problem_id int64 1 978 | question stringlengths 86 2.11k | source stringlengths 19 76 | solution stringlengths 94 14.7k | asymptote_code stringlengths 44 17.8k | solution_image_url stringclasses 16
values |
|---|---|---|---|---|---|
448 | 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 Problems/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"... | [] |
449 | 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 Problems/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... | [] |
450 | 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 Problems/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... | [] |
451 | Triangle $ABC$ has $\angle C = 60^{\circ}$ and $BC = 4$. Point $D$ is the midpoint of $BC$. What is the largest possible value of $\tan{\angle BAD}$?
$\mathrm{(A)}\ \frac{\sqrt{3}}{6}\qquad\mathrm{(B)}\ \frac{\sqrt{3}}{3}\qquad\mathrm{(C)}\ \frac{\sqrt{3}}{2\sqrt{2}}\qquad\mathrm{(D)}\ \frac{\sqrt{3}}{4\sqrt{2}-3}\q... | 2008 AMC 12A Problems/Problem 24 | Let $x = CA$. Then $\tan\theta = \tan(\angle BAF - \angle DAE)$, and since $\tan\angle BAF = \frac{2\sqrt{3}}{x-2}$ and $\tan\angle DAE = \frac{\sqrt{3}}{x-1}$, we have
\[\tan\theta = \frac{\frac{2\sqrt{3}}{x-2} - \frac{\sqrt{3}}{x-1}}{1 + \frac{2\sqrt{3}}{x-2}\cdot\frac{\sqrt{3}}{x-1}}= \frac{x\sqrt{3}}{x^2-3x+8}\]
... | // Block 1
unitsize(12mm);
pair C=(0,0), B=(4 * dir(60)), A = (8,0), D=(2 * dir(60));
pair E=(1,0), F=(2,0);
draw(C--B--A--C);
draw(A--D);draw(D--E);draw(B--F);
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);
label("\(C\)",C,SW);
label("\(B\)",B,N);
label("\(A\)",A,SE);
label("\(D\)",D,NW);
label("\(E\)",E,S);
label("\(F\)"... | [] |
452 | In order to earn her vacation spending money, Alexis helped her mother remove weeds from the garden. When she was done, she came into the house to put away her gardening gloves and change into clean clothes.
On her way to her room she notices Joshua with his face to the floor in the family room, looking pretty silly. ... | 2008 iTest Problems/Problem 24 | Note that because the unknown length is the furthest diagonal, draw the diagram as shown because it maximizes the unknown length. By the Pythagorean Theorem,
\[a^2 + b^2 = 81\]
\[a^2 + d^2 = 225\]
\[b^2 + c^2 = 256\]
Adding the second and third equations results in
\[a^2 + b^2 + c^2 + d^2 = 481\]
By substitution,
\[81... | // Block 1
draw((0,0)--(50,0)--(50,60)--(0,60)--(0,0));
dot((16,32));
draw((16,32)--(0,60));
label("9",(8,46),SW);
draw((16,32)--(0,0));
label("15",(8,16),NW);
draw((16,32)--(50,60));
label("16",(33,46),NW);
draw((16,32)--(50,0));
draw((0,32)--(50,32),dotted);
label("$a$",(8,32),S);
label("$c$",(33,32),S);
draw((16,0)... | [] |
453 | Wendy takes Honors Biology at school, a smallish class with only fourteen students (including Wendy) who sit around a circular table. Wendy’s friends Lucy, Starling, and Erin are also in that class. Last Monday none of the fourteen students were absent from class. Before the teacher arrived, Lucy and Starling stretched... | 2008 iTest Problems/Problem 49 | First, let Wendy pick a seat, and let that point be point $W$. Also, pick three arbitrary points clockwise from Wendy. The four points form a cyclic quadrilateral, and the only way for two lines to intersect strictly inside the circle is if the two lines are diagonals. Given one point, only one of the three points m... | // Block 1
draw(circle((0,0),65));
dot((39,52));
dot((-52,39));
dot((25,-60));
dot((-60,-25));
label("W",(-52,39),NW);
// Block 2
draw(circle((0,0),65)); dot((39,52)); dot((-52,39)); dot((25,-60)); dot((-60,-25)); label("W",(-52,39),NW); | [] |
454 | Michael draws $\triangle ABC$ in the sand such that $\angle ACB=90^\circ$ and $\angle CBA=15^\circ$. He then picks a point at random from within the triangle and labels it point $M$. Next, he draws a segment from $A$ to $BC$ that passes through $M$, hitting $BC$ at a point he labels $D$. Just then, a wave washes over h... | 2008 iTest Problems/Problem 66 | Let $D'$ be a point on $BC$ such that $AD' = 2\sqrt3-2$ feet. By using the definitions of sine and cosine, $AC = \sqrt6 - \sqrt2$ feet and $BC = \sqrt6 + \sqrt2$ feet. From the Pythagorean Theorem, $D'C = \sqrt6 - \sqrt2$ feet.
