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
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80 | Three of the edges of a cube are $\overline{AB}, \overline{BC},$ and $\overline{CD},$ and $\overline{AD}$ is an interior diagonal. Points $P, Q,$ and $R$ are on $\overline{AB}, \overline{BC},$ and $\overline{CD},$ respectively, so that $AP = 5, PB = 15, BQ = 15,$ and $CR = 10.$ What is the area of the polygon that is... | 1998 AIME Problem 11 | "For non-asymptote version of image, see Image:1998_AIME-11.png"
This approach uses analytic geometry. Let $A$ be at the origin, $B$ at $(20,0,0)$, $C$ at $(20,0,20)$, and $D$ at $(20,20,20)$. Thus, $P$ is at $(5,0,0)$, $Q$ is at $(20,0,15)$, and $R$ is at $(20,10,20)$.
Let the plane $PQR$ have the equation $ax + b... | // Block 1
import three;
size(280); defaultpen(linewidth(0.6)+fontsize(9));
currentprojection=perspective(30,-60,40);
triple A=(0,0,0),B=(20,0,0),C=(20,0,20),D=(20,20,20);
triple P=(5,0,0),Q=(20,0,15),R=(20,10,20),Pa=(15,20,20),Qa=(0,20,5),Ra=(0,10,0);
draw(box((0,0,0),(20,20,20)));
draw(P--Q--R--Pa--Qa--Ra--cycle,line... | [] |
81 | Given that $\sum_{k=1}^{35}\sin 5k=\tan \frac mn,$ where angles are measured in degrees, and $m_{}$ and $n_{}$ are relatively prime positive integers that satisfy $\frac mn<90,$ find $m+n.$ | 1999 AIME Problem 11 | This solution will work entirely in degrees. Let
\[z = e^{5^\circ i}.\]
For each integer $k$, consider the unit vector in the complex plane
\[v_k = z^k = e^{5k^\circ i}.\]
Geometrically, $v_k$ is a unit vector making an angle of $5k ^\circ$ with the positive real axis. Now draw each vector $v_k$ for $1 \leq k \leq 3... | // Block 1
pathpen = black + linewidth(0.5); pointpen = black;
unitsize(0.4cm);
pair O=(0,0);
real s=0.9;
pair[] P;
P[0]=O;
for(int k=1;k<=35;++k) P[k]=P[k-1]+s*dir(5*k);
for(int k=1;k<=35;++k) draw(P[k-1]--P[k],pathpen,Arrow(TeXHead));
int it=0;
for(int k=1;k<=35;++k) if(P[k].y>P[it].y) it=k;
pair B=P[it];
pair A... | [] |
82 | The inscribed circle of triangle $ABC$ is tangent to $\overline{AB}$ at $P_{},$ and its radius is $21$. Given that $AP=23$ and $PB=27,$ find the perimeter of the triangle.
Contents
1 Problem
2 Solution
2.1 Solution 1
2.2 Solution 2
3 Solution 3
4 Solution 4 (Trig Bash)
5 See also | 1999 AIME Problem 12 | Solution 1
Let $Q$ be the tangency point on $\overline{AC}$, and $R$ on $\overline{BC}$. By the Two Tangent Theorem, $AP = AQ = 23$, $BP = BR = 27$, and $CQ = CR = x$. Using $rs = A$, where $s = \frac{27 \cdot 2 + 23 \cdot 2 + x \cdot 2}{2} = 50 + x$, we get $(21)(50 + x) = A$. By Heron's formula, $A = \sqrt{s(s-a)(s-b... | // Block 1
pathpen = black + linewidth(0.65); pointpen = black;
pair A=(0,0),B=(50,0),C=IP(circle(A,23+245/2),circle(B,27+245/2)), I=incenter(A,B,C);
path P = incircle(A,B,C);
D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle);D(P);
D(MP("P",IP(A--B,P)));
pair Q=IP(C--A,P),R=IP(B--C,P);
D(MP("R",R,NE));D(MP("Q",Q,NW));
MP("23... | [] |
82 | The inscribed circle of triangle $ABC$ is tangent to $\overline{AB}$ at $P_{},$ and its radius is $21$. Given that $AP=23$ and $PB=27,$ find the perimeter of the triangle.
Contents
1 Problem
2 Solution
2.1 Solution 1
2.2 Solution 2
3 Solution 3
4 Solution 4 (Trig Bash)
5 See also | 1999 AIME Problem 12 | Let \(CR = CQ = x\). Using the inradius formula for the area of a triangle, we have
\[[ABC] = 1050 + 21x,\]
where \([ABC]\) denotes the area of \(\triangle ABC\).
Next, we find \(\sin \angle CBA\). Note that
\[\angle CBA = \angle RBP = 2\angle OBP.\]
First, we find the hypotenuse \(OB\) using the Pythagorean Theorem:... | // Block 1
import olympiad;
pathpen = black + linewidth(0.65); pointpen = black;
pair A=(0,0), B=(50,0);
pair C=IP(circle(A,23+245/2),circle(B,27+245/2));
pair O = incenter(A,B,C);
path P_circle = incircle(A,B,C);
pair P = foot(O, A, B);
pair Q = foot(O, A, C);
pair R = foot(O, B, C);
D(MP("A",A, SW)--MP("B",B, SE)--M... | [] |
83 | Point $P_{}$ is located inside triangle $ABC$ so that angles $PAB, PBC,$ and $PCA$ are all congruent. The sides of the triangle have lengths $AB=13, BC=14,$ and $CA=15,$ and the tangent of angle $PAB$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$
Contents
1 Problem
2 Solution... | 1999 AIME Problem 14 | Solution 1
Drop perpendiculars from $P$ to the three sides of $\triangle ABC$ and let them meet $\overline{AB}, \overline{BC},$ and $\overline{CA}$ at $D, E,$ and $F$ respectively.
Let $BE = x, CF = y,$ and $AD = z$. We have that
\begin{align*}DP&=z\tan\theta\\ EP&=x\tan\theta\\ FP&=y\tan\theta\end{align*}
We can ... | // Block 1
real theta = 29.66115; /* arctan(168/295) to five decimal places .. don't know other ways to construct Brocard */
pathpen = black +linewidth(0.65); pointpen = black;
pair A=(0,0),B=(13,0),C=IP(circle(A,15),circle(B,14));
D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle);
/* constructing P, C is there as check */... | [] |
83 | Point $P_{}$ is located inside triangle $ABC$ so that angles $PAB, PBC,$ and $PCA$ are all congruent. The sides of the triangle have lengths $AB=13, BC=14,$ and $CA=15,$ and the tangent of angle $PAB$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$
Contents
1 Problem
2 Solution... | 1999 AIME Problem 14 | Plot $\triangle ABC$ on the coordinate plane with \(B=(0,0), C=(14,0), A=(5,12).\)
Let $\angle PAB=\angle PBC=\angle PCA=a$, and let $P=(x,y)$. For convenience, define $\tan a = k$.
Since $\angle PBC=a$, the slope of $PB$ is $k$, so \(PB:\ y=kx.\)
Let $\angle ADB$ be the angle that line $AB$ makes with the $x$-axis. ... | // Block 1
import olympiad;
real theta = 29.66115; /* Brocard angle */
pathpen = black + linewidth(0.65);
pointpen = black;
pair B=(0,0), C=(14,0), A=(5,12);
pair R1=(0,12), R2=(14,12); // Corners of the bounding rectangle
D(B--R1--R2--C, dashed);
D(A--B--C--cycle);
pair Bb = B + (C-B)*dir(theta);
pair Cc = C + (A-C)*... | [] |
84 | Consider the paper triangle whose vertices are $(0,0), (34,0),$ and $(16,24).$ The vertices of its midpoint triangle are the midpoints of its sides. A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle. What is the volume of this pyramid? | 1999 AIME Problem 15 | As shown in the image above, let $D$, $E$, and $F$ be the midpoints of $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively. Suppose $P$ is the apex of the tetrahedron, and let $O$ be the foot of the altitude from $P$ to $\triangle ABC$. The crux of this problem is the following lemma.
Lemma: The poin... | // Block 1
defaultpen(fontsize(9)+linewidth(0.63)); pair A=(0,0), B=(16,24), C=(34,0), P=(8,12), Q=(25,12), R=(17,0); draw(A--B--C--A);draw(P--Q--R--P); draw(A--foot(A,B,C));draw(B--foot(B,A,C));draw(C--foot(C,A,B)); label("\(A\)",A,SW);label("\(B\)",B,NW);label("\(C\)",C,SE); label("\(D\)",foot(A,B,C),NE);label("\(E\)... | [] |
85 | Let $u$ and $v$ be integers satisfying $0 < v < u$. Let $A = (u,v)$, let $B$ be the reflection of $A$ across the line $y = x$, let $C$ be the reflection of $B$ across the y-axis, let $D$ be the reflection of $C$ across the x-axis, and let $E$ be the reflection of $D$ across the y-axis. The area of pentagon $ABCDE$ is $... | 2000 AIME I Problem 2 | Solution 1
Since $A = (u,v)$, we can find the coordinates of the other points: $B = (v,u)$, $C = (-v,u)$, $D = (-v,-u)$, $E = (v,-u)$. If we graph those points, we notice that since the latter four points are all reflected across the x/y-axis, they form a rectangle, and $ABE$ is a triangle. The area of $BCDE$ is $(2u)... | pointpen = black; pathpen = linewidth(0.7) + black; size(180); pair A=(11,10), B=(10,11), C=(-10, 11), D=(-10, -11), E=(10, -11); D(D(MP("A\ (u,v)",A,(1,0)))--D(MP("B",B,N))--D(MP("C",C,N))--D(MP("D",D))--D(MP("E",E))--cycle); D((-15,0)--(15,0),linewidth(0.6),Arrows(5)); D((0,-15)--(0,15),linewidth(0.6),Arrows(5)); D((... | [] |
86 | In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at $50$ miles per hour along the highways and at $14$ miles per hour across the prairie. Consider the set of points that can be reached by the firetruck within six minutes. The area of... | 2000 AIME I Problem 13 | Let the intersection of the highways be at the origin $O$, and let the highways be the x and y axes. We consider the case where the truck moves in the positive x direction.
After going $x$ miles, $t=\frac{d}{r}=\frac{x}{50}$ hours has passed. If the truck leaves the highway it can travel for at most $t=\frac{1}{10}-\... | // Block 1
pair truck(pair P){
pair Q = IP(P--P+(7/10,24/10),(35/31,35/31)--(5,0));
D(P--Q,EndArrow(5)); D(CP(P,Q),linewidth(0.5));
return Q;
}
pointpen = black; pathpen = black+linewidth(0.7); size(250);
pair B=(5,0), C=(35/31,35/31);
D(D(B)--D(C)--D(B*dir(90))--D(C*dir(90))--D(B*dir(180))--D(C*dir(180))--D(B*dir(... | [] |
87 | In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent. Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$.
Contents
1 Problem
2 Offi... | 2000 AIME I Problem 14 | Let point $R$ be in $\triangle ABC$ such that $QB = BR = RP$. Then $PQBR$ is a rhombus, so $AB \parallel PR$ and $APRB$ is an isosceles trapezoid. Since $\overline{PB}$ bisects $\angle QBR$, it follows by symmetry in trapezoid $APRB$ that $\overline{RA}$ bisects $\angle BAC$. Thus $R$ lies on the perpendicular bisector... | defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(60); draw(A--B--C--A);draw(P--Q);draw(A--R--B);draw(P--R);D(R--C,dashed); label("\(A\)",A,(0,1));label("\(B\)",B,(-1,-1));label("\(C\)",C,(1,-1));label("\(P\)",P,(1,1)); la... | [] |
87 | In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent. Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$.
Contents
1 Problem
2 Offi... | 2000 AIME I Problem 14 | Again, construct $R$ as above.
Let $\angle BAC = \angle QBR = \angle QPR = 2x$ and $\angle ABC = \angle ACB = y$, which means $x + y = 90$.
$\triangle QBC$ is isosceles with $QB = BC$, so $\angle BCQ = 90 - \frac {y}{2}$.
Let $S$ be the intersection of $QC$ and $BP$. Since $\angle BCQ = \angle BQC = \angle BRS$, $BCR... | defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(60), S; S=intersectionpoint(Q--C,P--B); draw(A--B--C--A);draw(B--P--Q--C--R--Q);draw(A--R--B);draw(P--R--S); label("\(A\)",A,(0,1));label("\(B\)",B,(-1,-1));label("\(C\)",C... | [] |
87 | In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent. Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$.
Contents
1 Problem
2 Offi... | 2000 AIME I Problem 14 | Reflect $\triangle ABC$ over $BC$ and translate it to attach side $AB$ onto $AC$, mapping $\triangle ABC$ to $\triangle CAD$. Point $P$ maps to $R$, and $Q$ maps to $P$. Then we have that $BC=BQ=QP=PA=AD=PR=RC$. Notice how $BQ=RC$ and $BQ\parallel RC$, so $BQRC$ is a parallelogram and $QR=BC$. But $BC=QP=PR$, so $\tria... | defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(0)+20*dir(80), D=20*dir(0)+20*dir(80)+20*dir(60)+20*dir(100); draw(Q--A--D--C--B--Q--P--R--Q); draw(A--C); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,NE... | [] |
88 | One base of a trapezoid is $100$ units longer than the other base. The segment that joins the midpoints of the legs divides the trapezoid into two regions whose areas are in the ratio $2: 3$. Let $x$ be the length of the segment joining the legs of the trapezoid that is parallel to the bases and that divides the trapez... | 2000 AIME II Problem 6 | Let the shorter base have length $b$ (so the longer has length $b+100$), and let the height be $h$. The length of the midline of the trapezoid is the average of its bases, which is $\frac{b+b+100}{2} = b+50$. The two regions which the midline divides the trapezoid into are two smaller trapezoids, both with height $h/2$... | // Block 1
pathpen = linewidth(0.7); pen d = linetype("4 4") + linewidth(0.7);
pair A=(0,0),B=(175,0),C=(105,100),D=(30,100);
D(A--B--C--D--cycle); D((A+D)/2 -- (B+C)/2, d); MP("b",(C+D)/2,N);MP("b+100",(A+B)/2);
// Block 2
pathpen = linewidth(0.7); pen d = linetype("4 4") + linewidth(0.7);
pair A=(0,0),B=(175,0),C=(10... | [] |
89 | In trapezoid $ABCD$, leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD}$, and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001}$, find $BC^2$. | 2000 AIME II Problem 8 | Solution 1
Let $x = BC$ be the height of the trapezoid, and let $y = CD$. Since $AC \perp BD$, it follows that $\triangle BAC \sim \triangle CBD$, so $\frac{x}{\sqrt{11}} = \frac{y}{x} \Longrightarrow x^2 = y\sqrt{11}$.
