((0,0)--(4,4),red);
draw((0,0)--(4,-4),red);
// Block 2
draw((0,0)--(4,0)); dot((4,0),red); draw((0,0)--(-4,0)); draw((0,0)--(0,-4)); draw((0,0)--(-4... | [] |
211 | A frog is positioned at the origin of the coordinate plane. From the point $(x, y)$, the frog can jump to any of the points $(x + 1, y)$, $(x + 2, y)$, $(x, y + 1)$, or $(x, y + 2)$. Find the number of distinct sequences of jumps in which the frog begins at $(0, 0)$ and ends at $(4, 4)$. | 2018 AIME II Problem 8 | We solve this problem by working backwards. Notice, the only points the frog can be on to jump to $(4,4)$ in one move are $(2,4),(3,4),(4,2),$ and $(4,3)$. This applies to any other point, thus we can work our way from $(0,0)$ to $(4,4)$, recording down the number of ways to get to each point recursively.
$(0,0): 1$
... | // Block 1
import graph;
add(shift(0,0)*grid(4,4));
label((0,0), "1", SW);
label((1,0), "1", SW);
label((2,0), "2", SW);
label((3,0), "3", SW);
label((4,0), "5", SW);
label((0,1), "1", SW);
label((1,1), "2", SW);
label((2,1), "5", SW);
label((3,1), "10", SW);
label((4,1), "20", SW);
label((0,2), "2", SW);
label((1,2)... | [] |
212 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problem 6 | NOTE: this solution is wrong. The equation is correct due to similar triangles as described in solution 8, not PoP.
Because $\angle KLN = \angle KMN = 90^{\circ}$, $KLMN$ is a cyclic quadrilateral. (THE FOLLOWING SENTENCE IS WRONG) Hence, by Power of Point, \[KO\cdot KM = KL^2 \implies KM=\dfrac{28^2}{8}=98 \implies... | // Block 1
size(250);
real h = sqrt(98^2+65^2);
real l = sqrt(h^2-28^2);
pair K = (0,0);
pair N = (h, 0);
pair M = ((98^2)/h, (98*65)/h);
pair L = ((28^2)/h, (28*l)/h);
pair P = ((28^2)/h, 0);
pair O = ((28^2)/h, (8*65)/h);
draw(K--L--N);
draw(K--M--N--cycle);
draw(L--M);
label("K", K, SW);
label("L", L, NW);
label("M"... | [] |
212 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problem 6 | First, let $P$ be the intersection of $LO$ and $KN$ as shown above. Note that $m\angle KPL = 90^{\circ}$ as given in the problem. Since $\angle KPL \cong \angle KLN$ and $\angle PKL \cong \angle LKN$, $\triangle PKL \sim \triangle LKN$ by AA similarity. Similarly, $\triangle KMN \sim \triangle KPO$. Using these similar... | // Block 1
size(250);
real h = sqrt(98^2+65^2);
real l = sqrt(h^2-28^2);
pair K = (0,0);
pair N = (h, 0);
pair M = ((98^2)/h, (98*65)/h);
pair L = ((28^2)/h, (28*l)/h);
pair P = ((28^2)/h, 0);
pair O = ((28^2)/h, (8*65)/h);
draw(K--L--N);
draw(K--M--N--cycle);
draw(L--M);
label("K", K, SW);
label("L", L, NW);
label("M"... | [] |
212 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problem 6 | Notice that $KLMN$ is inscribed in the circle with diameter $\overline{KN}$ and $XOMN$ is inscribed in the circle with diameter $\overline{ON}$. Furthermore, $(XLN)$ is tangent to $\overline{KL}$. Then, \[KO\cdot KM=KX\cdot KN=KL^2\implies KM=\frac{28^2}{8}=98,\]and $MO=KM-KO=\boxed{090}$.