If the number of feet in the length of $AD$ is less than $2\sqrt3-2,$ then point $D$ is... | // Block 1
pair C=(0,0), B=(3.864,0), A=(0,1.035);
draw(C--B--A--C);
dot(C);
label("C",C,SW);
dot(B);
label("B",B,SE);
dot(A);
label("A",A,NW);
dot((1.035,0));
label("$D'$",(1.035,0),S);
draw((1.035,0)--A);
label("4",(1.932,0.518),NE);
draw(anglemark(A,B,C,20));
label("$15^\circ$",(2.6,0),NW);
draw((0,0.25)--(0.25,0.... | [] |
455 | As the Kubiks head homeward, away from the beach in the family van, Jerry decides to take a different route away from the beach than the one they took to get there. The route involves lots of twists and turns, prompting Hannah to wonder aloud if Jerry's "shortcut" will save any time at all.
Michael offers up a problem... | 2008 iTest Problems/Problem 73 | First, we'll show that the farthest distance Jerry can travel with $n$ left turns and $n$ right turns is $n\sqrt{2}$.
Lemma 1: Farthest distance with $n$ left turns and $n$ right turns is $n\sqrt{2}$
Note that all of Jared's turns are perpendicular and occur at every mile. Thus, he travels north or south for $n$ mile... | // Block 1
draw((0,0)--(0,2)--(2,2)--(2,0)--(0,0),dotted);
draw((0,0)--(0,1)--(1,1)--(1,2)--(2,2));
draw((0,0)--(2,2));
// Block 2
draw((0,0)--(0,2)--(2,2)--(2,0)--(0,0),dotted); draw((0,0)--(0,1)--(1,1)--(1,2)--(2,2)); draw((0,0)--(2,2)); | [] |
456 | Points $C$ and $D$ lie on opposite sides of line $\overline{AB}$. Let $M$ and $N$ be the centroids of $\triangle ABC$ and $\triangle ABD$ respectively. If $AB=841$, $BC=840$, $AC=41$, $AD=609$, and $BD=580$, find the sum of the numerator and denominator of the value of $MN$ when expressed as a fraction in lowest terms. | 2008 iTest Problems/Problem 74 | Notice that $41^2 + 840^2 = 841^2$ and $580^2 + 609^2 = 841^2$, so $\triangle ABC$ and $\triangle ABD$ are right triangles. Since $\angle ACB = \angle ADB = 90^\circ$, quadrilateral $ACBD$ is cyclic.
Let $O$ be the midpoint of $AB$. Since $M$ and $N$ are centroids, $CM : MO = DN : DO = 2 : 1$. Thus, by SAS Similarit... | // Block 1
pair A=(0,0),C=(1681/841,34440/841),B=(841,0),D=(370881/841,-353220/841);
draw(A--C--B--D--A);
dot(A);
label("A",A,W);
dot(C);
label("C",C,NW);
dot(B);
label("B",B,E);
dot(D);
label("D",D,S);
draw(circle((841/2,0),841/2));
dot((841/2,0));
draw(C--(841/2,0)--D);
pair M=(1417924/5046,11480/841),n=(1282/3,-14... | [] |
457 | Two perpendicular planes intersect a sphere in two circles. These circles intersect in two points, $A$ and $B$, such that $AB=42$. If the radii of the two circles are $54$ and $66$, find $R^2$, where $R$ is the radius of the sphere. | 2008 iTest Problems/Problem 89 | Let $O$ be the center of the sphere, $X$ be the center of the circle radius $54$, $Y$ be the center of the circle radius $66$, and $M$ be the midpoint of $AB$. Since $XM \perp AB$ and $YM \perp AB$, by the Pythagorean Theorem, $XM = \sqrt{54^2 - 21^2} = 15\sqrt{11}$ and $YM = \sqrt{66^2 - 21^2} = 3\sqrt{435}$.