Let $E$ be the foot of the altitude from $A$ to $\overline{CD}$. Then $AE = x$, and $ADE$ is a righ... | size(200); pathpen = linewidth(0.7); pair C=(0,0),B=(0,110^.5),A=(11^.5,B.y),D=(10*11^.5,0),E=foot(A,C,D); D(MP("A",A,(2,.5))--MP("B",B,W)--MP("C",C)--MP("D",D)--cycle); D(A--C);D(B--D);D(A--E,linetype("4 4") + linewidth(0.7)); MP("\sqrt{11}",(A+B)/2,N);MP("\sqrt{1001}",(A+D)/2,NE);MP("\sqrt{1001}",(A+D)/2,NE);MP("x",(... | [] |
90 | The coordinates of the vertices of isosceles trapezoid $ABCD$ are all integers, with $A=(20,100)$ and $D=(21,107)$. The trapezoid has no horizontal or vertical sides, and $\overline{AB}$ and $\overline{CD}$ are the only parallel sides. The sum of the absolute values of all possible slopes for $\overline{AB}$ is $m/n$, ... | 2000 AIME II Problem 11 | For simplicity, we translate the points so that $A$ is on the origin and $D = (1,7)$. Suppose $B$ has integer coordinates; then $\overrightarrow{AB}$ is a vector with integer parameters (vector knowledge is not necessary for this solution). We construct the perpendicular from $A$ to $\overline{CD}$, and let $D' = (a,b)... | // Block 1
pathpen = linewidth(0.7);
pair A=(0,0), D=(1,7), Da = MP("D'",D((-7,1)),N), B=(-8,-6), C=B+Da, F=foot(A,C,D);
D(MP("A",A)--MP("B",B)--MP("C",C,N)--MP("D",D,N)--cycle); D(F--A--Da,linetype("4 4"));
// Block 2
pathpen = linewidth(0.7); pair A=(0,0), D=(1,7), Da = MP("D'",D((-7,1)),N), B=(-8,-6), C=B+Da, F=foot... | [] |
91 | In triangle $ABC$, angles $A$ and $B$ measure $60$ degrees and $45$ degrees, respectively. The bisector of angle $A$ intersects $\overline{BC}$ at $T$, and $AT=24$. The area of triangle $ABC$ can be written in the form $a+b\sqrt{c}$, where $a$, $b$, and $c$ are positive integers, and $c$ is not divisible by the square ... | 2001 AIME I Problem 4 | First, draw a good diagram.
We realize that $\angle C = 75^\circ$, and $\angle CAT = 30^\circ$. Therefore, $\angle CTA = 75^\circ$ as well, making $\triangle CAT$ an isosceles triangle. $AT$ and $AC$ are congruent, so $AC=24$. We now drop an altitude from $C$, and call the foot this altitude point $D$.
By 30-60-... | // Block 1
size(200);
defaultpen(linewidth(0.4)+fontsize(8));
pair A,B,C,D,T,F;
A = origin;
T = scale(24)*dir(30);
C = scale(24)*dir(60);
B = extension(C,T,A,(1,0));
F = foot(T,A,B);
D = foot(C,A,B);
draw(A--B--C--A--T, black+0.8);
draw(C--D, dashed);
label(rotate(degrees(T-A))*"$24$", A--T, N);
label(rotate(degrees(C... | [] |
92 | An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$. One vertex of the triangle is $(0,1)$, one altitude is contained in the y-axis, and the square of the length of each side is $\sqrt{\frac{m}{n}}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Contents
1 Proble... | 2001 AIME I Problem 5 | Solution 1
Denote the vertices of the triangle $A,B,$ and $C,$ where $B$ is in quadrant 4 and $C$ is in quadrant $3.$
Note that the slope of $\overline{AC}$ is $\tan 60^\circ = \sqrt {3}.$ Hence, the equation of the line containing $\overline{AC}$ is
\[y = x\sqrt {3} + 1.\]
This will intersect the ellipse when
\begin{... | // Block 1
pointpen = black; pathpen = black + linewidth(0.7);
path e = xscale(2)*unitcircle; real x = -8/13*3^.5;
D((-3,0)--(3,0)); D((0,-2)--(0,2)); /* axes */
D(e); D(D((0,1))--(x,x*3^.5+1)--(-x,x*3^.5+1)--cycle);
// Block 2
pointpen = black; pathpen = black + linewidth(0.7); path e = xscale(2)*unitcircle; real x = ... | [] |
93 | Triangle $ABC$ has $AB=21$, $AC=22$ and $BC=20$. Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$, respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$. Then $DE=m/n$, where $m$ and $n$ are relatively prime positive... | 2001 AIME I Problem 7 | Let $I$ be the incenter of $\triangle ABC$, so that $BI$ and $CI$ are angle bisectors of $\angle ABC$ and $\angle ACB$ respectively. Then, $\angle BID = \angle CBI = \angle DBI,$ so $\triangle BDI$ is isosceles, and similarly $\triangle CEI$ is isosceles. It follows that $DE = DB + EC$, so the perimeter of $\triangle A... | // Block 1
pointpen = black; pathpen = black+linewidth(0.7);
pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C);
D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle); D(MP("I",I,NE)); D(MP("E",E,NE)--MP("D",D,NW));
// D((A.x,0)--A,linetype("4 4")+linewi... | [] |
93 | Triangle $ABC$ has $AB=21$, $AC=22$ and $BC=20$. Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$, respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$. Then $DE=m/n$, where $m$ and $n$ are relatively prime positive... | 2001 AIME I Problem 7 | The semiperimeter of $ABC$ is $s = \frac{20 + 21 + 22}{2} = \frac{63}{2}$. By Heron's formula, the area of the whole triangle is $A = \sqrt{s(s-a)(s-b)(s-c)} = \frac{21\sqrt{1311}}{4}$. Using the formula $A = rs$, we find that the inradius is $r = \frac{A}{s} = \frac{\sqrt{1311}}6$. Since $\triangle ADE \sim \triangle ... | // Block 1
pointpen = black; pathpen = black+linewidth(0.7);
pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C);
D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle); D(incircle(A,B,C)); D(MP("I",I,NE)); D(MP("E",E,NE)--MP("D",D,NW));
D((A.x,0)--A,linet... | [] |
93 | Triangle $ABC$ has $AB=21$, $AC=22$ and $BC=20$. Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$, respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$. Then $DE=m/n$, where $m$ and $n$ are relatively prime positive... | 2001 AIME I Problem 7 | Let $P$ be the incenter; then it is be the intersection of all three angle bisectors. Draw the bisector $AP$ to where it intersects $BC$, and name the intersection $F$.
Using the angle bisector theorem, we know the ratio $BF:CF$ is $21:22$, thus we shall assign a weight of $22$ to point $B$ and a weight of $21$ to po... | pointpen = black; pathpen = black+linewidth(0.7); pen d = linewidth(0.7) + linetype("4 4"); pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C); D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle); D(incircle(A,B,C)); D(MP("P",I,(1,2))); D(MP("E",E,NE)-... | [] |
93 | Triangle $ABC$ has $AB=21$, $AC=22$ and $BC=20$. Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$, respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$. Then $DE=m/n$, where $m$ and $n$ are relatively prime positive... | 2001 AIME I Problem 7 | Diagram borrowed from Solution 3.
Let the angle bisector of $\angle{A}$ intersects $BC$ at $F$.
Applying the Angle Bisector Theorem on $\angle{A}$ we have
\[\frac{AB}{BF}=\frac{AC}{CF}\]
\[BF=BC\cdot(\frac{AB}{AB+AC})\]
\[BF=\frac{420}{43}\]
Since $BP$ is the angle bisector of $\angle{B}$, we can once again apply th... | // Block 1
pointpen = black; pathpen = black+linewidth(0.7); pen d = linewidth(0.7) + linetype("4 4");
pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C);
D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle); D(incircle(A,B,C)); D(MP("P",I,(1,2))); D(MP... | [] |
94 | In triangle $ABC$, $AB=13$, $BC=15$ and $CA=17$. Point $D$ is on $\overline{AB}$, $E$ is on $\overline{BC}$, and $F$ is on $\overline{CA}$. Let $AD=p\cdot AB$, $BE=q\cdot BC$, and $CF=r\cdot CA$, where $p$, $q$, and $r$ are positive and satisfy $p+q+r=2/3$ and $p^2+q^2+r^2=2/5$. The ratio of the area of triangle $DEF$ ... | 2001 AIME I Problem 9 | Solution 1
We let $[\ldots]$ denote area; then the desired value is
$\frac mn = \frac{[DEF]}{[ABC]} = \frac{[ABC] - [ADF] - [BDE] - [CEF]}{[ABC]}$
Using the formula for the area of a triangle $\frac{1}{2}ab\sin C$, we find that
$\frac{[ADF]}{[ABC]} = \frac{\frac 12 \cdot p \cdot AB \cdot (1-r) \cdot AC \cdot \sin \... | /* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */ real p = 0.5, q = 0.1, r = 0.05; /* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */ pointpen = black; pathpen = linewidth(0.7) + black; pair A=(0,0),B=(13,0),C=IP(CR(A,17),CR(B,15))... | [] |
95 | A sphere is inscribed in the tetrahedron whose vertices are $A = (6,0,0), B = (0,4,0), C = (0,0,2),$ and $D = (0,0,0).$ The radius of the sphere is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | 2001 AIME I Problem 12 | The center $I$ of the insphere must be located at $(r,r,r)$ where $r$ is the sphere's radius.
$I$ must also be a distance $r$ from the plane $ABC$
The signed distance between a plane and a point $I$ can be calculated as $\frac{(I-G) \cdot P}{|P|}$, where G is any point on the plane, and P is a vector perpendicular to ... | // Block 1
import three;
currentprojection = perspective(-2,9,4);
triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0);
triple E = (2/3,0,0), F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0);
triple I = (2/3,2/3,2/3);
triple J = (6/7,20/21,26/21);
draw(C--A--D--C--B--D--B--A--C);
dr... | [] |
96 | The numbers $1, 2, 3, 4, 5, 6, 7,$ and $8$ are randomly written on the faces of a regular octahedron so that each face contains a different number. The probability that no two consecutive numbers, where $8$ and $1$ are considered to be consecutive, are written on faces that share an edge is $m/n,$ where $m$ and $n$ ar... | 2001 AIME I Problem 15 | The probability is equivalent to counting the number of Hamiltonian cycles in this 3D graph over $7!.$ This is because each Hamiltonian cycle corresponds to eight unique ways to label the faces. Label the vertices $AR,BR,CR,DR,AX,BX,CX,DX$ where vertices $ab$ and $cd$ are connected if $a=c$ or $b=d.$
Case 1: Four of t... | // Block 1
import three;
draw((0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0));
draw((1,1,0)--(1,1,1));
draw((0,1,0)--(0,1,1));
draw((0,0,0)--(1,0,0));
draw((0,0,1)--(1,0,1));
for(int i = 0; i < 2; ++i) {
for(int j = 0; j < 2; ++j) {
for(int k = 0; k < 2; ++k) {
dot((i,j,... | [] |
97 | Let $R = (8,6)$. The lines whose equations are $8y = 15x$ and $10y = 3x$ contain points $P$ and $Q$, respectively, such that $R$ is the midpoint of $\overline{PQ}$. The length of $PQ$ equals $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 2001 AIME II Problem 4 | The coordinates of $P$ can be written as $\left(8a, 15a\right)$ and the coordinates of point $Q$ can be written as $\left(10b, 3b\right)$. By the midpoint formula, we have $8a+10b=16$ and $15a+3b=12$. Substituting $b=4-5a$ we derive $a = \frac{4}{7}$ and $b = \frac{8}{7}$. Thus $P$ is $\left(\frac{32}7, \frac{60}7\righ... | // Block 1
pointpen = black; pathpen = black+linewidth(0.7);
pair R = (8,6), P = (32,60)/7, Q= (80,24)/7;
D((0,0)--MP("x",(13,0),E),EndArrow(6)); D((0,0)--MP("y",(0,10),N),EndArrow(6));
D((0,0)--(10/(15/8),10),EndArrow(6)); D((0,0)--(13,13 * 3/10),EndArrow(6));
D(D(MP("P",P,NW))--D(MP("Q",Q),SE),linetype("4 4")); D(MP(... | [] |
98 | Square $ABCD$ is inscribed in a circle. Square $EFGH$ has vertices $E$ and $F$ on $\overline{CD}$ and vertices $G$ and $H$ on the circle. If the area of square $ABCD$ is $1$, then the area of square $EFGH$ can be expressed as $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers and $m < n$. Find $10n... | 2001 AIME II Problem 6 | Begin by drawing another square congruent to $EFGH$ but on side $AB$ instead of side $CD$. Call this square $IJKL$, such that $I$ lies on $\overline{AB}$ and $\overline{IJ}$ lies on line $FG$. Let the side length of $EFGH$ and $IJKL$ be $s$ and the side length of $ABCD$ be $1$.
By Power of a Point on point $F$, we fi... | // Block 1
size(150); pointpen=black; pathpen=black+linewidth(0.7);
pair A=(-1,-1), B=(-1,1), C=(1,1), D=(1,-1), O=(0,0);
pair E=(1,-0.2), F=(1,0.2), G=(1.4,0.2), H=(1.4,-0.2);
pair I=(-1,0.2), J=(-1.4,0.2), K=(-1.4,-0.2), L=(-1,-0.2);
draw(A--B--C--D--cycle);
draw(circle(O,sqrt(2)));
draw(E--F--G--H--cycle);
draw(I--... | [] |
98 | Square $ABCD$ is inscribed in a circle. Square $EFGH$ has vertices $E$ and $F$ on $\overline{CD}$ and vertices $G$ and $H$ on the circle. If the area of square $ABCD$ is $1$, then the area of square $EFGH$ can be expressed as $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers and $m < n$. Find $10n... | 2001 AIME II Problem 6 | Let $O$ be the center of the circle, $2a$ be the side length of square $ABCD$, and $2b$ be the side length of square $EFGH$. By symmetry, the horizontal and vertical displacements of $C$ from $O$ are both $\frac{2a}{2} = a$, so by the Pythagorean theorem, the radius of the circle is $OC = \sqrt{a^2+a^2} = a\sqrt{2}$.