(Solution by TheUltimate123)... | // Block 1
size(8cm);
pair K, L, M, NN, X, O;
K=(-sqrt(98^2+65^2)/2, 0);
NN=(sqrt(98^2+65^2)/2, 0);
L=sqrt(98^2+65^2)/2*dir(180-2*aSin(28/sqrt(98^2+65^2)));
M=sqrt(98^2+65^2)/2*dir(2*aSin(65/sqrt(98^2+65^2)));
X=foot(L, K, NN);
O=extension(L, X, K, M);
draw(K -- L -- M -- NN -- K -- M); draw(L -- NN); draw(arc((K+NN)/2... | [] |
212 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problem 6 | (Diagram by vedadehhc)
Call the base of the altitude from $L$ to $NK$ point $P$. Let $PO=x$. Now, we have that $KP=\sqrt{64-x^2}$ by the Pythagorean Theorem. Once again by Pythagorean, $LO=\sqrt{720+x^2}-x$. Using Power of a Point, we have
\[(KO)(OM)=(LO)(OQ)\] ($Q$ is the intersection of $OL$ with the circle $\neq ... | // Block 1
size(250);
real h = sqrt(98^2+65^2);
real l = sqrt(h^2-28^2);
pair K = (0,0);
pair N = (h, 0);
pair M = ((98^2)/h, (98*65)/h);
pair L = ((28^2)/h, (28*l)/h);
pair P = ((28^2)/h, 0);
pair O = ((28^2)/h, (8*65)/h);
draw(K--L--N);
draw(K--M--N--cycle);
draw(L--M);
label("K", K, SW);
label("L", L, NW);
label("M"... | [] |
212 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problem 6 | Note that since $\angle KLN = \angle KMN$, quadrilateral $KLMN$ is cyclic. Therefore, we have \[\angle LMK = \angle LNK = 90^{\circ} - \angle LKN = \angle KLP,\]so $\triangle KLO \sim \triangle KML$, giving \[\frac{KM}{28} = \frac{28}{8} \implies KM = 98.\] Therefore, $OM = 98-8 = \boxed{90}$. | // Block 1
size(250);
real h = sqrt(98^2+65^2);
real l = sqrt(h^2-28^2);
pair K = (0,0);
pair N = (h, 0);
pair M = ((98^2)/h, (98*65)/h);
pair L = ((28^2)/h, (28*l)/h);
pair P = ((28^2)/h, 0);
pair O = ((28^2)/h, (8*65)/h);
draw(K--L--N);
draw(K--M--N--cycle);
draw(L--M);
label("K", K, SW);
label("L", L, NW);
label("M"... | [] |
212 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problem 6 | By Pythagorean Theorem, $KM^2+65^2 = KN^2 = 28^2 + LN^2$. Thus, $LN^2 = KM^2 + 65^2 - 28^2$.
By Pythagorean Theorem, $KP^2 + LP^2 = 28^2$, and $PN^2 + LP^2 = LN^2$.
\[PN^2 = (KN - KP)^2 = (\sqrt{KM^2 + 65^2} - KP)^2\]
It follows that
\[(\sqrt{KM^2 + 65^2} - KP)^2 + LP^2 = KM^2 + 65^2 - 28^2\]
\[KM^2 + 65^2 - 2\sqrt... | // Block 1
size(250);
real h = sqrt(98^2+65^2);
real l = sqrt(h^2-28^2);
pair K = (0,0);
pair N = (h, 0);
pair M = ((98^2)/h, (98*65)/h);
pair L = ((28^2)/h, (28*l)/h);
pair P = ((28^2)/h, 0);
pair O = ((28^2)/h, (8*65)/h);
draw(K--L--N);
draw(K--M--N--cycle);
draw(L--M);
label("K", K, SW);
label("L", L, NW);
label("M"... | [] |
213 | In $\triangle ABC$, the sides have integer lengths and $AB=AC$. Circle $\omega$ has its center at the incenter of $\triangle ABC$. An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that ... | 2019 AIME I Problem 11 | First, assume $BC=2$ and $AB=AC=x$. The triangle can be scaled later if necessary. Let $I$ be the incenter and let $r$ be the inradius. Let the points at which the incircle intersects $AB$, $BC$, and $CA$ be denoted $M$, $N$, and $O$, respectively.