Addit... | // Block 1
draw(circle((0,0),100));
draw((-100,0)--(10,99.499));
draw((95,-31.225)--(-20,97.980));
dot((0,0));
label("O",(0,0),S);
pair X=(-45,49.750),Y=(37.5,33.376);
draw(X--(0,0)--Y);
dot(X);
label("X",X,NW);
dot(Y);
label("Y",Y,NE);
draw((0,0)--(-100,0),dotted);
// Block 2
draw(circle((0,0),100)); draw((-100,0)... | [] |
458 | In regular hexagon $ABCDEF$ with side length $1$, $AD$ intersects $BF$ at $G$, and $BD$ intersects $EC$ at $H$. Compute the length of $GH$. | 2008 Mock ARML 1 Problems/Problem 3 | Let $H'$ be the foot of the perpendicular from $H$ to $\overline{AD}$. Since $\angle CDA$ is an inscribed angle with measure $\frac{120}{2} = 60^{\circ}$, it follows that $\triangle CDH'$ is a $30-60-90 \triangle$, and $DH' = \frac{1}{2}$ and $CH' = BG = \frac{\sqrt{3}}{2}$. Also, $H'G = CB = 1$. Note that $\triangle D... | // Block 1
pointpen = black; pathpen = black + linewidth(0.62);
pair v(int n){ return dir(n * 60); }
D(MP("A",v(0))--MP("B",v(1),N)--MP("C",v(2),N)--MP("D",v(3),SW)--MP("E",v(4))--MP("F",v(5))--cycle);
D(v(0)--v(3));D(v(1)--v(5));D(v(1)--v(3));D(v(2)--v(4));
D(D(MP("G",IP(v(0)--v(3),v(1)--v(5)),NE))--D(MP("H",IP(v(1)-... | [] |
459 | Square $ABCD$ has side length $2$. $M$ is the midpoint of $CD$, and $N$ is the midpoint of $BC$. $P$ is on $MN$ such that $N$ is between $M$ and $P$, and $m\angle MAN = m\angle NAP$. Compute the length of $AP$. | 2008 Mock ARML 1 Problems/Problem 6 | By the Pythagorean Theorem, $MA = NA = \sqrt{5}$ and $MN = \sqrt{2}$. Let $\theta = \angle MAN = \angle NAP$. By the Law of Cosines on $\triangle MAN$,
\[\left(\sqrt{2}\right)^2 = \left(\sqrt{5}\right)^2 + \left(\sqrt{5}\right)^2 - 2 \cdot \left(\sqrt{5}\right) \cdot \left(\sqrt{5}\right) \cos \theta \Longrightarrow \c... | // Block 1
pointpen=black;pathpen=black+linewidth(0.7);
pair A=(0,2), D=(0,0), C=(2,0), B=(2,2), M=(C+D)/2, N=(B+C)/2, P=8/3*(N-M)+M;
D(MP("A",A,(0,1))--MP("B",B,(0,1))--MP("C",C)--MP("D",D)--cycle); D(D(MP("M",M))--D(MP("N",N,E))--D(MP("P",P,E)));
D(M--A--N--A--P); D(anglemark(M,A,P)); MP("2",(A+D)/2,W);
// Block 2
p... | [] |
460 | $ABCD$ is a convex quadrilateral such that $|AB| = 5$, $|BC| = 17$, $|CD| = 7$, and $|DA| = 25$. Given that $m\angle{ABC} + m\angle{BCD} = 270^{\circ}$, find the area of $ABCD$. | 2008 Mock ARML 2 Problems/Problem 1 | Note that $m\angle{BAD} + m\angle{ADC} = 360 - 270 = 90^{\circ}$. Thus, if we let $E$ be the intersection of the extensions of $\overline{AB}$ and $\overline{CD}$, it follows that $\triangle EAD$ is a right triangle. Immediately we notice that $\triangle ECB$ is a $8-15-17,\,\triangle$ and that $\triangle EAD$ is a $15... | // Block 1
pointpen = black; pathpen = black + linewidth(0.7);
pair A=(0,0), D=(25,0), F=(16,12), B=(3*A + F)/4, C=(8*D+7*F)/15;
D(MP("A",A,SW)--MP("B",B,NW)--MP("C",C,NE)--MP("D",D,SE)--cycle); D(B--MP("E",F,N)--C,linetype("4 4")); D(rightanglemark(A,F,D,30),linetype("4 4"));
// Block 2
pointpen = black; pathpen = bla... | [] |
461 | Equilateral triangle $ABC$ has a side length of $7$. A ball begins at vertex $A$, rolls through the interior of the triangle, bounces off side $BC$, and settles at point $P$. Given that $BP = 3$ and $CP = 5$, find the total distance that the ball travels. | 2008 Mock ARML 2 Problems/Problem 4 | Reflect $P$ across $\overline{BC}$ to point $P'$; since the ball travels in a straight path, it follows that the distance the ball traveled is $AP'$. By symmetry, $BP' = 3, CP' = 5$. By the Law of Cosines on $\triangle BCP'$,
$7^2 = 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cos \theta \Longrightarrow \theta = 120^{\circ}.$
Since... | // Block 1
pointpen = black; pathpen = black + linewidth(0.7);
pair A=(0,0),B=(7,0),C=7*expi(pi/3);
pair Q=IP(CR(B,3),CR(C,5)), P=OP(CR(B,3),CR(C,5)),D=IP(A--Q,B--C);
D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle);
D(A--D--D(MP("P",P)));D(D--D(MP("P'",Q,NE))--C--Q--B,linetype("5 5"));
MP("7",(A+B)/2); MP("3",(B+Q)/2,SE);... | [] |
462 | 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 Problems/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 = ... | [] |
463 | 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 Problems/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,... | [] |
464 | 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 Problems/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,... | [] |
465 | 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 Problems/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);... | [] |
466 | 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 Problems/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)... | [] |
467 | A square is divided into three pieces of
equal area by two parallel lines as shown.