... | // Block 1
size(150); pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("4 4") + blue + linewidth(0.7);
pair C=(1,1), D=(1,-1), B=(-1,1), A=(-1,-1), E= (1, -0.2), F=(1, 0.2), G=(1.4, 0.2), H=(1.4, -0.2);
D(MP("A",A)--MP("B",B,N)--MP("C",C,N)--MP("D",D)--cycle); D(MP("E",E,SW)--MP("F",F,NW)--MP("G",G,NE... | [] |
99 | Let $\triangle{PQR}$ be a right triangle with $PQ = 90$, $PR = 120$, and $QR = 150$. Let $C_{1}$ be the inscribed circle. Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$, such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$. Construct $\overline{UV}$ with... | 2001 AIME II Problem 7 | Solution 1 (analytic)
Let $P = (0,0)$ be at the origin. Using the formula $A = rs$ on $\triangle PQR$, where $r_{1}$ is the inradius (similarly define $r_2, r_3$ to be the radii of $C_2, C_3$), $s = \frac{PQ + QR + RP}{2} = 180$ is the semiperimeter, and $A = \frac 12 bh = 5400$ is the area, we find $r_{1} = \frac As ... | // Block 1
pointpen = black; pathpen = black + linewidth(0.7); pair P = (0,0), Q = (90, 0), R = (0, 120), S=(0, 60), T=(45, 60), U = (60,0), V=(60, 40), O1 = (30,30), O2 = (15, 75), O3 = (70, 10); D(MP("P",P)--MP("Q",Q)--MP("R",R,W)--cycle); D(MP("S",S,W) -- MP("T",T,NE)); D(MP("U",U) -- MP("V",V,NE)); D(O2 -- O3, rgb... | [] |
100 | In quadrilateral $ABCD$, $\angle{BAD}\cong\angle{ADC}$ and $\angle{ABD}\cong\angle{BCD}$, $AB = 8$, $BD = 10$, and $BC = 6$. The length $CD$ may be written in the form $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 2001 AIME II Problem 13 | Notice that $\angle BAD = \angle ADC$ and $\angle ABD = \angle DCB$, which motivates us to find similar triangles. We extend $CB$ and $DA$ to point $E$. We know that $\triangle ABD \sim \triangle DCE$. Hence $\angle ADB = \angle DEC$, and $\triangle BDE$ is isosceles. Then $BD = BE = 10$.
Using the similarity, we ha... | // Block 1
/* We arbitrarily set AD = x */
real x = 60^.5, anglesize = 28;
pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("6 6")+linewidth(0.7);
pair A=(0,0), D=(x,0), B=IP(CR(A,8),CR(D,10)), E=(-3x/5,0), C=IP(CR(E,16),CR(D,64/5));
D(MP("A",A)--MP("B",B,NW)--MP("C",C,NW)--MP("D",D)--cycle); D(B--D)... | [] |
101 | Let $EFGH$, $EFDC$, and $EHBC$ be three adjacent square faces of a cube, for which $EC = 8$, and let $A$ be the eighth vertex of the cube. Let $I$, $J$, and $K$ be the points on $\overline{EF}$, $\overline{EH}$, and $\overline{EC}$, respectively, so that $EI = EJ = EK = 2$. A solid $S$ is obtained by drilling a tunnel ... | 2001 AIME II Problem 15 | Set the coordinate system so that vertex $E$, where the drilling starts, is at $(8,8,8)$. Using a little visualization (involving some similar triangles, because we have parallel lines) shows that the tunnel meets the bottom face (the xy plane one) in the line segments joining $(1,0,0)$ to $(2,2,0)$, and $(0,1,0)$ to $... | // Block 1
import three; currentprojection = orthographic(camera=(1/4,2,3/4)); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype("10 2");
triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8);
draw((1,0,0)--(8,0,0)--(8,0,8)--(0,0,8)--(0,0,1)); draw((1,0,0)--(8,0,0)--(8,8,0)--(0,8,0)--(0,1... | [] |
102 | In triangle $ABC$ the medians $\overline{AD}$ and $\overline{CE}$ have lengths $18$ and $27$, respectively, and $AB=24$. Extend $\overline{CE}$ to intersect the circumcircle of $ABC$ at $F$. The area of triangle $AFB$ is $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any ... | 2002 AIME I Problem 13 | Applying Stewart's Theorem to medians $AD, CE$, we have:
\begin{align*} BC^2 + 4 \cdot 18^2 &= 2\left(24^2 + AC^2\right) \\ 24^2 + 4 \cdot 27^2 &= 2\left(AC^2 + BC^2\right) \end{align*}
Substituting the first equation into the second and simplification yields $24^2 = 2\left(3AC^2 + 2 \cdot 24^2 - 4 \cdot 18^2\righ... | // Block 1
size(150); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(8); pair A=(0,0), B=(24,0), E=(A+B)/2, C=IP(CR(A,3*70^.5),CR(E,27)), D=(B+C)/2, F=IP(circumcircle(A,B,C),E--C+2*(E-C));
D(D(MP("A",A))--D(MP("B",B))--D(MP("C",C,NW))--cycle); D(circumcircle(A,B,C)); D(MP("F",F)); D(A--D); D(C--F); D(A--F... | [] |
103 | Polyhedron $ABCDEFG$ has six faces. Face $ABCD$ is a square with $AB = 12;$ face $ABFG$ is a trapezoid with $\overline{AB}$ parallel to $\overline{GF},$ $BF = AG = 8,$ and $GF = 6;$ and face $CDE$ has $CE = DE = 14.$ The other three faces are $ADEG, BCEF,$ and $EFG.$ The distance from $E$ to face $ABCD$ is 12. Give... | 2002 AIME I Problem 15 | Let's put the polyhedron onto a coordinate plane. For simplicity, let the origin be the center of the square: $A(-6,6,0)$, $B(-6,-6,0)$, $C(6,-6,0)$ and $D(6,6,0)$. Since $ABFG$ is an isosceles trapezoid and $CDE$ is an isosceles triangle, we have symmetry about the $xz$-plane.
Therefore, the $y$-component of $E$ is 0... | // Block 1
size(200);
import three;
import graph;
defaultpen(linewidth(0.7)+fontsize(8));
currentprojection=orthographic(-30,50,40);
triple A=(-6,-6,0), B = (-6,6,0), C = (6,6,0), D = (6,-6,0), E = (2,0,12), H=(-6+2*sqrt(19),0,12), H1=(-6-2*sqrt(19),0,12), F, G, E1 = (6,0,12);
F = 1/2*H+1/2*B;
G = 1/2*H+1/2*A;
draw((A-... | [] |
104 | A basketball player has a constant probability of $.4$ of making any given shot, independent of previous shots. Let $a_n$ be the ratio of shots made to shots attempted after $n$ shots. The probability that $a_{10} = .4$ and $a_n\le.4$ for all $n$ such that $1\le n\le9$ is given to be $p^aq^br/\left(s^c\right)$ where $p... | 2002 AIME II Problem 12 | Solution 1
We graph the $10$ shots on a grid. Suppose that a made shot is represented by a step of $(0,1)$, and a missed shot is represented by $(1,0)$. Then the basketball player's shots can be represented by the number of paths from $(0,0)$ to $(6,4)$ that always stay below the line $y=\frac{2x}{3}$. We can find the ... | size(150); for (int i=0;i<7;++i) {draw((i,0)--(i,4));} for (int i=0;i<5;++i) {draw((0,i)--(6,i));} draw((0,0)--(6,4),dashed); label("$1$",(0,0),SE,fontsize(8)); label("$1$",(1,0),SE,fontsize(8)); label("$1$",(2,0),SE,fontsize(8)); label("$1$",(2,1),SE,fontsize(8)); label("$1$",(3,0),SE,fontsize(8)); label("$2$",(3,1),S... | [] |
105 | In triangle $ABC,$ point $D$ is on $\overline{BC}$ with $CD = 2$ and $DB = 5,$ point $E$ is on $\overline{AC}$ with $CE = 1$ and $EA = 3,$ $AB = 8,$ and $\overline{AD}$ and $\overline{BE}$ intersect at $P.$ Points $Q$ and $R$ lie on $\overline{AB}$ so that $\overline{PQ}$ is parallel to $\overline{CA}$ and $\overline{P... | 2002 AIME II Problem 13 | Let $X$ be the intersection of $\overline{CP}$ and $\overline{AB}$.
Since $\overline{PQ} \parallel \overline{CA}$ and $\overline{PR} \parallel \overline{CB}$, $\angle CAB = \angle PQR$ and $\angle CBA = \angle PRQ$. So $\Delta ABC \sim \Delta QRP$, and thus, $\frac{[\Delta PQR]}{[\Delta ABC]} = \left(\frac{PX}{CX}\... | // Block 1
size(10cm);
pair A,B,C,D,E,X,P,Q,R;
A=(0,0);
B=(8,0);
C=(1.9375,3.4994);
D=(3.6696,2.4996);
E=(1.4531,2.6246);
X=(4.3636,0);
P=(2.9639,2.0189);
Q=(1.8462,0);
R=(6.4615,0);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(X);
dot(P);
dot(Q);
dot(R);
label("$A$",A,WSW);
label("$B$",B,ESE);
label("$C$",C,N... | [] |
105 | In triangle $ABC,$ point $D$ is on $\overline{BC}$ with $CD = 2$ and $DB = 5,$ point $E$ is on $\overline{AC}$ with $CE = 1$ and $EA = 3,$ $AB = 8,$ and $\overline{AD}$ and $\overline{BE}$ intersect at $P.$ Points $Q$ and $R$ lie on $\overline{AB}$ so that $\overline{PQ}$ is parallel to $\overline{CA}$ and $\overline{P... | 2002 AIME II Problem 13 | First draw $\overline{CP}$ and extend it so that it meets with $\overline{AB}$ at point $X$.
We have that $[ABC]=\frac{1}{2}\cdot AC \cdot BC\sin{C}=\frac{1}{2}\cdot 4\cdot {7}\sin{C}=14\sin{C}$
By Ceva's, \[3\cdot{\frac{2}{5}}\cdot{\frac{BX}{AX}}=1\implies BX=\frac{5\cdot AX}{6}\] That means that \[\frac{11\cdot {... | // Block 1
size(10cm);
pair A,B,C,D,E,X,P,Q,R;
A=(0,0);
B=(8,0);
C=(1.9375,3.4994);
D=(3.6696,2.4996);
E=(1.4531,2.6246);
X=(4.3636,0);
P=(2.9639,2.0189);
Q=(1.8462,0);
R=(6.4615,0);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(X);
dot(P);
dot(Q);
dot(R);
label("$A$",A,WSW);
label("$B$",B,ESE);
label("$C$",C,N... | [] |
105 | In triangle $ABC,$ point $D$ is on $\overline{BC}$ with $CD = 2$ and $DB = 5,$ point $E$ is on $\overline{AC}$ with $CE = 1$ and $EA = 3,$ $AB = 8,$ and $\overline{AD}$ and $\overline{BE}$ intersect at $P.$ Points $Q$ and $R$ lie on $\overline{AB}$ so that $\overline{PQ}$ is parallel to $\overline{CA}$ and $\overline{P... | 2002 AIME II Problem 13 | Use the mass of point. Denoting the mass of $C=15,B=6,A=5,D=21,E=20$, we can see that the mass of $P$ is $26$, hence we know that $\frac{BP}{PE}=\frac{10}{3}$, now we can find that $\frac{PQ}{AE}=\frac{10}{13}$ which implies $PQ=\frac{30}{13}$, it is obvious that $\triangle{PQR}$ is similar to $\triangle{ACB}$ so we ne... | // Block 1
size(10cm);
pair A,B,C,D,E,P,Q,R;
A=(0,0);
B=(8,0);
C=(1.9375,3.4994);
D=(3.6696,2.4996);
E=(1.4531,2.6246);
P=(2.9639,2.0189);
Q=(1.8462,0);
R=(6.4615,0);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(P);
dot(Q);
dot(R);
label("$A$",A,WSW);
label("$B$",B,ESE);
label("$C$",C,NNW);
label("$D$",D,NE);
l... | [] |
105 | In triangle $ABC,$ point $D$ is on $\overline{BC}$ with $CD = 2$ and $DB = 5,$ point $E$ is on $\overline{AC}$ with $CE = 1$ and $EA = 3,$ $AB = 8,$ and $\overline{AD}$ and $\overline{BE}$ intersect at $P.$ Points $Q$ and $R$ lie on $\overline{AB}$ so that $\overline{PQ}$ is parallel to $\overline{CA}$ and $\overline{P... | 2002 AIME II Problem 13 | Construct $\overline{CP}$ and extend it to line $\overline{AB}$ at point $F$.
Using Ceva's Theorem on triangle $ABC$ and point $P$, we get \[\frac{AF}{BF} \times \frac{BD}{CD} \times \frac{CE}{AE}=1\]
Thus, $\frac{AF}{BF}=\frac{6}{5}$.