Next, we calculate $r$ in terms of $x$. Note the right triangle form... | // Block 1
unitsize(1cm);
var x = 9;
pair A = (0,sqrt(x^2-1));
pair B = (-1,0);
pair C = (1,0);
dot(Label("$A$",A,NE),A);
dot(Label("$B$",B,SW),B);
dot(Label("$C$",C,SE),C);
draw(A--B--C--cycle);
var r = sqrt((x-1)/(x+1));
pair I = (0,r);
dot(Label("$I$",I,SE),I);
draw(circle(I,r));
draw(Label("$r$"),I--I+r*SSW... | [] |
213 | In $\triangle ABC$, the sides have integer lengths and $AB=AC$. Circle $\omega$ has its center at the incenter of $\triangle ABC$. An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that ... | 2019 AIME I Problem 11 | Before we start thinking about the problem, let’s draw it out;
For the sake of space, I've drawn only 2 of the 3 excircles because the third one looks the same as the second large one because the triangle is isosceles. By the incenter-excenter lemma, $AII_A$ and $BII_B$ are collinear, $E$ is the tangent of circle $I... | // Block 1
unitsize(1cm);
var x = 9;
pair A = (0,sqrt(x^2-1));
pair B = (-1,0);
pair C = (1,0);
dot(Label("$A$",A,NE),A);
dot(Label("$B$",B,SW),B);
dot(Label("$C$",C,SE),C);
draw(A--B--C--cycle);
var r = sqrt((x-1)/(x+1));
pair I = (0,r);
dot(Label("$I$",I,SE),I);
draw(circle(I,r));
pair G = intersectionpoint... | [] |
214 | Triangle $ABC$ has side lengths $AB=4$, $BC=5$, and $CA=6$. Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$. The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$. Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$, where $a$, $b$... | 2019 AIME I Problem 13 | Notice that \[\angle DFE=\angle CFE-\angle CFD=\angle CBE-\angle CAD=180-B-A=C.\]By the Law of Cosines, \[\cos C=\frac{AC^2+BC^2-AB^2}{2\cdot AC\cdot BC}=\frac34.\]Then, \[DE^2=DF^2+EF^2-2\cdot DF\cdot EF\cos C=32\implies DE=4\sqrt2.\]Let $X=\overline{AB}\cap\overline{CF}$, $a=XB$, and $b=XD$. Then, \[XA\cdot XD=XC\cdo... | // Block 1
unitsize(20);
pair A, B, C, D, E, F, X, O1, O2;
A = (0, 0); B = (4, 0);
C = intersectionpoints(circle(A, 6), circle(B, 5))[0];
D = B + (5/4 * (1 + sqrt(2)), 0); E = D + (4 * sqrt(2), 0);
F = intersectionpoints(circle(D, 2), circle(E, 7))[1];
X = extension(A, E, C, F);
O1 = circumcenter(C, A, D);
O2 = circumc... | [] |
215 | Let $\overline{AB}$ be a chord of a circle $\omega$, and let $P$ be a point on the chord $\overline{AB}$. Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$. Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$. Circles $\omega_1$ and $\omega_2$ intersect at... | 2019 AIME I Problem 15 | Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$, respectively. There is a homothety at $A$ sending $\omega$ to $\omega_1$ that sends $B$ to $P$ and $O$ to $O_1$, so $\overline{OO_2}\parallel\overline{O_1P}$. Similarly, $\overline{OO_1}\parallel\overline{O_2P}$, so $OO_1PO_2$ is a parallelogram. Moreover... | // Block 1
size(8cm);
pair O, A, B, P, O1, O2, Q, X, Y;
O=(0, 0);
A=dir(140); B=dir(40);
P=(3A+5B)/8;
O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);
O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O);
Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1];
X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1... | [] |