If the distance between the two parallel
lines is $8$ what is the area of the square? | 2009 UNCO Math Contest II Problems/Problem 9 | In $\triangle$ADC, $\frac {AE}{ED}$ = $\frac {1}{2}$ because $\frac {[ACE]}{[CED]}$ = $\frac {1}{2}$ (According to question)
$[ECD]$ = $2[EFC]$ (AECF is a parallelogram and $2[EFC]$ = $[AECF]$)
$2[EFC]$ = $[EDC]$ (same reason as before)
$[EFC]$ = $\frac {1}{2}$ * 8 * EC
$[EDC]$ = $\frac{1}{2}$ * EC * (Perpendicular... | // Block 1
unitsize(135);
defaultpen(linewidth(.8pt)+fontsize(10pt));
pair A, B, C, D, E, F;
A=(0, 0); B=(1, 0); C=(1, -1); D=(0,-1); E=(0,-1/3); F=(1,-2/3);
draw(A--B--C--D--cycle);
label("$A$",A,N); label("$B$",B,N); label("$C$",C,E); label("$D$",D,S); label("$E$",E,W); label("$F$",F,SE);
draw(A--F);
draw(C--E);
draw... | [] |
468 | 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 Problems/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... | [] |
469 | Rectangle $ABCD$ and a semicircle with diameter $AB$ are coplanar and have nonoverlapping interiors. Let $\mathcal{R}$ denote the region enclosed by the semicircle and the rectangle. Line $\ell$ meets the semicircle, segment $AB$, and segment $CD$ at distinct points $N$, $U$, and $T$, respectively. Line $\ell$ divides ... | 2010 AIME I Problems/Problem 13 | Diagram
Solution 1
The center of the semicircle is also the midpoint of $AB$. Let this point be O. Let $h$ be the length of $AD$.
Rescale everything by 42, so $AU = 2, AN = 3, UB = 4$. Then $AB = 6$ so $OA = OB = 3$.
Since $ON$ is a radius of the semicircle, $ON = 3$. Thus $OAN$ is an equilateral triangle.
Let $X$,... | /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */ import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(500); pen zzttqq = rgb(0.6,0.2,0); pen xdxdff = rgb(0.4902,0.4902,1); /* segments and figures */ draw((0,-154.31785)--(0,0)); draw((0,0)--(252,0)); draw((0,0)--(126,0),zzttqq); draw((12... | [] |
470 | In $\triangle{ABC}$ with $AB = 12$, $BC = 13$, and $AC = 15$, let $M$ be a point on $\overline{AC}$ such that the incircles of $\triangle{ABM}$ and $\triangle{BCM}$ have equal radii. Then $\frac{AM}{CM} = \frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$. | 2010 AIME I Problems/Problem 15 | Solution 1
Let $AM = x$, then $CM = 15 - x$. Also let $BM = d$ Clearly, $\frac {[ABM]}{[CBM]} = \frac {x}{15 - x}$. We can also express each area by the rs formula. Then $\frac {[ABM]}{[CBM]} = \frac {p(ABM)}{p(CBM)} = \frac {12 + d + x}{28 + d - x}$. Equating and cross-multiplying yields $25x + 2dx = 15d + 180$ or $d ... | // Block 1
/* from geogebra: see azjps, userscripts.org/scripts/show/72997 */
import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(200);
/* segments and figures */
draw((0,0)--(15,0));
draw((15,0)--(6.66667,9.97775));
draw((6.66667,9.97775)--(0,0));
draw((7.33333,0)--(6.66667,9.97775));
draw(circle((4.66667,2... | [] |
471 | Triangle $ABC$ with right angle at $C$, $\angle BAC < 45^\circ$ and $AB = 4$. Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$. The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$, where $p$, $q$, $r$ are positive integers and $r$ is not divisible by the squa... | 2010 AIME II Problems/Problem 14 | Let $O$ be the circumcenter of $ABC$ and let the intersection of $CP$ with the circumcircle be $D$. It now follows that $\angle{DOA} = 2\angle ACP = \angle{APC} = \angle{DPB}$. Hence $ODP$ is isosceles and $OD = DP = 2$.
Denote $E$ the projection of $O$ onto $CD$. Now $CD = CP + DP = 3$. By the Pythagorean Theorem, $... | // Block 1
/* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */
import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(250); real lsf = 0.5; /* changes label-to-point distance */
pen xdxdff = rgb(0.49,0.49,1); pen qqwuqq = rgb(0,0.39,0); pen fftttt = rgb(1,0.2,0.2);
/* segments and figures */... | [] |
472 | A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$. The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\frac{r-\sqrt{s}}{t}$, where $r$, $s$, and $t$ are positive integ... | 2011 AIME I Problems/Problem 13 | First, shift the cube down by 11 so the vertices adjacent to $A$ are $-1$, $0$, and $1$ above the plane.
Consider the two points 1 above and 1 below the plane:
Now we rotate the cube so that all 3 vertices adjacent to $A$ are on the plane. We can calculate $A$ to be $\frac{10\sqrt3}{3}$ below the plane.
Notice that... | // Block 1
defaultpen(fontsize(10)+0.8); size(300);
pen p=fontsize(9)+linewidth(3);
// Define points
pair A=(-7,0), B=(0,0), C=(7,0);
pair D=(-7,-1), E=(7,1);
// Draw segments
draw(A--B--C, linewidth(1.2));
draw(A--D, linewidth(1.2));
draw(C--E, linewidth(1.2));
draw(D--E, linewidth(1.2));
// Label lengths
label("$7... | [] |
473 | In triangle $ABC$, $AB=125$, $AC=117$ and $BC=120$. The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$, and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$. Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$, respectively. Find ... | 2011 AIME I Problems/Problem 4 | Extend ${CM}$ and ${CN}$ such that they intersect line ${AB}$ at points $P$ and $Q$, respectively.