Then, using this info, we apply Menelaus on triangle $ACF$ and line $BE$, obtaini... | // Block 1
size(10cm);
pair A,B,C,D,E,F,P,Q,R;
A=(0,0);
B=(8,0);
C=(1.9375,3.4994);
D=(3.6696,2.4996);
E=(1.4531,2.6246);
F=(4.3636,0);
P=(2.9639,2.0189);
Q=(1.8462,0);
R=(6.4615,0);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(F);
dot(P);
dot(Q);
dot(R);
label("$A$",A,WSW);
label("$B$",B,ESE);
label("$C$",C,N... | [] |
106 | Circles $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ intersect at two points, one of which is $(9,6)$, and the product of the radii is $68$. The x-axis and the line $y = mx$, where $m > 0$, are tangent to both circles. It is given that $m$ can be written in the form $a\sqrt {b}/c$, where $a$, $b$, and $c$ are positive integ... | 2002 AIME II Problem 15 | Let $r_1$ and $r_2$ be the radii of the circles. Then the centers of the circles are of the form $(kr_1,r_1)$ and $(kr_2,r_2)$ for the same constant $k,$ since the two centers are collinear with the origin. Since $(9,6)$ lies on both circles,
\[(kr_i - 9)^2 + (r_i - 6)^2 = r^2,\]where $r_i$ represents either radius. Ex... | // Block 1
unitsize(0.25 cm);
pair[] O;
real[] r;
pair P;
r[1] = 4.096;
r[2] = 16.6;
O[1] = (r[1]/(2/3*sqrt(17/13)),r[1]);
O[2] = (r[2]/(2/3*sqrt(17/13)),r[2]);
P = reflect(O[1],O[2])*(9,6);
draw(Circle(O[1],r[1]));
//draw(Circle(O[2],r[2]));
draw(arc(O[2],r[2],130,300));
draw((0,0)--(8,12*sqrt(221)/49*8));
draw((0,... | [] |
107 | Consider the set of points that are inside or within one unit of a rectangular parallelepiped (box) that measures $3$ by $4$ by $5$ units. Given that the volume of this set is $\frac{m + n\pi}{p},$ where $m, n,$ and $p$ are positive integers, and $n$ and $p$ are relatively prime, find $m + n + p.$ | 2003 AIME I Problem 5 | The set can be broken into several parts: the big $3\times 4 \times 5$ parallelepiped, $6$ external parallelepipeds that each share a face with the large parallelepiped and have a height of $1$, the $1/8$ spheres (one centered at each vertex of the large parallelepiped), and the $1/4$ cylinders connecting each adjacent... | // Block 1
size(220);
import three; currentprojection = perspective(5,4,3); defaultpen(linetype("8 8")+linewidth(0.6));
draw(box((0,-.1,0),(0.4,0.6,0.3)));
draw(box((-.1,0,0),(0.5,0.5,0.3)));
draw(box((0,0,-.1),(0.4,0.5,0.4)));
draw(box((0,0,0),(0.4,0.5,0.3)),linewidth(1.2)+linetype("1"));
// Block 2
size(220); import ... | [] |
108 | The sum of the areas of all triangles whose vertices are also vertices of a $1$ by $1$ by $1$ cube is $m + \sqrt{n} + \sqrt{p},$ where $m, n,$ and $p$ are integers. Find $m + n + p.$ | 2003 AIME I Problem 6 | Since there are $8$ vertices of a cube, there are ${8 \choose 3} = 56$ total triangles to consider. They fall into three categories: there are those which are entirely contained within a single face of the cube (whose sides are two edges and one face diagonal), those which lie in a plane perpendicular to one face of... | // Block 1
size(120);
import three;
unitsize(1cm);
draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1));
draw((0,1,1)--(0,1,0)); draw((1,1,1)--(1,1,0)); draw((1,0,1)--(1,0,0));
draw((1,1,1)--(1,0,0)--(1,1,0)--(1,1,1), blue);
// Block 2
size(120);
import three;
unitsize(1cm);
dr... | [] |
109 | Point $B$ is on $\overline{AC}$ with $AB = 9$ and $BC = 21.$ Point $D$ is not on $\overline{AC}$ so that $AD = CD,$ and $AD$ and $BD$ are integers. Let $s$ be the sum of all possible perimeters of $\triangle ACD$. Find $s.$ | 2003 AIME I Problem 7 | Denote the height of $\triangle ACD$ as $h$, $x = AD = CD$, and $y = BD$. Using the Pythagorean theorem, we find that $h^2 = y^2 - 6^2$ and $h^2 = x^2 - 15^2$. Thus, $y^2 - 36 = x^2 - 225 \Longrightarrow x^2 - y^2 = 189$. The LHS is difference of squares, so $(x + y)(x - y) = 189$. As both $x,\ y$ are integers, $x+y,\ ... | // Block 1
size(220);
pointpen = black; pathpen = black + linewidth(0.7);
pair O=(0,0),A=(-15,0),B=(-6,0),C=(15,0),D=(0,8);
D(D(MP("A",A))--D(MP("C",C))--D(MP("D",D,NE))--cycle);
D(D(MP("B",B))--D); D((0,-4)--(0,12),linetype("4 4")+linewidth(0.7));
MP("6",B/2); MP("15",C/2); MP("9",(A+B)/2);
// Block 2
size(220); point... | [] |
110 | Triangle $ABC$ is isosceles with $AC = BC$ and $\angle ACB = 106^\circ.$ Point $M$ is in the interior of the triangle so that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ.$ Find the number of degrees in $\angle CMB.$
Contents
1 Problem
2 Solutions
2.1 Solution 1
2.2 Solution 2
2.3 Solution 3
2.4 Solution 4... | 2003 AIME I Problem 10 | Solution 1
Take point $N$ inside $\triangle ABC$ such that $\angle CBN = 7^\circ$ and $\angle BCN = 23^\circ$.
$\angle MCN = 106^\circ - 2\cdot 23^\circ = 60^\circ$. Also, since $\triangle AMC$ and $\triangle BNC$ are congruent (by ASA), $CM = CN$. Hence $\triangle CMN$ is an equilateral triangle, so $\angle CNM = 60... | // Block 1
pointpen = black; pathpen = black+linewidth(0.7); size(220);
/* We will WLOG AB = 2 to draw following */
pair A=(0,0), B=(2,0), C=(1,Tan(37)), M=IP(A--(2Cos(30),2Sin(30)),B--B+(-2,2Tan(23)));
D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(A--D(MP("M",M))--B); D(C--M);
// Block 2
pointpen = black; pathpen ... | [] |
111 | In convex quadrilateral $ABCD, \angle A \cong \angle C, AB = CD = 180,$ and $AD \neq BC.$ The perimeter of $ABCD$ is $640$. Find $\lfloor 1000 \cos A \rfloor.$ (The notation $\lfloor x \rfloor$ means the greatest integer that is less than or equal to $x.$) | 2003 AIME I Problem 12 | Solution 1
By the Law of Cosines on $\triangle ABD$ at angle $A$ and on $\triangle BCD$ at angle $C$ (note $\angle C = \angle A$),
\[180^2 + AD^2 - 360 \cdot AD \cos A = 180^2 + BC^2 - 360 \cdot BC \cos A\]
\[(AD^2 - BC^2) = 360(AD - BC) \cos A\]
\[(AD - BC)(AD + BC) = 360(AD - BC) \cos A\]
\[(AD + BC) = 360 \cos A\]... | // Block 1
real x = 1.60; /* arbitrary */ pointpen = black; pathpen = black+linewidth(0.7); size(180); real BD = x*x + 1.80*1.80 - 2 * 1.80 * x * 7 / 9; pair A=(0,0),B=(1.8,0),D=IP(CR(A,x),CR(B,BD)),C=OP(CR(D,1.8),CR(B,2.80 - x)); D(MP("A",A)--MP("B",B)--MP("C",C)--MP("D",D,N)--B--A--D); MP("180",(A+B)/2); MP("180",(C... | [] |
112 | In $\triangle ABC, AB = 360, BC = 507,$ and $CA = 780.$ Let $M$ be the midpoint of $\overline{CA},$ and let $D$ be the point on $\overline{CA}$ such that $\overline{BD}$ bisects angle $ABC.$ Let $F$ be the point on $\overline{BC}$ such that $\overline{DF} \perp \overline{BD}.$ Suppose that $\overline{DF}$ meets $\overl... | 2003 AIME I Problem 15 | In the following, let the name of a point represent the mass located there. Since we are looking for a ratio, we assume that $AB=120$, $BC=169$, and $CA=260$ in order to simplify our computations.
First, reflect point $F$ over angle bisector $BD$ to a point $F'$.
As $BD$ is an angle bisector of both triangles $BAC$ ... | // Block 1
size(400); pointpen = black; pathpen = black+linewidth(0.7);
pair A=(0,0),C=(7.8,0),B=IP(CR(A,3.6),CR(C,5.07)), M=(A+C)/2, Da = bisectorpoint(A,B,C), D=IP(B--B+(Da-B)*10,A--C), F=IP(D--D+10*(B-D)*dir(270),B--C), E=IP(B--M,D--F);pair Fprime=2*D-F; /* scale down by 100x */
D(MP("A",A,NW)--MP("B",B,N)--MP("C",... | [] |
113 | In triangle $ABC,$ $AB = 13,$ $BC = 14,$ $AC = 15,$ and point $G$ is the intersection of the medians. Points $A',$ $B',$ and $C',$ are the images of $A,$ $B,$ and $C,$ respectively, after a $180^\circ$ rotation about $G.$ What is the area of the union of the two regions enclosed by the triangles $ABC$ and $A'B'C'?$ | 2003 AIME II Problem 6 | Since a $13-14-15$ triangle is a $5-12-13$ triangle and a $9-12-15$ triangle "glued" together on the $12$ side, $[ABC]=\frac{1}{2}\cdot12\cdot14=84$.
There are six points of intersection between $\Delta ABC$ and $\Delta A'B'C'$. Connect each of these points to $G$.
There are $12$ smaller congruent triangles which... | // Block 1
size(8cm);
pair A,B,C,G,D,E,F,A_1,A_2,B_1,B_2,C_1,C_2;
B=(0,0);
A=(5,12);
C=(14,0);
E=(12.6667,8);
D=(7.6667,-4);
F=(-1.3333,8);
G=(6.3333,4);
B_1=(4.6667,0);
B_2=(1.6667,4);
A_1=(3.3333,8);
A_2=(8,8);
C_1=(11,4);
C_2=(9.3333,0);
dot(A);
dot(B);
dot(C);
dot(G);
dot(D);
dot(E);
dot(F);
dot(A_1);
dot(B_1);
dot... | [] |
114 | Let $A = (0,0)$ and $B = (b,2)$ be points on the coordinate plane. Let $ABCDEF$ be a convex equilateral hexagon such that $\angle FAB = 120^\circ,$ $\overline{AB}\parallel \overline{DE},$ $\overline{BC}\parallel \overline{EF,}$ $\overline{CD}\parallel \overline{FA},$ and the y-coordinates of its vertices are distinct e... | 2003 AIME II Problem 14 | From this image, we can see that the y-coordinate of F is 4, and from this, we can gather that the coordinates of E, D, and C, respectively, are 8, 10, and 6.
Let the angle between the $x$-axis and segment $AB$ be $\theta$, as shown above. Thus, as $\angle FAB=120^\circ$, the angle between the $x$-axis and segment ... | // Block 1
size(200);
draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle);
dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4));
label("$A (0,0)$",(0,0),S);label("$B (b,2)$",(10/sqrt(3),2),SE);label("$C$",(18/sqrt(3),6),E);label("$... | [] |
114 | Let $A = (0,0)$ and $B = (b,2)$ be points on the coordinate plane. Let $ABCDEF$ be a convex equilateral hexagon such that $\angle FAB = 120^\circ,$ $\overline{AB}\parallel \overline{DE},$ $\overline{BC}\parallel \overline{EF,}$ $\overline{CD}\parallel \overline{FA},$ and the y-coordinates of its vertices are distinct e... | 2003 AIME II Problem 14 | First, we see that the y-coordinates of F, E, D, and C must be $4$, $8$, and $10$, and $6$, respectively, as in the previous solutions. We can draw a rectangle around the hexagon ABCDEF and use negative space to find the area of the hexagon. If we call the distance from the foot of the perpendiculars of B and F to A $x... | size(200); draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle); dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4)); label("$A (0,0)$",(0,0),SE);label("$B (b,2)$",(10/sqrt(3),2),SE);label("$C$",(18/sqrt(3),6),E);label("$D$",(10/sq... | [] |
115 | Square $ABCD$ has sides of length 2. Set $S$ is the set of all line segments that have length 2 and whose endpoints are on adjacent sides of the square. The midpoints of the line segments in set $S$ enclose a region whose area to the nearest hundredth is $k$. Find $100k$. | 2004 AIME I Problem 4 | Without loss of generality, let $(0,0)$, $(2,0)$, $(0,2)$, and $(2,2)$ be the vertices of the square. Suppose the endpoints of the segment lie on the two sides of the square determined by the vertex $(0,0)$. Let the two endpoints of the segment have coordinates $(x,0)$ and $(0,y)$. Because the segment has length 2, $... | // Block 1
size(100);
pointpen=black;pathpen = black+linewidth(0.7);
pair A=(0,0),B=(2,0),C=(2,2),D=(0,2);
D(A--B--C--D--A);
picture p;
draw(p,CR(A,1));draw(p,CR(B,1));draw(p,CR(C,1));draw(p,CR(D,1));
clip(p,A--B--C--D--cycle);
add(p);
// Block 2
size(100); pointpen=black;pathpen = black+linewidth(0.7); pair A=(0,0),B... | [] |
115 | Square $ABCD$ has sides of length 2. Set $S$ is the set of all line segments that have length 2 and whose endpoints are on adjacent sides of the square. The midpoints of the line segments in set $S$ enclose a region whose area to the nearest hundredth is $k$. Find $100k$. | 2004 AIME I Problem 4 | If we imagine an arbitrary line with length $2$ connecting two sides of the square, we can draw the rectangle formed by drawing a perpendicular from where that line touches the square.