215 | Let $\overline{AB}$ be a chord of a circle $\omega$, and let $P$ be a point on the chord $\overline{AB}$. Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$. Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$. Circles $\omega_1$ and $\omega_2$ intersect at... | 2019 AIME I Problem 15 | $PX \cdot PY=AP \cdot PB=5 \cdot 3=15$ by power of a point. Also, $PX+PY=XY=11$, so $PX$ and $PY$ are solutions to the quadratic $x^2-11x+15=0$ so $PX$ and $PY$ is $\frac{11\pm\sqrt{61}}{2}$ in some order. Now, because we want $PQ^2$ and it is known to be rational, we can guess that $PQ$ is irrational or the problem wo... | // Block 1
size(8cm);
pair O, A, B, P, O1, O2, Q, X, Y;
O=(0, 0);
A=dir(140); B=dir(40);
P=(3A+5B)/8;
O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);
O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O);
Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1];
X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1... | [] |
216 | Two different points, $C$ and $D$, lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$, $BC=AD=10$, and $CA=DB=17$. The intersection of these two triangular regions has area $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2019 AIME II Problem 1 | - Diagram by Brendanb4321
Extend $AB$ to form a right triangle with legs $6$ and $8$ such that $AD$ is the hypotenuse and connect the points $CD$ so
that you have a rectangle. (We know that $\triangle ADE$ is a $6-8-10$, since $\triangle DEB$ is an $8-15-17$.) The base $CD$ of the rectangle will be $9+6+6=21$. Now,... | // Block 1
unitsize(10);
pair A = (0,0);
pair B = (9,0);
pair C = (15,8);
pair D = (-6,8);
pair E = (-6,0);
draw(A--B--C--cycle);
draw(B--D--A);
label("$A$",A,dir(-120));
label("$B$",B,dir(-60));
label("$C$",C,dir(60));
label("$D$",D,dir(120));
label("$E$",E,dir(-135));
label("$9$",(A+B)/2,dir(-90));
label("$10$",(D+A)... | [] |
216 | Two different points, $C$ and $D$, lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$, $BC=AD=10$, and $CA=DB=17$. The intersection of these two triangular regions has area $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2019 AIME II Problem 1 | - Diagram by Brendanb4321 extended by Duoquinquagintillion
Begin with the first step of solution 1, seeing $AD$ is the hypotenuse of a $6-8-10$ triangle and calling the intersection of $DB$ and $AC$ point $E$. Next, notice $DB$ is the hypotenuse of an $8-15-17$ triangle. Drop an altitude from $E$ with length $h$, so t... | // Block 1
unitsize(10);
pair A = (0,0);
pair B = (9,0);
pair C = (15,8);
pair D = (-6,8);
draw(A--B--C--cycle);
draw(B--D--A);
label("$A$",A,dir(-120));
label("$B$",B,dir(-60));
label("$C$",C,dir(60));
label("$D$",D,dir(120));
label("$9$",(A+B)/2,dir(-90));
label("$10$",(D+A)/2,dir(-150));
label("$10$",(C+B)/2,dir(-30... | [] |
216 | Two different points, $C$ and $D$, lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$, $BC=AD=10$, and $CA=DB=17$. The intersection of these two triangular regions has area $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2019 AIME II Problem 1 | First, let's define H as the intersection of CB and DA, and define G as the midpoint of AB. Next, let E and F be the feet of the perpendicular lines from C and D to line AB respectively. We get a pleasing line of symmetry HG where A maps to B, C maps to D, and E maps to F. We notice that 8-15-17 and 8-6-10 are both pyt... | //Made by Afly. I used some resources. //Took me 10 min to get everything right. import olympiad; unitsize(18); pair A = (0,0); pair B = (0,8); pair C = (6,0); pair D = (15,0); pair E = (21,0); pair F = (21,8); pair G = (21/2,0); pair H = intersectionpoints(B--D,C--F)[0]; pen dash1 = linetype(new real [] {9,9})+linewid... | [] |