Since ${BM}$ is the angle bisector of angle $B$ and ${CM}$ is perpendicular to ${BM}$, $\triangle BCP$ must be an isoceles triangle, so $BP=BC=120$, and $M$ is the midpoint of ${CP}$. For the same reason, $AQ=AC=117$, a... | // Block 1
defaultpen(fontsize(10)+0.8); size(200);
pen p=fontsize(9)+linewidth(3);
pair A,B,C,D,K,L,M,N,P,Q;
A=origin; B=(125,0); C=IP(CR(A,117),CR(B,120)); L=extension(B,C,A,bisectorpoint(B,A,C)); K=extension(A,C,B,bisectorpoint(C,B,A)); M=foot(C,B,K); N=foot(C,A,L);
draw(A--B--C--A); draw(A--L^^B--K, gray+dashed+0.5... | [] |
474 | On square $ABCD$, point $E$ lies on side $AD$ and point $F$ lies on side $BC$, so that $BE=EF=FD=30$. Find the area of the square $ABCD$. | 2011 AIME II Problems/Problem 2 | Drawing the square and examining the given lengths,
you find that the three segments cut the square into three equal horizontal sections. Therefore, ($x$ being the side length), $\sqrt{x^2+(x/3)^2}=30$, or $x^2+(x/3)^2=900$. Multiplying both sides by $9$ and simplifying, we find that $10x^2=8100$. Dividing by ten give... | // Block 1
size(2inch, 2inch);
currentpen = fontsize(8pt);
pair A = (0, 0); dot(A); label("$A$", A, plain.SW);
pair B = (3, 0); dot(B); label("$B$", B, plain.SE);
pair C = (3, 3); dot(C); label("$C$", C, plain.NE);
pair D = (0, 3); dot(D); label("$D$", D, plain.NW);
pair E = (0, 1); dot(E); label("$E$", E, plain.W);
pa... | [] |
475 | In triangle $ABC$, $AB=20$ and $AC=11$. The angle bisector of $\angle A$ intersects $BC$ at point $D$, and point $M$ is the midpoint of $AD$. Let $P$ be the point of the intersection of $AC$ and $BM$. The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive ... | 2011 AIME II Problems/Problem 4 | Let $D'$ be on $\overline{AC}$ such that $BP \parallel DD'$. It follows that $\triangle BPC \sim \triangle DD'C$, so \[\frac{PC}{D'C} = 1 + \frac{BD}{DC} = 1 + \frac{AB}{AC} = \frac{31}{11}\] by the Angle Bisector Theorem. Similarly, we see by the Midline Theorem that $AP = PD'$. Thus, \[\frac{CP}{PA} = \frac{1}{\frac{... | // Block 1
pointpen = black; pathpen = linewidth(0.7);
pair A = (0,0), C= (11,0), B=IP(CR(A,20),CR(C,18)), D = IP(B--C,CR(B,20/31*abs(B-C))), M = (A+D)/2, P = IP(M--2*M-B, A--C), D2 = IP(D--D+P-B, A--C);
D(MP("A",D(A))--MP("B",D(B),N)--MP("C",D(C))--cycle); D(A--MP("D",D(D),NE)--MP("D'",D(D2))); D(B--MP("P",D(P))); D... | [] |
476 | Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$, $BC=7$, and $AC=3$. The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $\overline{DE}$. Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$. Then $AF^2 = ... | 2012 AIME II Problems/Problem 15 | Use the angle bisector theorem to find $CD=\tfrac{21}{8}$, $BD=\tfrac{35}{8}$, and use Stewart's Theorem to find $AD=\tfrac{15}{8}$. Use Power of Point $D$ to find $DE=\tfrac{49}{8}$, and so $AE=8$. Use law of cosines to find $\angle CAD = \tfrac{\pi} {3}$, hence $\angle BAD = \tfrac{\pi}{3}$ as well, and $\triangle BC... | // Block 1
size(150);
defaultpen(fontsize(9pt));
picture pic;
pair A,B,C,D,E,F,W;
B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D... | [] |
476 | Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$, $BC=7$, and $AC=3$. The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $\overline{DE}$. Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$. Then $AF^2 = ... | 2012 AIME II Problems/Problem 15 | Let $a = BC$, $b = CA$, $c = AB$ for convenience. Let $M$ be the midpoint of segment $BC$. We claim that $\angle MAD=\angle DAF$.