Drawing the other diagonal of the rectangle, it also has length two, and it bisects with the original line. Since their intersection ... | // Block 1
size(100);
pointpen=black;pathpen = black+linewidth(0.7);
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);
draw(arc((0,2),1,270,360));
draw((0,1)--(1.7,2)); draw((0,2)--(1.7,1)); draw((0,1)--(1.7,1)--(1.7,2));
// Block 2
size(100); pointpen=black;pathpen = black+linewidth(0.7); draw((0,0)--(2,0)--(2,2)--(0,2)--cycl... | [] |
116 | Let $ABC$ be a triangle with sides 3, 4, and 5, and $DEFG$ be a 6-by-7 rectangle. A segment is drawn to divide triangle $ABC$ into a triangle $U_1$ and a trapezoid $V_1$ and another segment is drawn to divide rectangle $DEFG$ into a triangle $U_2$ and a trapezoid $V_2$ such that $U_1$ is similar to $U_2$ and $V_1$ is s... | 2004 AIME I Problem 9 | We let $AB=3, AC=4, DE=6, DG=7$ for the purpose of labeling. Clearly, the dividing segment in $DEFG$ must go through one of its vertices, without loss of generality $D$. The other endpoint ($D'$) of the segment can either lie on $\overline{EF}$ or $\overline{FG}$. $V_2$ is a trapezoid with a right angle then, from whic... | // Block 1
defaultpen(linewidth(0.7));
pair A=(0,0),B=(0,3),C=(4,0);
draw(MP("A",A)--MP("B",B,N)--MP("C",C)--cycle);
draw((9*4/14,0)--(9*4/14,5*3/14),dashed);
label("\(V_1\)",(1,1.2)); label("\(U_1\)",(3,0.3));
// Block 2
defaultpen(linewidth(0.7)); pointpen = black;
pair D=(0,0),E=(0,6),F=(7,6),G=(7,0),H=(4.5,6);
dra... | [] |
117 | A unicorn is tethered by a $20$-foot silver rope to the base of a magician's cylindrical tower whose radius is $8$ feet. The rope is attached to the tower at ground level and to the unicorn at a height of $4$ feet. The unicorn has pulled the rope taut, the end of the rope is $4$ feet from the nearest point on the tower... | 2004 AIME I Problem 14 | Looking from an overhead view, call the center of the circle $O$, the tether point to the unicorn $A$ and the last point where the rope touches the tower $B$. $\triangle OAB$ is a right triangle because $OB$ is a radius and $BA$ is a tangent line at point $B$. We use the Pythagorean Theorem to find the horizontal com... | // Block 1
/* Settings */
import three; defaultpen(fontsize(10)+linewidth(0.62));
currentprojection = perspective(-2,-50,15); size(200);
/* Variables */
real x = 20 - ((750)^.5)/3, CE = 8*(6^.5) - 4*(5^.5), CD = 8*(6^.5), h = 4*CE/CD;
pair Cxy = 8*expi((3*pi)/2-CE/8);
triple Oxy = (0,0,0), A=(4*5^.5,-8,4), B=(0,-8,... | [] |
118 | $ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD.$ The crease is $EF,$ where $E$ is on $AB$ and $F$ is on $CD.$ The dimensions $AE=8, BE=17,$ and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n,$ where $m$ and $n$ are relatively prime posi... | 2004 AIME II Problem 7 | Solution 1 (Synthetic)
Since $EF$ is the perpendicular bisector of $\overline{BB'}$, it follows that $BE = B'E$ (by SAS). By the Pythagorean Theorem, we have $AB' = 15$. Similarly, from $BF = B'F$, we have
\begin{align*} BC^2 + CF^2 = B'D^2 + DF^2 &\Longrightarrow BC^2 + 9 = (BC - 15)^2 + 484 \\ BC &= \frac{70}{3} \e... | pointpen = black; pathpen = black +linewidth(0.7); pair A=(0,0),B=(0,25),C=(70/3,25),D=(70/3,0),E=(0,8),F=(70/3,22),G=(15,0); D(MP("A",A)--MP("B",B,N)--MP("C",C,N)--MP("D",D)--cycle); D(MP("E",E,W)--MP("F",F,(1,0))); D(B--G); D(E--MP("B'",G)--F--B,dashed); MP("8",(A+E)/2,W);MP("17",(B+E)/2,W);MP("22",(D+F)/2,(1,0)); | [] |
119 | Let $ABCD$ be an isosceles trapezoid, whose dimensions are $AB = 6, BC=5=DA,$and $CD=4.$ Draw circles of radius 3 centered at $A$ and $B,$ and circles of radius 2 centered at $C$ and $D.$ A circle contained within the trapezoid is tangent to all four of these circles. Its radius is $\frac{-k+m\sqrt{n}}p,$ where $k, m, ... | 2004 AIME II Problem 12 | Let the radius of the center circle be $r$ and its center be denoted as $O$.
Clearly line $AO$ passes through the point of tangency of circle $A$ and circle $O$. Let $y$ be the height from the base of the trapezoid to $O$. From the Pythagorean Theorem,
\[3^2 + y^2 = (r + 3)^2 \Longrightarrow y = \sqrt {r^2 + 6r}.\]
... | // Block 1
pointpen = black; pathpen = black+linewidth(0.7); pen d = linewidth(0.7) + linetype("4 4"); pen f = fontsize(8);
real r = (-60 + 48 * 3^.5)/23;
pair A=(0,0), B=(6,0), D=(1, 24^.5), C=(5,D.y), O = (3,(r^2 + 6*r)^.5);
D(MP("A",A)--MP("B",B)--MP("C",C,N)--MP("D",D,N)--cycle); D(CR(A,3));D(CR(B,3));D(CR(C,2)... | [] |
120 | Let $ABCDE$ be a convex pentagon with $AB \parallel CE, BC \parallel AD, AC \parallel DE, \angle ABC=120^\circ, AB=3, BC=5,$ and $DE = 15.$ Given that the ratio between the area of triangle $ABC$ and the area of triangle $EBD$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$ | 2004 AIME II Problem 13 | Let the intersection of $\overline{AD}$ and $\overline{CE}$ be $F$. Since $AB \parallel CE, BC \parallel AD,$ it follows that $ABCF$ is a parallelogram, and so $\triangle ABC \cong \triangle CFA$. Also, as $AC \parallel DE$, it follows that $\triangle ABC \sim \triangle EFD$.
By the Law of Cosines, $AC^2 = 3^2 + 5^2... | // Block 1
pointpen = black; pathpen = black+linewidth(0.7);
pair D=(0,0), E=(15,0), F=IP(CR(D, 75/7), CR(E, 45/7)), A=D+ (5+(75/7))/(75/7) * (F-D), C = E+ (3+(45/7))/(45/7) * (F-E), B=IP(CR(A,3), CR(C,5));
D(MP("A",A,(1,0))--MP("B",B,N)--MP("C",C,NW)--MP("D",D)--MP("E",E)--cycle); D(D--A--C--E); D(MP("F",F)); MP("5"... | [] |
121 | In quadrilateral $ABCD,\ BC=8,\ CD=12,\ AD=10,$ and $m\angle A= m\angle B = 60^\circ.$ Given that $AB = p + \sqrt{q},$ where $p$ and $q$ are positive integers, find $p+q.$
Contents
1 Problem
2 Solution
2.1 Solution 1
2.2 Solution 2
2.3 Solution 3
2.4 Solution 4
2.5 Solution 5 (EASY EASY VERY TREMENDOUSLY EASY)
3 S... | 2005 AIME I Problem 7 | Solution 1
Draw line segment $DE$ such that line $DE$ is concurrent with line $BC$. Then, $ABED$ is an isosceles trapezoid so $AD=BE=10$, and $BC=8$ and $EC=2$. We are given that $DC=12$. Since $\angle CED = 120^{\circ}$, using Law of Cosines on $\bigtriangleup CED$ gives \[12^2=DE^2+4-2(2)(DE)(\cos 120^{\circ})\] wh... | draw((0,0)--(20.87,0)--(15.87,8.66)--(5,8.66)--cycle); draw((5,8.66)--(5,0)); draw((15.87,8.66)--(15.87,0)); draw((5,8.66)--(16.87,6.928)); label("$A$",(0,0),SW); label("$B$",(20.87,0),SE); label("$E$",(15.87,8.66),NE); label("$D$",(5,8.66),NW); label("$P$",(5,0),S); label("$Q$",(15.87,0),S); label("$C$",(16.87,7),E); ... | ["https://artofproblemsolving.com/wiki/images/4/41/AIME_2005I_Solution_7_1.png", "https://artofproblemsolving.com/wiki/images/2/2f/AIME_2005I_Solution_7_2.png"] |
122 | A particle moves in the Cartesian plane according to the following rules:
From any lattice point $(a,b),$ the particle may only move to $(a+1,b), (a,b+1),$ or $(a+1,b+1).$
There are no right angle turns in the particle's path.
How many different paths can the particle take from $(0,0)$ to $(5,5)$?
Contents
1 Problem... | 2005 AIME I Problem 13 | Solution 1
The length of the path (the number of times the particle moves) can range from $l = 5$ to $9$; notice that $d = 10-l$ gives the number of diagonals. Let $R$ represent a move to the right, $U$ represent a move upwards, and $D$ to be a move that is diagonal. Casework upon the number of diagonal moves:
Case $... | defaultpen(fontsize(8)+0.8+heavyblue); size(250); for(int i = 0; i<6; ++i) { draw((0,i)--(5,i)^^(i,0)--(i,5), gray+0.25); } label("$(0,0,0)$", (0,0), N); for(int i = 1; i<6; ++i) { label("$(0,0,1)$", (0,i), N); label("$(1,0,0)$", (i,0), N); label("$({"+string(i-1)+"},1,0)$", (i,1), N); label("$(0,1,{"+string(i-1)+"})$... | [] |
123 | Triangle $ABC$ has $BC=20.$ The incircle of the triangle evenly trisects the median $AD.$ If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n.$ | 2005 AIME I Problem 15 | Let $E$, $F$ and $G$ be the points of tangency of the incircle with $BC$, $AC$ and $AB$, respectively. Without loss of generality, let $AC < AB$, so that $E$ is between $D$ and $C$. Let the length of the median be $3m$. Then by two applications of the Power of a Point Theorem, $DE^2 = 2m \cdot m = AF^2$, so $DE = AF... | // Block 1
size(300); pointpen=black;pathpen=black+linewidth(0.65); pen s = fontsize(10);
pair A=(0,0),B=(26,0),C=IP(circle(A,10),circle(B,20)),D=(B+C)/2,I=incenter(A,B,C);
path cir = incircle(A,B,C); pair E1=IP(cir,B--C),F=IP(cir,A--C),G=IP(cir,A--B),P=IP(A--D,cir),Q=OP(A--D,cir);
D(MP("A",A,s)--MP("B",B,s)--MP("C",C,... | [] |
124 | Circles $C_1$ and $C_2$ are externally tangent, and they are both internally tangent to circle $C_3.$ The radii of $C_1$ and $C_2$ are 4 and 10, respectively, and the centers of the three circles are all collinear. A chord of $C_3$ is also a common external tangent of $C_1$ and $C_2.$ Given that the length of the chor... | 2005 AIME II Problem 8 | Let $O_1, O_2, O_3$ be the centers and $r_1 = 4, r_2 = 10,r_3 = 14$ the radii of the circles $C_1, C_2, C_3$. Let $T_1, T_2$ be the points of tangency from the common external tangent of $C_1, C_2$, respectively, and let the extension of $\overline{T_1T_2}$ intersect the extension of $\overline{O_1O_2}$ at a point $H$.... | // Block 1
pointpen = black;
pathpen = black + linewidth(0.7);
size(200);
pair C1 = (-10,0), C2 = (4,0), C3 = (0,0), H = (-10-28/3,0), T = 58/7*expi(pi-acos(3/7));
path cir1 = CR(C1,4.01), cir2 = CR(C2,10), cir3 = CR(C3,14), t = H--T+2*(T-H);
pair A = OP(cir3, t), B = IP(cir3, t), T1 = IP(cir1, t), T2 = IP(cir2, t)... | [] |
125 | Given that $O$ is a regular octahedron, that $C$ is the cube whose vertices are the centers of the faces of $O,$ and that the ratio of the volume of $O$ to that of $C$ is $\frac mn,$ where $m$ and $n$ are relatively prime integers, find $m+n.$
Contents
1 Problem
2 Solutions
2.1 Solution 1
2.2 Solution 2
3 See als... | 2005 AIME II Problem 10 | Solution 1
Let the side of the octahedron be of length $s$. Let the vertices of the octahedron be $A, B, C, D, E, F$ so that $A$ and $F$ are opposite each other and $AF = s\sqrt2$. The height of the square pyramid $ABCDE$ is $\frac{AF}2 = \frac s{\sqrt2}$ and so it has volume $\frac 13 s^2 \cdot \frac s{\sqrt2} = \fr... | // Block 1
import three; currentprojection = perspective(4,-15,4); defaultpen(linewidth(0.7));
draw(box((-1,-1,-1),(1,1,1)));
draw((-3,0,0)--(0,0,3)--(0,-3,0)--(-3,0,0)--(0,0,-3)--(0,-3,0)--(3,0,0)--(0,0,-3)--(0,3,0)--(0,0,3)--(3,0,0)--(0,3,0)--(-3,0,0));
// Block 2
import three; currentprojection = perspective(4,-15,4... | [] |
126 | Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p+q+r.$
Contents
1 Problem
2 Solutions
2.1 So... | 2005 AIME II Problem 12 | Solution 1 (trigonometry)
Let $G$ be the foot of the perpendicular from $O$ to $AB$. Denote $x = EG$ and $y = FG$, and $x > y$ (since $AE < BF$ and $AG = BG$). Then $\tan \angle EOG = \frac{x}{450}$, and $\tan \angle FOG = \frac{y}{450}$.
By the tangent addition rule $\left( \tan (a + b) = \frac{\tan a + \tan b}{1 - ... | // Block 1
size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), G=(4.5,9), O=(4.5,4.5); draw(A--B--C--D--A);draw(E--O--F);draw(G--O); dot(A^^B^^C^^D^^E^^F^^G^^O); label("\(A\)",A,(-1,1));label("\(B\)",B,(1,1));label("\(C\)",C,(1,-1));... | [] |
127 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | 2005 AIME II Problem 14 | By the Law of Sines and since $\angle BAE = \angle CAD, \angle BAD = \angle CAE$, we have
\begin{align*} \frac{CD \cdot CE}{AC^2} &= \frac{\sin CAD}{\sin ADC} \cdot \frac{\sin CAE}{\sin AEC} \\ &= \frac{\sin BAE \sin BAD}{\sin ADB \sin AEB} \\ &= \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{B... | // Block 1
import olympiad; import cse5; import geometry; size(150);
defaultpen(fontsize(10pt));
defaultpen(0.8);
dotfactor = 4;
pair A = origin;
pair C = rotate(15,A)*(A+dir(-50));
pair B = rotate(15,A)*(A+dir(-130));
pair D = extension(A,A+dir(-68),B,C);
pair E = extension(A,A+dir(-82),B,C);
label("$A$",A,N);
label("... | [] |
127 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | 2005 AIME II Problem 14 | Note: We first recall some helpful tips regarding 13, 14, 15 triangles. Drawing an altitude H from B to AC results in AHB being a 5-12-13 right triangle and CHB being a 3-4-5 (9-12-15) right triangle.
Now we start by drawing altitudes from D and E onto AC, labeling them as F and G, and labelling $\angle DAG = \alpha... | // Block 1
import olympiad; import cse5; import geometry; size(300);
defaultpen(fontsize(10pt));
defaultpen(0.8);
dotfactor = 4;
pair A = origin;
pair C = rotate(15,A)*(A+dir(-50));
pair B = rotate(15,A)*(A+dir(-130));
pair D = extension(A,A+dir(-68),B,C);
pair E = extension(A,A+dir(-82),B,C);
label("$A$",A,N);
label("... | [] |
127 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | 2005 AIME II Problem 14 | Diagram borrowed from Solution 1.