217 | Triangle $ABC$ has side lengths $AB=120,BC=220$, and $AC=180$. Lines $\ell_A,\ell_B$, and $\ell_C$ are drawn parallel to $\overline{BC},\overline{AC}$, and $\overline{AB}$, respectively, such that the intersections of $\ell_A,\ell_B$, and $\ell_C$ with the interior of $\triangle ABC$ are segments of lengths $55,45$, an... | 2019 AIME II Problem 7 | Let's squish a triangle with side lengths 15, 22.5, and 27.5 into a equilateral triangle with side length 1. Then, the original triangle gets turned into a equilateral triangle with side length 8. Since 15 is one eighth of 120, it has a length of one. Since 45 and 55 are one fourth of 180 and 220 respectively, they are... | // Block 1
pair A,B,C; B = (0,0); C = (1,0); A = intersectionpoints(circle(B,3/2),circle(C,11/6))[0]; draw(A--B--C--cycle); draw((3/2,3/4)--(5/2,3/4)); draw((3/2,1/4)--(5/2,1/4)); draw((9/4,1)--(11/4,1/2)--(9/4,0)); draw(shift(dir(0)*13/4)*shift(dir(30))*polygon(3));
// Block 2
for (int i=0; i<8; ++i) { for (int j=0; ... | [] |
218 | Triangle $ABC$ has side lengths $AB=7, BC=8,$ and $CA=9.$ Circle $\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\omega_1$ and $\omega_2$ not equal to $A.$ Then $AK=\tfrac mn,$ where $m$ an... | 2019 AIME II Problem 11 | -Diagram by Brendanb4321
Note that from the tangency condition that the supplement of $\angle CAB$ with respects to lines $AB$ and $AC$ are equal to $\angle AKB$ and $\angle AKC$, respectively, so from tangent-chord, \[\angle AKC=\angle AKB=180^{\circ}-\angle BAC\] Also note that $\angle ABK=\angle KAC$$^{(*)}$, so ... | // Block 1
unitsize(20);
pair B = (0,0);
pair A = (2,sqrt(45));
pair C = (8,0);
draw(circumcircle(A,B,(-17/8,0)),rgb(.7,.7,.7));
draw(circumcircle(A,C,(49/8,0)),rgb(.7,.7,.7));
draw(B--A--C--cycle);
label("$A$",A,dir(105));
label("$B$",B,dir(-135));
label("$C$",C,dir(-75));
dot((2.68,2.25));
label("$K$",(2.68,2.25),2*d... | [] |
219 | Regular octagon $A_1A_2A_3A_4A_5A_6A_7A_8$ is inscribed in a circle of area $1.$ Point $P$ lies inside the circle so that the region bounded by $\overline{PA_1},\overline{PA_2},$ and the minor arc $\widehat{A_1A_2}$ of the circle has area $\tfrac{1}{7},$ while the region bounded by $\overline{PA_3},\overline{PA_4},$ an... | 2019 AIME II Problem 13 | Instead of considering the actual values of the areas, consider only the changes in the areas that result from moving point $P$ from the center of the circle. We will proceed by coordinates. Set the origin at the center of the circle and refer to the following diagram, where the octagon is oriented so as $A_1A_2$ is ho... | // Block 1
size(7cm);
draw(Circle((0,0),1));
pair P = (0.1,-0.15);
filldraw(P--dir(112.5)--dir(112.5-45)--cycle,yellow,red);
filldraw(P--dir(112.5-90)--dir(112.5-135)--cycle,yellow,red);
filldraw(P--dir(112.5-225)--dir(112.5-270)--cycle,green,red);
dot(P);
for(int i=0; i<8; ++i)
{
draw(dir(22.5+45i)--dir(67.5+45i)... | [] |
220 | In $\triangle ABC$ with $AB=AC,$ point $D$ lies strictly between $A$ and $C$ on side $\overline{AC},$ and point $E$ lies strictly between $A$ and $B$ on side $\overline{AB}$ such that $AE=ED=DB=BC.$ The degree measure of $\angle ABC$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n... | 2020 AIME I Problem 1 | If we set $\angle{BAC}$ to $x$, we can find all other angles through these two properties:
1. Angles in a triangle sum to $180^{\circ}$.
2. The base angles of an isosceles triangle are congruent.
Now we angle chase. $\angle{ADE}=\angle{EAD}=x$, $\angle{AED} = 180-2x$, $\angle{BED}=\angle{EBD}=2x$, $\angle{EDB} = 180-4... | // Block 1
size(10cm);
pair A, B, C, D, F;
A = (0, tan(3 * pi / 7));
B = (1, 0);
C = (-1, 0);
F = rotate(90/7, A) * (A - (0, 2));
D = rotate(900/7, F) * A;
draw(A -- B -- C -- cycle);
draw(F -- D);
draw(D -- B);
label("$A$", A, N);
label("$B$", B, E);
label("$C$", C, W);
label("$D$", D, W);
label("$E$", F, E);
// Blo... | [] |
220 | In $\triangle ABC$ with $AB=AC,$ point $D$ lies strictly between $A$ and $C$ on side $\overline{AC},$ and point $E$ lies strictly between $A$ and $B$ on side $\overline{AB}$ such that $AE=ED=DB=BC.$ The degree measure of $\angle ABC$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n... | 2020 AIME I Problem 1 | Let $x = \angle ABC = \angle ACB$. Because $\triangle BCD$ is isosceles, $\angle CBD = 180^\circ - 2x$. Then
\[\angle DBE = x - \angle CBD = x - (180^\circ - 2x) = 3x - 180^\circ\!.\]Because $\triangle EDA$ and $\triangle DBE$ are also isosceles,
\[\angle BAC =\frac12(\angle EAD + \angle ADE) = \frac12(\angle BED)= \f... | // Block 1
unitsize(4 cm);
pair A, B, C, D, E;
real a = 180/7;
A = (0,0);
B = dir(180 - a/2);
C = dir(180 + a/2);
D = extension(B, B + dir(270 + a), A, C);
E = extension(D, D + dir(90 - 2*a), A, B);
draw(A--B--C--cycle);
draw(B--D--E);
label("$A$", A, dir(0));
label("$B$", B, NW);
label("$C$", C, SW);
label("$D$", ... | [] |
221 | Six cards numbered $1$ through $6$ are to be lined up in a row. Find the number of arrangements of these six cards where one of the cards can be removed leaving the remaining five cards in either ascending or descending order. | 2020 AIME I Problem 5 | We start with five cards in ascending order, then insert the sixth card to obtain a valid arrangement.
Based on the card to be inserted, we have six cases. As shown below, the red squares indicate the possible positions to insert the sixth card.
Note that all arrangements of the six cards are distinct.
There are $26... | // Block 1
/* Made by MRENTHUSIASM */
unitsize(7mm);
void drawSquare(real x, real y)
{
draw((x+0.5,y+0.5)--(x-0.5,y+0.5)--(x-0.5,y-0.5)--(x+0.5,y-0.5)--cycle,red);
}
label("$2$",(0.5,8));
label("$3$",(2.5,8));
label("$4$",(4.5,8));
label("$5$",(6.5,8));
label("$6$",(8.5,8));
label("Insert $1.$",(13.5,8),red);
drawSqu... | [] |
222 | A flat board has a circular hole with radius $1$ and a circular hole with radius $2$ such that the distance between the centers of the two holes is $7.$ Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is $\tfrac{m}{n},$ where $m$... | 2020 AIME I Problem 6 | Set the common radius to $r$. First, take the cross section of the sphere sitting in the hole of radius $1$. If we draw the perpendicular bisector of the chord (the hole) through the circle, this line goes through the center. Connect the center also to where the chord hits the circle, for a right triangle with hypotenu... | // Block 1
size(10cm);
pair A, B, C, D, O, P, H, L, X, Y;
A = (-1, 0);
B = (1, 0);
H = (0, 0);
C = (5, 0);
D = (9, 0);
L = (7, 0);
O = (0, sqrt(160/13 - 1));
P = (7, sqrt(160/13 - 4));
X = (0, sqrt(160/13 - 4));
Y = (O + P) / 2;
draw(A -- O -- B -- cycle);
draw(C -- P -- D -- cycle);
draw(B -- C);
draw(O -- P);
draw(P... | [] |
223 | A club consisting of $11$ men and $12$ women needs to choose a committee from among its members so that the number of women on the committee is one more than the number of men on the committee. The committee could have as few as $1$ member or as many as $23$ members. Let $N$ be the number of such committees that can be... | 2020 AIME I Problem 7 | Applying [1] by setting $m=12$, $n=11$, and $r=11$, we obtain $\binom{23}{11}\implies$ $\boxed{081}$.