$\textit{Proof}$. Since $AE$ is the angle bisector, it follows that $EB = EC$ and consequently $EM\perp BC$. Therefore, $M\in \gamma$. Now let $X = FD\cap \omega$. Since $\angle EFX=90^\ci... | // Block 1
size(175);
defaultpen(fontsize(10pt));
picture pic;
pair A,B,C,D,E,F,W;
B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(... | [] |
476 | Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$, $BC=7$, and $AC=3$. The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $\overline{DE}$. Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$. Then $AF^2 = ... | 2012 AIME II Problems/Problem 15 | Use the angle bisector theorem to find $CD=\tfrac{21}{8}$, $BD=\tfrac{35}{8}$, and use Stewart's Theorem to find $AD=\tfrac{15}{8}$. Use Power of Point $D$ to find $DE=\tfrac{49}{8}$, and so $AE=8$. Then use the Extended Law of Sine to find that the length of the circumradius of $\triangle ABC$ is $\tfrac{7\sqrt{3}}{3}... | // Block 1
size(175);
defaultpen(fontsize(9pt));
pair A,B,C,D,E,F,W;
B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D... | [] |
477 | A circle of radius $1$ is externally tangent to a circle of radius $3$
and both circles are tangent to a line. Find the area of the shaded
region that lies between the two circles and the line. | 2012 UNCO Math Contest II Problems/Problem 7 | If we draw in the second tangent between these circles, we get a point where the tangents intersect. Drawing a line through the centers of the circle and the point where the tangents meet, we get two similar triangles in ratio $1:3$. Let the hypotenuse of the smaller triangle be $x$. Since the distance between the cent... | // Block 1
draw((-1,-1)--(7,-1),black);
filldraw((0,-1)--(0,0)--(2*sqrt(3),2)--(2*sqrt(3),-1)--cycle,grey);
filldraw(circle((0,0),1),white);
filldraw(circle((2*sqrt(3),2),3),white);
// Block 2
draw((-1,-1)--(7,-1),black); filldraw((0,-1)--(0,0)--(2*sqrt(3),2)--(2*sqrt(3),-1)--cycle,grey); filldraw(circle((0,0),1),white... | [] |
478 | In $\triangle ABC$, $AC = BC$, and point $D$ is on $\overline{BC}$ so that $CD = 3\cdot BD$. Let $E$ be the midpoint of $\overline{AD}$. Given that $CE = \sqrt{7}$ and $BE = 3$, the area of $\triangle ABC$ can be expressed in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the ... | 2013 AIME II Problems/Problem 13 | Let $X$ be the foot of the altitude from $C$ with other points labelled as shown below.
Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. Th... | // Block 1
size(200);
pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7;
draw(A--B--C--cycle);draw(A--D^^B--L^^C--M);
label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE);label("$L$",L,NW);label("$M$",M,S);
pair X=foot(C,A,B), Y=foot(L,A,B);
pair EE=D/2;
label("$X$",X,S);label("$E$"... | [] |
478 | In $\triangle ABC$, $AC = BC$, and point $D$ is on $\overline{BC}$ so that $CD = 3\cdot BD$. Let $E$ be the midpoint of $\overline{AD}$. Given that $CE = \sqrt{7}$ and $BE = 3$, the area of $\triangle ABC$ can be expressed in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the ... | 2013 AIME II Problems/Problem 13 | Let $BD = x$. Then $CD = 3x$ and $AC = 4x$. Also, let $AE = ED = l$. Using Stewart's Theorem on $\bigtriangleup CEB$ gives us the equation $(x)(3x)(4x) + (4x)(l^2) = 27x + 7x$ or, after simplifying, $4l^2 = 34 - 12x^2$. We use Stewart's again on $\bigtriangleup CAD$: $(l)(l)(2l) + 7(2l) = (16x^2)(l) + (9x^2)(l)$, which... | // Block 1
size(200);
pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7;
draw(A--B--C--cycle);draw(A--D^^B--L^^C--M);
label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE);label("$L$",L,NW);label("$M$",M,S);
pair EE=D/2;
label("$\sqrt{7}$", C--EE, W); label("$x$", D--B, E); label("$3... | [] |
478 | In $\triangle ABC$, $AC = BC$, and point $D$ is on $\overline{BC}$ so that $CD = 3\cdot BD$. Let $E$ be the midpoint of $\overline{AD}$. Given that $CE = \sqrt{7}$ and $BE = 3$, the area of $\triangle ABC$ can be expressed in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the ... | 2013 AIME II Problems/Problem 13 | Let $AB = 2x$ and let $y = BD.$ Then $CD = 3y$ and $AC = 4y.$
By the Law of Cosines on triangle $ABC,$
\[\cos C = \frac{16y^2 + 16y^2 - 4x^2}{2 \cdot 4y \cdot 4y} = \frac{32y^2 - 4x^2}{32y^2} = \frac{8y^2 - x^2}{8y^2}.\]Then by the Law of Cosines on triangle $ACD,$
\begin{align*}
AD^2 &= 16y^2 + 9y^2 - 2 \cdot 4y \c... | // Block 1
unitsize(1.5 cm);
pair A, B, C, D, E;
A = (-sqrt(7),0);
B = (sqrt(7),0);
C = (0,3);
D = interp(B,C,1/4);
E = (A + D)/2;
draw(A--B--C--cycle);
draw(A--D);
draw(B--E--C);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("$D$", D, NE);
label("$E$", E, NW);
label("$2x$", (A + B)/2, S);
la... | [] |
479 | In acute triangle $ABC$, the orthocenter $H$ lies on the line connecting the midpoint of segment $AB$ to the midpoint of segment $BC$. If $AC=24$, and the altitude from $B$ has length $14$, find $AB\cdot BC$. | 2013 Mock AIME I Problems/Problem 12 | Toss on the coordinate plane with $A=(0,0)$, $B=(t,14)$, and $C=(24,0)$, where $t$ is a real number and $0<t<24$.