Applying Law of Cosines on $\bigtriangleup ABC$ with respect to $\angle C$ we have
\[AB^2=AC^2+BC^2-2(AC)(BC)\cos C\]
Solving gets $\cos C=\frac{3}{5}$, which implies that
\[\sin C=\sqrt{1-\cos C}=\frac{4}{5}\]
Applying Stewart's Theorem with cevian $AD$ we have
\[(BC)(AD)^2+(BC)(BD... | // Block 1
import olympiad; import cse5; import geometry; size(150);
defaultpen(fontsize(10pt));
defaultpen(0.8);
dotfactor = 4;
pair A = origin;
pair C = rotate(15,A)*(A+dir(-50));
pair B = rotate(15,A)*(A+dir(-130));
pair D = extension(A,A+dir(-68),B,C);
pair E = extension(A,A+dir(-82),B,C);
label("$A$",A,N);
label("... | [] |
127 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | 2005 AIME II Problem 14 | We begin by drawing a line through point $B$ parallel to side $AC$. Extend Lines $AE$ and $AD$ to meet the new parallel line at points $F$ and $G$, respectively. This will help us create a lot of similar triangles.
The first pair of similar triangles that is easy to spot is triangle $ACD$ and triangle $GBD$ by AA sim... | // Block 1
// 13–14–15 triangle with AB = 13, AC = 14, BC = 15;
// BF→G extended; “9-x” hugging ED and nudged right; “6” nudged up;
// B and C inverted (B at top, C on right); “13” label shifted left.
import olympiad;
size(350);
pair A = (0,0);
pair B = (5,12); // now AB = 13 (5-0,12-0)
pair C = (14,0); // n... | [] |
128 | Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and ... | 2005 AIME II Problem 15 | Rewrite the given equations as $(x+5)^2 + (y-12)^2 = 256$ and $(x-5)^2 + (y-12)^2 = 16$.
Let $w_3$ have center $(x,y)$ and radius $r$. Now, if two circles with radii $r_1$ and $r_2$ are externally tangent, then the distance between their centers is $r_1 + r_2$, and if they are internally tangent, it is $|r_1 - r_2|$.... | // Block 1
size(220); pointpen = black; pen d = linewidth(0.7); pathpen = d;
pair A = (-5, 12), B = (5, 12), C = (0, 0);
D(CR(A,16));D(CR(B,4));D(shift((0,12)) * yscale(3^.5 / 2) * CR(C,10), linetype("2 2") + d + red);
D((0,30)--(0,-10),Arrows(4));D((15,0)--(-25,0),Arrows(4));D((0,0)--MP("y=ax",(14,14 * (69/100)^.5),E... | [] |
128 | Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and ... | 2005 AIME II Problem 15 | As above, we rewrite the equations as $(x+5)^2 + (y-12)^2 = 256$ and $(x-5)^2 + (y-12)^2 = 16$. Let $F_1=(-5,12)$ and $F_2=(5,12)$. If a circle with center $C=(a,b)$ and radius $r$ is externally tangent to $w_2$ and internally tangent to $w_1$, then $CF_1=16-r$ and $CF_2=4+r$. Therefore, $CF_1+CF_2=20$. In particular, ... | // Block 1
size(220);
pair F1 = (-5, 12), F2 = (5, 12),C=(0,12);
draw(circle(F1,16));
draw(circle(F2,4));
draw(ellipse(C,10,5*sqrt(3)));
xaxis("$x$",Arrows);
yaxis("$y$",Arrows);
dot(F1^^F2^^C);
real l(real x) {return sqrt(69)*x/10;}
path g=graph(l,-7,14);
draw(g);
draw(reflect((0,0),(10,l(10)))*ellipse(C,10,5*sqrt(3... | [] |
129 | In quadrilateral $ABCD$, $\angle B$ is a right angle, diagonal $\overline{AC}$ is perpendicular to $\overline{CD}$, $AB=18$, $BC=21$, and $CD=14$. Find the perimeter of $ABCD$. | 2006 AIME I Problem 1 | We construct the following diagram:
Using the Pythagorean Theorem, we get the following two equations:
\[AD^2 = AC^2 + CD^2\]
\[AC^2 = AB^2 + BC^2\]
Substituting $AB^2 + BC^2$ for $AC^2$ gives us $AD^2 = AB^2 + BC^2 + CD^2$. Plugging in the given information, we get $AD^2 = 18^2 + 21^2 + 14^2 = 961 \implies AD = 31$,... | // Block 1
pathpen = black;
pair C=(0,0),D=(0,-14),A=(-sqrt(765),0),B=IP(circle(C,21),circle(A,18));
D(MP("A",A,W)--MP("B",B,N)--MP("C",C,E)--MP("D",D,E)--A--C);
D(rightanglemark(A,C,D,40));
D(rightanglemark(A,B,C,40));
// Block 2
pathpen = black; pair C=(0,0),D=(0,-14),A=(-sqrt(765),0),B=IP(circle(C,21),circle(A,18));... | [] |
130 | Hexagon $ABCDEF$ is divided into five rhombuses, $\mathcal{P, Q, R, S,}$ and $\mathcal{T,}$ as shown. Rhombuses $\mathcal{P, Q, R,}$ and $\mathcal{S}$ are congruent, and each has area $\sqrt{2006}.$ Let $K$ be the area of rhombus $\mathcal{T}$. Given that $K$ is a positive integer, find the number of possible values f... | 2006 AIME I Problem 8 | To determine the possible values of $[GIHJ],$ we must determine the maximum and minimum possible areas.
In the case where the $4$ rhombi are squares, we have $[GIHJ]=0,$ implying the minimum possible positive-integer-valued area is $1.$
Denote the length $HC=a$ and $KH=b.$ We have
\[KI=\sqrt{a^2-b^2}\]
by the Pythago... | // Block 1
size(8cm);
pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6);
draw(A--B--C--D--EE--F--cycle);
label("$A$",A,2*N);
label("$B$",B,2*N);
label("$C$",C,2*E);
label("$D$",D,2*S);
label("$E$",EE,2*S);
label("$F$",F... | [] |
131 | Eight circles of diameter 1 are packed in the first quadrant of the coordinate plane as shown. Let region $\mathcal{R}$ be the union of the eight circular regions. Line $l,$ with slope 3, divides $\mathcal{R}$ into two regions of equal area. Line $l$'s equation can be expressed in the form $ax=by+c,$ where $a, b,$ and ... | 2006 AIME I Problem 10 | Solution 1
The line passing through the tangency point of the bottom left circle and the one to its right and through the tangency of the top circle in the middle column and the one beneath it is the line we are looking for: a line passing through the tangency of two circles cuts congruent areas, so our line cuts throu... | // Block 1
size(150);defaultpen(linewidth(0.7));
draw((6.5,0)--origin--(0,6.5), Arrows(5));
int[] array={3,3,2};
int i,j;
for(i=0; i<3; i=i+1) {
for(j=0; j<array[i]; j=j+1) {
draw(Circle((1+2*i,1+2*j),1));
}}
label("x", (7,0));
label("y", (0,7));
draw((5/3,0)--(23/6,6.5),blue);
// Block 2
size(150);defaultpen(linewidth... | [] |
132 | A tripod has three legs each of length $5$ feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is $4$ feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let $h$ be the height in feet of the top o... | 2006 AIME I Problem 14 | We will use $[...]$ to denote volume (four letters), area (three letters) or length (two letters).
Let $T$ be the top of the tripod, $A,B,C$ are end points of three legs. Let $S$ be the point on $TA$ such that $[TS] = 4$ and $[SA] = 1$. Let $O$ be the center of the base equilateral triangle $ABC$. Let $M$ be the midpo... | // Block 1
size(200);
import three; pointpen=black;pathpen=black+linewidth(0.65);pen ddash = dashed+linewidth(0.65);
currentprojection = perspective(1,-10,3.3);
triple O=(0,0,0),T=(0,0,5),C=(0,3,0),A=(-3*3^.5/2,-3/2,0),B=(3*3^.5/2,-3/2,0);
triple M=(B+C)/2,S=(4*A+T)/5;
draw(T--S--B--T--C--B--S--C);draw(B--A--C--A--S,dd... | [] |
132 | A tripod has three legs each of length $5$ feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is $4$ feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let $h$ be the height in feet of the top o... | 2006 AIME I Problem 14 | Diagram borrowed from Solution 1
Apply Pythagorean Theorem on $\bigtriangleup TOB$ yields
\[BO=\sqrt{TB^2-TO^2}=3\]
Since $\bigtriangleup ABC$ is equilateral, we have $\angle MOB=60^{\circ}$ and
\[BC=2BM=2(OB\sin MOB)=3\sqrt{3}\]
Apply Pythagorean Theorem on $\bigtriangleup TMB$ yields
\[TM=\sqrt{TB^2-BM^2}=\sqrt{5^... | // Block 1
size(200);
import three; pointpen=black;pathpen=black+linewidth(0.65);pen ddash = dashed+linewidth(0.65);
currentprojection = perspective(1,-10,3.3);
triple O=(0,0,0),T=(0,0,5),C=(0,3,0),A=(-3*3^.5/2,-3/2,0),B=(3*3^.5/2,-3/2,0);
triple M=(B+C)/2,S=(4*A+T)/5;
draw(T--S--B--T--C--B--S--C);draw(B--A--C--A--S,dd... | [] |
133 | In convex hexagon $ABCDEF$, all six sides are congruent, $\angle A$ and $\angle D$ are right angles, and $\angle B, \angle C, \angle E,$ and $\angle F$ are congruent. The area of the hexagonal region is $2116(\sqrt{2}+1).$ Find $AB$. | 2006 AIME II Problem 1 | Because $\angle B$, $\angle C$, $\angle E$, and $\angle F$ are congruent, the degree-measure of each of them is ${{720-2\cdot90}\over4}= 135$. Lines $BF$ and $CE$ divide the hexagonal region into two right triangles and a rectangle. Let $AB=x$. Then $BF=x\sqrt2$. Thus
\begin{align*} 2116(\sqrt2+1)&=[ABCDEF]\\ &=2\cdot ... | // Block 1
pair A,B,C,D,E,F;
A=(0,0);
B=(7,0);
C=(13,6);
E=(6,13);
D=(13,13);
F=(0,7);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(F);
draw(A--B--C--D--E--F--cycle,linewidth(0.7));
label("{\tiny $A$}",A,S);
label("{\tiny $B$}",B,S);
label("{\tiny $C$}",C,S);
label("{\tiny $D$}",D,N);
label("{\tiny $E$}",E,N);
label("{\... | [] |
134 | Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral. A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqrt... | 2006 AIME II Problem 6 | Call the vertices of the new square A', B', C', and D', in relation to the vertices of $ABCD$, and define $s$ to be one of the sides of that square. Since the sides are parallel, by corresponding angles and AA~ we know that triangles $AA'D'$ and $D'C'E$ are similar. Thus, the sides are proportional: $\frac{AA'}{A'D'} =... | // Block 1
unitsize(32mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=3;
pair B = (0, 0), C = (1, 0), D = (1, 1), A = (0, 1);
pair Ep = (2 - sqrt(3), 0), F = (1, sqrt(3) - 1);
pair Ap = (0, (3 - sqrt(3))/6);
pair Cp = ((3 - sqrt(3))/6, 0);
pair Dp = ((3 - sqrt(3))/6, (3 - sqrt(3))/6);
pair[] dots = {A, B, C... | [] |
134 | Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral. A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqrt... | 2006 AIME II Problem 6 | Since $AEF$ is equilateral, $AE=EF$. Let $BE=x$. By the Pythagorean theorem, $1+x^2=2(1-x)^2$. Simplifying, we get $x^2-4x+1=0$. By the quadratic formula, the roots are $2 \pm \sqrt{3}$. Since $x<1$, we discard the root with the "+", giving $x=2-\sqrt{3}$.
Let the side length of the square be s. Since $MEK$ is simila... | // Block 1
real n;
n=0.26794919243;
real m;
m=0.2113248654;
draw((0,0)--(0,n)--(1,0)--(0,0));
draw((0,m)--(m,m)--(m,0));
label((0,0), "$B$",SW);
label((0,n), "$E$",SW);
label((0,m), "$M$",SW);
label((1,0), "$A$",SW);
label((m,0), "$N$",SW);
label((m,m), "$K$",NE);
// Block 2
real n; n=0.26794919243; real m; m=0.2113248... | [] |
135 | Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$. Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overlin... | 2006 AIME II Problem 12 | Notice that $\angle{E} = \angle{BGC} = 120^\circ$ because $\angle{A} = 60^\circ$. Also, $\angle{GBC} = \angle{GAC} = \angle{FAE}$ because they both correspond to arc ${GC}$. So $\Delta{GBC} \sim \Delta{EAF}$.
\[[EAF] = \frac12 (AE)(EF)\sin \angle AEF = \frac12\cdot11\cdot13\cdot\sin{120^\circ} = \frac {143\sqrt3}4.\]... | // Block 1
size(250); pointpen = black; pathpen = black + linewidth(0.65); pen s = fontsize(8);
pair A=(0,0),B=(-3^.5,-3),C=(3^.5,-3),D=13*expi(-2*pi/3),E1=11*expi(-pi/3),F=E1+D;
path O = CP((0,-2),A); pair G = OP(A--F,O);
D(MP("A",A,N,s)--MP("B",B,W,s)--MP("C",C,E,s)--cycle);D(O);
D(B--MP("D",D,W,s)--MP("F",F,s)--MP("... | [] |
136 | In isosceles triangle $\triangle ABC$, $A$ is located at the origin and $B$ is located at $(20,0)$. Point $C$ is in the first quadrant with $AC = BC$ and angle $BAC = 75^{\circ}$. If triangle $ABC$ is rotated counterclockwise about point $A$ until the image of $C$ lies on the positive $y$-axis, the area of the region... | 2007 AIME I Problem 12 | Solution 1
Let the new triangle be $\triangle AB'C'$ ($A$, the origin, is a vertex of both triangles). Let $\overline{B'C'}$ intersect with $\overline{AC}$ at point $D$, $\overline{BC}$ intersect with $\overline{B'C'}$ at $E$, and $\overline{BC}$ intersect with $\overline{AB'}$ at $F$. The region common to both triangl... | // Block 1
defaultpen(fontsize(12)+0.6); size(300);
var theta=15;
pair A=origin, B=(20,0), C=extension(A,dir(75),B/2,bisectorpoint(A,B)), Cp=rotate(theta,A)*C, Bp=rotate(theta,A)*B, X=extension(A,Bp,B,C), Y=extension(B,C,Bp,Cp);
draw(A--B--C--A); draw(A--Bp--Cp--A, royalblue);
markscalefactor=0.1;
draw(rightanglemark(Y... | ["https://artofproblemsolving.com/wiki/images/thumb/2/25/I-07-12.png/250px-I-07-12.png"] |
137 | Square $ABCD$ has side length $13$, and points $E$ and $F$ are exterior to the square such that $BE=DF=5$ and $AE=CF=12$. Find $EF^{2}$.