~Lcz
Short Proof
Consider the following setup:
The dots to the left represent the men, and the dots to the right represent the women. Now, suppose we put a mark on $11$ people (the $*$). Those to the left of the dash... | // Block 1
size(1000, 100);
for(int i=0; i<23; ++i){
dot((i, 0));
}
draw((10.5, -1.5)--(10.5, 1.5), dashed);
// Block 2
size(1000, 100);
for(int i=0; i<23; ++i){
dot((i, 0));
}
for(int i=0; i<23; ++i){
if(i%2==0){
if(i >= 11){
draw(circle((i, 0), 0.25));
}
continue;
}
label("$*$", (i,0.5), N);
if(i < 11){
draw(circle((... | [] |
224 | A bug walks all day and sleeps all night. On the first day, it starts at point $O$, faces east, and walks a distance of $5$ units due east. Each night the bug rotates $60^\circ$ counterclockwise. Each day it walks in this new direction half as far as it walked the previous day. The bug gets arbitrarily close to the poi... | 2020 AIME I Problem 8 | We notice that the moves cycle every 6 moves, so we plot out the first 6 moves on the coordinate grid with point $O$ as the origin. We will keep a tally of the x-coordinate and y-coordinate separately. Then, we will combine them and account for the cycling using the formula for an infinite geometric series.
First move... | // Block 1
size(8cm);
pair O, A, B, C, D, F, G, H, I, P, X;
O = (0, 0);
A = (5, 0);
X = (8, 0);
P = (5, 5 / sqrt(3));
B = rotate(-120, A) * ((O + A) / 2);
C = rotate(-120, B) * ((A + B) / 2);
D = rotate(-120, C) * ((B + C) / 2);
F = rotate(-120, D) * ((C + D) / 2);
G = rotate(-120, F) * ((D + F) / 2);
H = rotate(-120, ... | [] |
225 | Let $S$ be the set of positive integer divisors of $20^9.$ Three numbers are chosen independently and at random with replacement from the set $S$ and labeled $a_1,a_2,$ and $a_3$ in the order they are chosen. The probability that both $a_1$ divides $a_2$ and $a_2$ divides $a_3$ is $\tfrac{m}{n},$ where $m$ and $n$ are ... | 2020 AIME I Problem 9 | First, prime factorize $20^9$ as $2^{18} \cdot 5^9$. Denote $a_1$ as $2^{b_1} \cdot 5^{c_1}$, $a_2$ as $2^{b_2} \cdot 5^{c_2}$, and $a_3$ as $2^{b_3} \cdot 5^{c_3}$.
In order for $a_1$ to divide $a_2$, and for $a_2$ to divide $a_3$, $b_1\le b_2\le b_3$, and $c_1\le c_2\le c_3$. We will consider each case separately. N... | // Block 1
size(12cm);
for (int x = 1; x < 18; ++x) {
draw((x, 0) -- (x, 9), dotted);
}
for (int y = 1; y < 9; ++y) {
draw((0, y) -- (18, y), dotted);
}
draw((0, 0) -- (18, 0) -- (18, 9) -- (0, 9) -- cycle);
pair b1, b2, b3;
pair c1, c2, c3;
pair a1, a2, a3;
b1 = (3, 0); b2 = (12, 0); b3 = (16, 0);
c1 = (0, 2... | [] |
226 | Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be writ... | 2020 AIME I Problem 13 | Let $M_A$, $M_B$, $M_C$ be the midpoints of arcs $BC$, $CA$, $AB$. By Fact 5, we know that $M_AB = M_AC = M_AI$, and so by Ptolmey's theorem, we deduce that \[AB\cdot M_AC + AC\cdot M_AB = BC\cdot M_AA \implies M_AA = 2M_AI.\]
In particular, we have $AI = IM_A$.
Now the key claim is that:
Claim: $\triangle DEF$ and... | // Block 1
defaultpen(fontsize(10pt));
size(200);
pair A, B, C, D, E, F, I, P, MA, MB, MC;
B = (0,0);
C = (5,0);
A = IP(Circle(B, 4), Circle(C, 6), 0);
I = incenter(A, B, C);
D = extension(A, I, B, C);
P = midpoint(A--D);
E = extension(P, rotate(90, P)*A, B, I);
F = extension(P, rotate(90, P)*A, C, I);
MA = IP(Line(A, ... | [] |
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