Then, the line connecting the midpoints of $AB$ and $BC$ runs from $\left(\frac{t}{2}, 7\right)$ to $\left(\frac{24+t}{2}, 7\right)$, or more simply the line $y=7$.
The orthocenter of $\triangle ABC$ will... | // Block 1
real t = 12 - sqrt(46);
pair A = (0, 0);
pair B = (t, 14);
pair C = (24, 0);
draw(A--B--C--cycle);
draw((A+B)/2--(B+C)/2, dashed);
draw(B--(t, 0), blue);
draw(A--(8.572, 11.55), red);
draw(C--(2.927, 7.854), green);
label("$A$", A, W);
label("$B$", B, N);
label("$C$", C, E);
real t = 12 + sqrt(46);
pair A = ... | [] |
479 | In acute triangle $ABC$, the orthocenter $H$ lies on the line connecting the midpoint of segment $AB$ to the midpoint of segment $BC$. If $AC=24$, and the altitude from $B$ has length $14$, find $AB\cdot BC$. | 2013 Mock AIME I Problems/Problem 12 | Let $M$ be the midpoint of $\overline{AB}$ and $N$ be the midpoint of $\overline{BC}$. Further let $D$ be the foot of the altitude from $A$, $E$ from $B$, and $F$ from $C$, as in the diagram.
Because $\overline{MN}$ is a midpoint connector of $\triangle ABC$ amd $H$ is on $\overline{MN}$ and $\overline{BE}$, we know t... | // Block 1
import geometry;
point A = origin;
point B = (12-sqrt(46),14);
point C = (24,0);
point M = midpoint(A--B);
point N = midpoint(B--C);
triangle t = triangle(A,B,C);
point H = orthocentercenter(t);
point D, E, F;
line a = altitude(t.BC);
line b = altitude(t.AC);
line c = altitude(t.AB);
// Triangle ABC and... | [] |
480 | In acute $\triangle ABC$, $H$ is the orthocenter, $G$ is the centroid, and $M$ is the midpoint of $BC$. It is obvious that $AM \ge GM$, but $GM \ge HM$ does not always hold. If $[ABC] = 162$, $BC=18$, then the value of $GM$ which produces the smallest value of $AB$ such that $GM \ge HM$ can be expressed in the form $a+... | 2013 Mock AIME I Problems/Problem 13 | Because $[\triangle ABC] = 168$ and $BC=18$, we know that the height from $A$ to $BC$ must be $18$. Thus, because the perpendicular is the shortest segment from a line to a point not on the line, we know that $AB \geq 18$. Thus, the minimum value of $AB$ is $18$, when $\overline{AB} \perp \overline{BC}$. This is techni... | // Block 1
import geometry;
point A = origin;
point B = (0,18);
point C = (18,18);
triangle t = triangle(A,B,C);
point M = midpoint(B--C);
circle c = circumcircle(t);
point O = circumcenter(t);
point G = centroid(t);
// Triangle and Circumcircle
draw(t);
draw(c);
// Median AM
draw(A--M);
// Labelling Points
dot... | [] |
481 | In quadrilateral $ABCD$, $AC\cap BD=M$. Also, $MA=6, MB=8, MC=4, MD=3$, and $BC=2CD$. The perimeter of $ABCD$ can be expressed in the form $\frac{p\sqrt{q}}{r}$ where $p$ and $r$ are relatively prime, and $q$ is not divisible by the square of any prime number. Find $p+q+r$. | 2013 Mock AIME I Problems/Problem 5 | Let $CD=x$, as in the diagram. Thus, from the problem, $BC=2x$. Because $AM \cdot MC = DM \cdot MB = 24$, by Power of a Point, we know that $ABCD$ is cyclic. Thus, we know that $\measuredangle DAC = \measuredangle DBC$, so, by the congruency of vertical angles and subsequently AA Similarity, we know that $\triangle AMD... | // Block 1
import geometry;
// Defining Points
point O = origin;
point B = (1,0);
point A = dir(115.583);
point C = dir(-115.583);
point D = dir(-165.638);
point M;
// Circle
draw(circle(O, 1));
// Quadrilateral and Diagonals
draw(A--B--C--D--cycle);
draw(A--C);
draw(B--D);
// Defining M
pair[] m = intersectionpoin... | [] |
482 | In $\triangle ABC$, $AB = 3$, $BC = 4$, and $CA = 5$. Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$, $\overline{BC}$ at $B$ and $D$, and $\overline{AC}$ at $F$ and $G$. Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4}$, length $DE=\frac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive inte... | 2014 AIME I Problems/Problem 15 | First we note that $\triangle DEF$ is an isosceles right triangle with hypotenuse $\overline{DE}$ the same as the diameter of $\omega$. We also note that $\triangle DGE \sim \triangle ABC$ since $\angle EGD$ is a right angle and the ratios of the sides are $3:4:5$.