Contents
1 Problem
2 Solution
2.1 Solution 1
2.2 Solution 2
2.3 Solution 3
2.4 Solution 4
2.5 Solution 5 (Ptolemy's Theorem)
2.6 Solution 6 (Coordinate Bash)
2.7 Solution 7 (Trig ... | 2007 AIME II Problem 3 | Solution 1
Let $\angle FCD = \alpha$, so that $FB = \sqrt{12^2 + 13^2 + 2\cdot12\cdot13\sin(\alpha)} = \sqrt{433}$. By the diagonal, $DB = 13\sqrt{2}, DB^2 = 338$.
The sum of the squares of the sides of a parallelogram is the sum of the squares of the diagonals.
\[EF^2 = 2\cdot(5^2 + 433) - 338 = 578.\]
Solution 2
E... | unitsize(0.25 cm); pair A, B, C, D, E, F, G, H; A = (0,13); B = (13,13); C = (13,0); D = (0,0); E = A + (12*12/13,5*12/13); F = D + (5*5/13,-5*12/13); G = rotate(90,(A + C)/2)*(E); H = rotate(90,(A + C)/2)*(F); draw(A--B--C--D--cycle); draw(E--G--F--H--cycle); dot("$A$", A, N); dot("$B$", B, dir(0)); dot("$C$", C, ... | ["https://artofproblemsolving.com/wiki/images/5/59/AIME_II_prob10_bruteforce.PNG"] |
138 | Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$, $\omega_{B}$ to $BC$ and $BA$, $\omega_{C}$ to $CA$ and $CB$, and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ a... | 2007 AIME II Problem 15 | First, apply Heron's formula to find that $[ABC] = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84$. The semiperimeter is $21$, so the inradius is $\frac{A}{s} = \frac{84}{21} = 4$.
Now consider the incenter $I$ of $\triangle ABC$. Let the radius of one of the small circles be $r$. Let the centers of the three little circles ... | // Block 1
defaultpen(fontsize(12)+0.8); size(350);
pair A,B,C,X,Y,Z,P,Q,R,Zp;
B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C);
real r=260/129;
Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A);
R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C); Zp=circumcent... | [] |
138 | Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$, $\omega_{B}$ to $BC$ and $BA$, $\omega_{C}$ to $CA$ and $CB$, and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ a... | 2007 AIME II Problem 15 | Let $O_A, O_B, O_C, O$ be the centers of $w_A, w_B, w_C, w$, respectively. Also, let $I$ be the incenter of $ABC$ and $r$ be the radius of circle $w$. Since $AB||O_AO_B$, $BC||O_BO_C$, and $CA||O_CO_A$, we know that
\[\angle BAI = \angle O_BO_AI, \angle CBI = \angle O_CO_BI, \angle ACI = \angle O_AO_CI \text{ and }\a... | // Block 1
defaultpen(fontsize(12)+0.8); size(300);
pair A,B,C,X,Y,Z,P,Q,R;
B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C);
real r=260/129;
Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A);
R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C);
draw(A--B--C--A)... | [] |
139 | Square $AIME$ has sides of length $10$ units. Isosceles triangle $GEM$ has base $EM$, and the area common to triangle $GEM$ and square $AIME$ is $80$ square units. Find the length of the altitude to $EM$ in $\triangle GEM$. | 2008 AIME I Problem 2 | Note that if the altitude of the triangle is at most $10$, then the maximum area of the intersection of the triangle and the square is $5\cdot10=50$.
This implies that vertex G must be located outside of square $AIME$.
Let $GE$ meet $AI$ at $X$ and let $GM$ meet $AI$ at $Y$. Clearly, $XY=6$ since the area of trapezoi... | // Block 1
pair E=(0,0), M=(10,0), I=(10,10), A=(0,10);
draw(A--I--M--E--cycle);
pair G=(5,25);
draw(G--E--M--cycle);
label("\(G\)",G,N);
label("\(A\)",A,NW);
label("\(I\)",I,NE);
label("\(M\)",M,NE);
label("\(E\)",E,NW);
label("\(10\)",(M+E)/2,S);
// Block 2
pair E=(0,0), M=(10,0), I=(10,10), A=(0,10); draw(A--I--M--E... | [] |
140 | Let $ABCD$ be an isosceles trapezoid with $\overline{AD}||\overline{BC}$ whose angle at the longer base $\overline{AD}$ is $\dfrac{\pi}{3}$. The diagonals have length $10\sqrt {21}$, and point $E$ is at distances $10\sqrt {7}$ and $30\sqrt {7}$ from vertices $A$ and $D$, respectively. Let $F$ be the foot of the altitud... | 2008 AIME I Problem 10 | Key observation. $AD = 20\sqrt{7}$.
Proof 1. By the triangle inequality, we can immediately see that $AD \geq 20\sqrt{7}$. However, notice that $10\sqrt{21} = 20\sqrt{7}\cdot\sin\frac{\pi}{3}$, so by the law of sines, when $AD = 20\sqrt{7}$, $\angle ACD$ is right and the circle centered at $A$ with radius $10\sqrt{21}... | // Block 1
size(300);
defaultpen(1);
pair A=(0,0), D=(4,0), B= A+2 expi(1/3*pi), C= D+2expi(2/3*pi), E=(-4/3,0), F=(3,0);
draw(F--C--B--A);
draw(E--A--D--C);
draw(A--C,dashed);
draw(circle(A,abs(C-A)),dotted);
label("\(A\)",A,S);
label("\(B\)",B,NW);
label("\(C\)",C,NE);
label("\(D\)",D,SE);
label("\(E\)",E,N);
label("... | [] |
141 | Let
\[p(x,y) = a_0 + a_1x + a_2y + a_3x^2 + a_4xy + a_5y^2 + a_6x^3 + a_7x^2y + a_8xy^2 + a_9y^3.\]
Suppose that
\[p(0,0) = p(1,0) = p( - 1,0) = p(0,1) = p(0, - 1)\\ = p(1,1) = p(1, - 1) = p(2,2) = 0.\]
There is a point $\left(\frac {a}{c},\frac {b}{c}\right)$ for which $p\left(\frac {a}{c},\frac {b}{c}\right) = 0... | 2008 AIME I Problem 13 | Solution 1
\begin{align*} p(0,0) &= a_0 \\ &= 0 \\ p(1,0) &= a_0 + a_1 + a_3 + a_6 \\ &= a_1 + a_3 + a_6 \\ &= 0 \\ p(-1,0) &= -a_1 + a_3 - a_6 \\ &= 0 \end{align*}
Adding the above two equations gives $a_3 = 0$, and so we can deduce that $a_6 = -a_1$.
Similarly, plugging in $(0,1)$ and $(0,-1)$ gives $a_5 = 0$ and $... | // Block 1
unitsize(1.2 cm); real upperhyper (real x) { return(sqrt((3*x^2 - 3*x + 2)/2)); } real lowerhyper (real x) { return(-sqrt((3*x^2 - 3*x + 2)/2)); } int i; for (i = -3; i <= 3; ++i) { draw((-3,i)--(3,i),gray(0.7)); draw((i,-3)--(i,3),gray(0.7)); } draw((0,-3)--(0,3),red); draw((1,-3)--(1,3),red); draw((-3... | [] |
142 | Let $\overline{AB}$ be a diameter of circle $\omega$. Extend $\overline{AB}$ through $A$ to $C$. Point $T$ lies on $\omega$ so that line $CT$ is tangent to $\omega$. Point $P$ is the foot of the perpendicular from $A$ to line $CT$. Suppose $\overline{AB} = 18$, and let $m$ denote the maximum possible length of segment ... | 2008 AIME I Problem 14 | Solution 1
Let $x = OC$. Since $OT, AP \perp TC$, it follows easily that $\triangle APC \sim \triangle OTC$. Thus $\frac{AP}{OT} = \frac{CA}{CO} \Longrightarrow AP = \frac{9(x-9)}{x}$. By the Law of Cosines on $\triangle BAP$,
\begin{align*}BP^2 = AB^2 + AP^2 - 2 \cdot AB \cdot AP \cdot \cos \angle BAP \end{align*}
w... | // Block 1
size(250); defaultpen(0.70 + fontsize(10)); import olympiad; pair O = (0,0), B = O - (9,0), A= O + (9,0), C=A+(18,0), T = 9 * expi(-1.2309594), P = foot(A,C,T); draw(Circle(O,9)); draw(B--C--T--O); draw(A--P); dot(A); dot(B); dot(C); dot(O); dot(T); dot(P); draw(rightanglemark(O,T,C,30)); draw(rightanglemark... | [] |
142 | Let $\overline{AB}$ be a diameter of circle $\omega$. Extend $\overline{AB}$ through $A$ to $C$. Point $T$ lies on $\omega$ so that line $CT$ is tangent to $\omega$. Point $P$ is the foot of the perpendicular from $A$ to line $CT$. Suppose $\overline{AB} = 18$, and let $m$ denote the maximum possible length of segment ... | 2008 AIME I Problem 14 | (Diagram Credits to Solution 2)
We can see that $AP = \frac{9x}{x+9}$. From the Law of Cosines, we can now find the length of BP.
$BP^2 = (\frac{9x}{x+9})^2 + 18^2 - 2(18)(\frac{9x}{x+9})(-\frac{9}{x+9})$
$BP^2 = 81(\frac{x^2}{(x+9)^2} + 4 + \frac{36x}{(x+9)^2})$
$BP^2 = 81(\frac{x^2+36x}{(x+9)^2}+4)$
From here, w... | // Block 1
size(250); defaultpen(0.70 + fontsize(10)); import olympiad;
pair O = (0,0), B = O - (9,0), A= O + (9,0), C=A+(18,0), T = 9 * expi(-1.2309594), P = foot(A,C,T);
draw(Circle(O,9)); draw(B--C--T--O); draw(A--P); dot(A); dot(B); dot(C); dot(O); dot(T); dot(P);
draw(rightanglemark(O,T,C,30)); draw(rightanglemark... | [] |
143 | A square piece of paper has sides of length $100$. From each corner a wedge is cut in the following manner: at each corner, the two cuts for the wedge each start at a distance $\sqrt{17}$ from the corner, and they meet on the diagonal at an angle of $60^{\circ}$ (see the figure below). The paper is then folded up along... | 2008 AIME I Problem 15 | In the original picture, let $P$ be the corner, and $M$ and $N$ be the two points whose distance is $\sqrt{17}$ from $P$. Also, let $R$ be the point where the two cuts intersect.
Using $\triangle{MNP}$ (a 45-45-90 triangle), $MN=MP\sqrt{2}\quad\Longrightarrow\quad MN=\sqrt{34}$. $\triangle{MNR}$ is equilateral, so $MR... | // Block 1
import cse5;
size(200);
pathpen=black;
real s=sqrt(17), r=(sqrt(51)+s)/(sqrt(2));
pair P=(0,0), N=(0,sqrt(17)), M=(sqrt(17),0), R=r*dir(45), G=((sqrt(51)+sqrt(17))/2,0);
D(2*N--P--2*M); D(N--R--M); D(P--R);
D((R.x,2*N.y)--R--(2*M.x,R.y));
MP("30^\circ",R-(0.25,1),SW);
MP("30^\circ",R-(1,0.5),SW);
MP("\sqrt{1... | [] |
144 | In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$, let $BC = 1000$ and $AD = 2008$. Let $\angle A = 37^\circ$, $\angle D = 53^\circ$, and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$, respectively. Find the length $MN$. | 2008 AIME II Problem 5 | Solution 1
Extend $\overline{AB}$ and $\overline{CD}$ to meet at a point $E$. Then $\angle AED = 180 - 53 - 37 = 90^{\circ}$.
As $\angle AED = 90^{\circ}$, note that the midpoint of $\overline{AD}$, $N$, is the center of the circumcircle of $\triangle AED$. We can do the same with the circumcircle about $\triangle BE... | // Block 1
size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); draw(A--B--C--D--cycle); draw(B--E--C,dashed); draw(M[0]--N); draw(N--E,dashed); draw(rightangle... | [] |
145 | The diagram below shows a $4\times4$ rectangular array of points, each of which is $1$ unit away from its nearest neighbors.
Define a growing path to be a sequence of distinct points of the array with the property that the distance between consecutive points of the sequence is strictly increasing. Let $m$ be the maxi... | 2008 AIME II Problem 10 | We label our points using coordinates $0 \le x,y \le 3$, with the bottom-left point being $(0,0)$. By the Pythagorean Theorem, the distance between two points is $\sqrt{d_x^2 + d_y^2}$ where $0 \le d_x, d_y \le 3$; these yield the possible distances (in decreasing order)
\[\sqrt{18},\ \sqrt{13},\ \sqrt{10},\ \sqrt{9},\... | // Block 1
unitsize(0.25inch);
defaultpen(linewidth(0.7)); dotfactor = 4; pen s = linewidth(4);
int i, j;
for(i = 0; i < 4; ++i)
for(j = 0; j < 4; ++j)
dot(((real)i, (real)j));
dot((0,0)^^(3,3),s); draw((0,0)--(3,3));
// Block 2
unitsize(0.25inch);
defaultpen(linewidth(0.7)); dotfactor = 4; pen s = linewidth(4); pe... | [] |
146 | In triangle $ABC$, $AB = AC = 100$, and $BC = 56$. Circle $P$ has radius $16$ and is tangent to $\overline{AC}$ and $\overline{BC}$. Circle $Q$ is externally tangent to $P$ and is tangent to $\overline{AB}$ and $\overline{BC}$. No point of circle $Q$ lies outside of $\triangle ABC$. The radius of circle $Q$ can be expr... | 2008 AIME II Problem 11 | Let $X$ and $Y$ be the feet of the perpendiculars from $P$ and $Q$ to $BC$, respectively. Let the radius of $\odot Q$ be $r$. We know that $PQ = r + 16$. From $Q$ draw segment $\overline{QM} \parallel \overline{BC}$ such that $M$ is on $PX$. Clearly, $QM = XY$ and $PM = 16-r$. Also, we know $QPM$ is a right triangle.
... | // Block 1
size(200);
pathpen=black;pointpen=black;pen f=fontsize(9);
real r=44-6*35^.5;
pair A=(0,96),B=(-28,0),C=(28,0),X=C-(64/3,0),Y=B+(4*r/3,0),P=X+(0,16),Q=Y+(0,r),M=foot(Q,X,P);
path PC=CR(P,16),QC=CR(Q,r);
D(A--B--C--cycle); D(Y--Q--P--X); D(Q--M); D(P--C,dashed);
D(PC); D(QC);
MP("A",A,N,f);MP("B",B,f);MP("C"... | [] |
147 | A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let $R$ be the region outside the hexagon, and let $S = \left\lbrace\frac{1}{z}|z \in R\right\rbrace$. Then the area of $S$ has the form $a\pi + \sqrt{b}$, wh... | 2008 AIME II Problem 13 | If a point $z = r\text{cis}\,\theta$ is in $R$, then the point $\frac{1}{z} = \frac{1}{r} \text{cis}\, \left(-\theta\right)$ is in $S$ (where cis denotes $\text{cis}\, \theta = \cos \theta + i \sin \theta$). Since $R$ is symmetric every $60^{\circ}$ about the origin, it suffices to consider the area of the result of th... | // Block 1
unitsize(1.5cm);
defaultpen(linewidth(0.7)); picture p; real max = .5 + 1/3^.5; pen d = linetype("4 4"); fill(1.5*expi(-pi/6)--arc((0,0),1,-30,30)--1.5*expi(pi/6)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--cycle,rgb(0.5,0.5,1));
draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);for(int i = 0; i < 6; ++i) a... | [] |
148 | There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that
\[\frac {z}{z + n} = 4i.\]
Find $n$. | 2009 AIME I Problem 2 | Below is an image of the complex plane. Let $\operatorname{Im}(z)$ denote the imaginary part of a complex number $z$.
$z$ must lie on the line $\operatorname{Im}(z)=164$. $z+n$ must also lie on the same line, since $n$ is real and does not affect the imaginary part of $z$.
Consider $z$ and $z+n$ in terms of their mag... | // Block 1
unitsize(1cm);
xaxis("Re",Arrows);
yaxis("Im",Arrows);
real f(real x) {return 164;}
pair F(real x) {return (x,f(x));}
draw(graph(f,-700,100),red,Arrows);
label("Im$(z)=164$",F(-330),N);
pair z = (-656,164);
dot(Label("$z$",align=N),z);
dot(Label("$z+n$",align=N),z+(697,0));
draw(Label("$4x$"),z--(0,0... | [] |
149 | In parallelogram $ABCD$, point $M$ is on $\overline{AB}$ so that $\frac {AM}{AB} = \frac {17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac {AN}{AD} = \frac {17}{2009}$. Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$. Find $\frac {AC}{AP}$. | 2009 AIME I Problem 4 | We approach the problem using mass points on triangle $ABD$ as displayed below.
But as $MN$ does not protrude from a vertex, we will have to "split the mass" at point $A$. First, we know that $DO$ is congruent to $BO$ because diagonals of parallelograms bisect each other. Therefore, we can assign equal masses to poi... | // Block 1
pair A=(0,0),B=(50,0),D=(10,40),C=B+D,M=(8,0),NN=(2,8);
draw(A--B--C--D--cycle);
draw(B--D^^A--C^^M--NN);
pair O=extension(A,C,B,D);
pair P=extension(A,C,M,NN);
dot(A);dot(B);dot(C);dot(D);dot(O);dot(M);dot(NN);dot(P);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,NW);
label("$M$",M,... | [] |
150 | Triangle $ABC$ has $AC = 450$ and $BC = 300$. Points $K$ and $L$ are located on $\overline{AC}$ and $\overline{AB}$ respectively so that $AK = CK$, and $\overline{CL}$ is the angle bisector of angle $C$. Let $P$ be the point of intersection of $\overline{BK}$ and $\overline{CL}$, and let $M$ be the point on line $BK$ f... | 2009 AIME I Problem 5 | Since $K$ is the midpoint of $\overline{PM}$ and $\overline{AC}$, quadrilateral $AMCP$ is a parallelogram, which implies $AM||LP$ and $\bigtriangleup{AMB}$ is similar to $\bigtriangleup{LPB}$
Thus,
\[\frac {AM}{LP}=\frac {AB}{LB}=\frac {AL+LB}{LB}=\frac {AL}{LB}+1\]
Now let's apply the angle bisector theorem.
\[\f... | // Block 1
import markers;
defaultpen(fontsize(8));
size(300);
pair A=(0,0), B=(30*sqrt(331),0), C, K, L, M, P;
C = intersectionpoints(Circle(A,450), Circle(B,300))[0];
K = midpoint(A--C);
L = (3*B+2*A)/5;
P = extension(B,K,C,L);
M = 2*K-P;
draw(A--B--C--cycle);
draw(C--L);draw(B--M--A);
markangle(n=1,radius=15,A,C,L,... | [] |
151 | In right $\triangle ABC$ with hypotenuse $\overline{AB}$, $AC = 12$, $BC = 35$, and $\overline{CD}$ is the altitude to $\overline{AB}$. Let $\omega$ be the circle having $\overline{CD}$ as a diameter. Let $I$ be a point outside $\triangle ABC$ such that $\overline{AI}$ and $\overline{BI}$ are both tangent to circle $\o... | 2009 AIME I Problem 12 | Let $AP=AD=x$, let $BQ=BD=y$, and let $IP=IQ=z$. Let $OD=r$. We find $AB=37$. Let $\alpha$, $\beta$, and $\theta$ be the angles $OAD$, $OBD$, and $OPI$ respectively. Then $\alpha + \beta + \theta = 90^\circ$, so \[\theta = 90^\circ - (\alpha+\beta).\]
The perimeter of $\triangle ABI$ is $2(x+y+z)=2(37+z)$. The desired... | // Block 1
size(300);
defaultpen(linewidth(0.4)+fontsize(10));
pen s = linewidth(0.8)+fontsize(8);
pair A,B,C,D,O,X;
C=origin;
A=(0,12);
B=(18,0);
D=foot(C,A,B);
O = (C+D)/2;
real r = length(D-C)/2;
path c = CR(O, r);
pair OA = (O+A)/2;
real rA = length(A-O)/2;
pair Ap = OP(CR(OA,rA), c);
pair OB = (O+B)/2;
real rB = ... | [] |
152 | In rectangle $ABCD$, $AB=100$. Let $E$ be the midpoint of $\overline{AD}$. Given that line $AC$ and line $BE$ are perpendicular, find the greatest integer less than $AD$. | 2009 AIME II Problem 3 | Solution 1
From the problem, $AB=100$ and triangle $FBA$ is a right triangle. As $ABCD$ is a rectangle, triangles $BCA$, and $ABE$ are also right triangles. By $AA$, $\triangle FBA \sim \triangle BCA$, and $\triangle FBA \sim \triangle ABE$, so $\triangle ABE \sim \triangle BCA$. This gives $\frac {AE}{AB}= \frac {AB}... | // Block 1
pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5); draw (A--B--C--D--cycle); pair E=(7,10); draw (B--E); draw (A--C); pair F=(6.7,6.7); label("\(E\)",E,N); label("\(A\)",A,NW); label("\(B\)",B,SW); label("\(C\)",C,SE); label("\(D\)",D,NE); label("\(F\)",F,W); label("\(100\)",Q,W);
// Block 2
pair A=(0,10)... | [] |
153 | Four lighthouses are located at points $A$, $B$, $C$, and $D$. The lighthouse at $A$ is $5$ kilometers from the lighthouse at $B$, the lighthouse at $B$ is $12$ kilometers from the lighthouse at $C$, and the lighthouse at $A$ is $13$ kilometers from the lighthouse at $C$. To an observer at $A$, the angle determined by ... | 2009 AIME II Problem 10 | Let $CD$ and $AB$ meet at $A'$; then $\overline{AD}$ is an angle bisector of isosceles $\triangle AA'C$. Then by the Angle Bisector Theorem, $A'D=\frac {10}{10+13} \cdot 13=\frac {130}{23}$, and $\cos \angle DA'A=\frac {5}{13}$. By the Law of Cosines on $\triangle AA'D$, we have \[AD^2=10^2+\left(\frac {130}{23}\right)... | size(120); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(10); pair B=(0,0), A=(5,0), C=(0,13), E=(-5,0), O = incenter(E,C,A), D=IP(A -- A+3*(O-A),E--C); D(A--B--C--cycle); D(A--D--C); D(D--E--B, linetype("4 4")); MP("5", (A+B)/2, f); MP("13", (A+C)/2, NE,f); MP("A",D(A),f); MP("B",D(B),f); MP("C",D(C),N,... | [] |
153 | Four lighthouses are located at points $A$, $B$, $C$, and $D$. The lighthouse at $A$ is $5$ kilometers from the lighthouse at $B$, the lighthouse at $B$ is $12$ kilometers from the lighthouse at $C$, and the lighthouse at $A$ is $13$ kilometers from the lighthouse at $C$. To an observer at $A$, the angle determined by ... | 2009 AIME II Problem 10 | Using the law of sines on $\bigtriangleup ADE$,
\begin{align*} AD &= AE \cdot \dfrac{\sin(2\alpha)}{\sin(180^\circ-3\alpha)}\\ &= AE \cdot \dfrac{\sin(2\alpha)}{\sin(\alpha+2\alpha)}\\ &= 10 \cdot \dfrac{\sin(2\alpha)}{\sin(\alpha)\cos(2\alpha)+\cos(\alpha)\sin(2\alpha)}\\ &= 10 \cdot \dfrac{\sin(2\alpha... | // Block 1
pathpen = linewidth(0.7);
pen f = fontsize(10);
size(144);
pair B=(0,0);
pair A=(5,0);
pair C=(0,12);
pair E=(-5,0);
pair abA = A+unit(C-A)+unit(B-A);
pair D = extension(A,abA,C,E);
D(A--C);
D(C--B);
D(A--B);
D(C--D);
D(A--D);
D(D--E,dashed);
D(B--E,dashed);
D(rightanglemark(C,B,A,20),red);
MP("A",D(A),pla... | [] |
154 | Let $\overline{MN}$ be a diameter of a circle with diameter 1. Let $A$ and $B$ be points on one of the semicircular arcs determined by $\overline{MN}$ such that $A$ is the midpoint of the semicircle and $MB=\frac{3}5$. Point $C$ lies on the other semicircular arc. Let $d$ be the length of the line segment whose endpoin... | 2009 AIME II Problem 15 | Solution 1 (Quick Calculus)
Let $V = \overline{NM} \cap \overline{AC}$ and $W = \overline{NM} \cap \overline{BC}$. Further more let $\angle NMC = \alpha$ and $\angle MNC = 90^\circ - \alpha$. Angle chasing reveals $\angle NBC = \angle NAC = \alpha$ and $\angle MBC = \angle MAC = 90^\circ - \alpha$. Additionally $NB = \... | unitsize(144); pair A, B, C, M, n; A = (0,1); B = (-7/25, 24/25); C=(1/7,-4*sqrt(3)/7); M = (-1,0); n = (1,0); pair [] D = intersectionpoints(A--C,M--n); pair [] e = intersectionpoints(B--C,M--n); draw(circle((0,0),1)); draw(M--n--B--M--A--n--C--A--B--C--cycle); label("$A$",A,N); label("$B$",B,NNW); label("$M$",M,W);... | [] |
155 | Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$, and suppose $P(11) = 181$. Find $P(16)$. | 2010 AIME I Problem 6 | Solution 1
Let $Q(x) = x^2 - 2x + 2$, $R(x) = 2x^2 - 4x + 3$. Completing the square, we have $Q(x) = (x-1)^2 + 1$, and $R(x) = 2(x-1)^2 + 1$, so it follows that $P(x) \ge Q(x) \ge 1$ for all $x$ (by the Trivial Inequality).
Also, $1 = Q(1) \le P(1) \le R(1) = 1$, so $P(1) = 1$, and $P$ obtains its minimum at the poi... | import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1); real P(real x) { return 8*(x-1)^2/5+1; } real Q(real x) { return (x-1)^2+1; } real R(real x) { return 2*(x-1)^2+1; } draw(graph(P,min,max),dark); draw(graph(Q,min,max),linetype("6 2")+linewidth(0.7)); draw(graph(R,min,max),linetype("6 2")+linewidth(0.7... | [] |
156 | For a real number $a$, let $\lfloor a \rfloor$ denote the greatest integer less than or equal to $a$. Let $\mathcal{R}$ denote the region in the coordinate plane consisting of points $(x,y)$ such that $\lfloor x \rfloor ^2 + \lfloor y \rfloor ^2 = 25$. The region $\mathcal{R}$ is completely contained in a disk of radiu... | 2010 AIME I Problem 8 | The desired region consists of 12 boxes, whose lower-left corners are integers solutions of $x^2 + y^2 = 25$, namely $(\pm5,0), (0,\pm5), (\pm3,\pm4), (\pm4,\pm3).$ Since the points themselves are symmetric about $(0,0)$, the boxes are symmetric about $\left(\frac12,\frac12\right)$. The distance from $\left(\frac12,\fr... | // Block 1
import graph; size(10.22cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.68,xmax=6.54,ymin=-5.52,ymax=6.5;
pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0);
filldraw((-3,4)--(-2,4)--(-2,5)--(-3,5)--cycle,evevff,blue); filldraw((... | [] |
157 | Let $\mathcal{R}$ be the region consisting of the set of points in the coordinate plane that satisfy both $|8 - x| + y \le 10$ and $3y - x \ge 15$. When $\mathcal{R}$ is revolved around the line whose equation is $3y - x = 15$, the volume of the resulting solid is $\frac {m\pi}{n\sqrt {p}}$, where $m$, $n$, and $p$ are... | 2010 AIME I Problem 11 | The inequalities are equivalent to $y \ge x/3 + 5, y \le 10 - |x - 8|$. We can set them equal to find the two points of intersection, $x/3 + 5 = 10 - |x - 8| \Longrightarrow |x - 8| = 5 - x/3$. This implies that one of $x - 8, 8 - x = 5 - x/3$, from which we find that $(x,y) = \left(\frac 92, \frac {13}2\right), \left(... | // Block 1
size(280);
import graph; real min = 2, max = 12; pen dark = linewidth(1);
real P(real x) { return x/3 + 5; }
real Q(real x) { return 10 - abs(x - 8); }
path p = (2,P(2))--(8,P(8))--(12,P(12)), q = (2,Q(2))--(12,Q(12));
pair A = (8,10), B = (4.5,6.5), C= (9.75,8.25), F=foot(A,B,C), G=2*F-A; fill(A--B--C--cy... | [] |
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