From congruent arc intersections, we know that $\ang... | // Block 1
pair A = (0,3);
pair B = (0,0);
pair C = (4,0);
draw(A--B--C--cycle);
dotfactor = 3;
dot("$A$",A,dir(135));
dot("$B$",B,dir(215));
dot("$C$",C,dir(305));
pair D = (2.21, 0);
pair E = (0, 1.21);
pair F = (1.71, 1.71);
pair G = (2, 1.5);
dot("$D$",D,dir(270));
dot("$E$",E,dir(180));
dot("$F$",F,dir(90));
dot("... | [] |
483 | In $\triangle RED$, $\measuredangle DRE=75^{\circ}$ and $\measuredangle RED=45^{\circ}$. $RD=1$. Let $M$ be the midpoint of segment $\overline{RD}$. Point $C$ lies on side $\overline{ED}$ such that $\overline{RC}\perp\overline{EM}$. Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA=AR$. Then $AE=\fr... | 2014 AIME II Problems/Problem 11 | Let $P$ be the foot of the perpendicular from $A$ to $\overline{CR}$, so $\overline{AP}\parallel\overline{EM}$. Since triangle $ARC$ is isosceles, $P$ is the midpoint of $\overline{CR}$, and $\overline{PM}\parallel\overline{CD}$. Thus, $APME$ is a parallelogram and $AE = PM = \frac{CD}{2}$. We can then use coordinates.... | // Block 1
unitsize(8cm);
pair a, o, d, r, e, m, cm, c,p;
o =(0,0);
d = (0.5, 0);
r = (0,sqrt(3)/2);
e = (-sqrt(3)/2,0);
m = midpoint(d--r);
draw(e--m);
cm = foot(r, e, m);
draw(L(r, cm,1, 1));
c = IP(L(r, cm, 1, 1), e--d);
clip(r--d--e--cycle);
draw(r--d--e--cycle);
draw(rightanglemark(e, cm, c, 1.5));
a = -(4sqrt(3)... | [] |
483 | In $\triangle RED$, $\measuredangle DRE=75^{\circ}$ and $\measuredangle RED=45^{\circ}$. $RD=1$. Let $M$ be the midpoint of segment $\overline{RD}$. Point $C$ lies on side $\overline{ED}$ such that $\overline{RC}\perp\overline{EM}$. Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA=AR$. Then $AE=\fr... | 2014 AIME II Problems/Problem 11 | Let $P$ be the foot of the perpendicular from $A$ to $\overline{CR}$, so $\overline{AP}\parallel\overline{EM}$. Since $\triangle ARC$ is isosceles, $P$ is the midpoint of $\overline{CR}$, and by midpoint theorem $\overline{PM}\parallel\overline{CD}$. Thus, $APME$ is a parallelogram and therefore $AE = PM = \tfrac 12 C... | // Block 1
unitsize(8cm); pair a, d, r, e, m, cm, c,p;
d=origin; r=dir(60); e=extension(d,left,r,r+dir(75)*(d-r)); m = midpoint(d--r); cm = foot(r, e, m); c=extension(r,cm,d,e); p=midpoint(r--c); a=p+(e-m);
draw(e--m); draw(L(r, cm,1, 1)); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)... | [] |
484 | In $\triangle{ABC}, AB=10, \angle{A}=30^\circ$ , and $\angle{C=45^\circ}$. Let $H, D,$ and $M$ be points on the line $BC$ such that $AH\perp{BC}$, $\angle{BAD}=\angle{CAD}$, and $BM=CM$. Point $N$ is the midpoint of the segment $HM$, and point $P$ is on ray $AD$ such that $PN\perp{BC}$. Then $AP^2=\dfrac{m}{n}$, where ... | 2014 AIME II Problems/Problem 14 | Draw the $45-45-90 \triangle AHC$. Now, take the perpendicular bisector of $BC$ to intersect the circumcircle of $\triangle ABC$ and $AC$ at $F, L, G$ as shown, and denote $O$ to be the circumcenter of $\triangle ABC$. It is not difficult to see by angle chasing that $AHBGO$ is cyclic, namely with diameter $AB$. Then, ... | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(15cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black... | [] |
485 | Circle $C$ with radius 2 has diameter $\overline{AB}$. Circle D is internally tangent to circle $C$ at $A$. Circle $E$ is internally tangent to circle $C$, externally tangent to circle $D$, and tangent to $\overline{AB}$. The radius of circle $D$ is three times the radius of circle $E$, and can be written in the form $... | 2014 AIME II Problems/Problem 8 | Using the diagram above, let the radius of $D$ be $3r$, and the radius of $E$ be $r$. Then, $EF=r$, and $CE=2-r$, so the Pythagorean theorem in $\triangle CEF$ gives $CF=\sqrt{4-4r}$. Also, $CD=CA-AD=2-3r$, so \[DF=DC+CF=2-3r+\sqrt{4-4r}.\] Noting that $DE=4r$, we can now use the Pythagorean theorem in $\triangle DEF$ ... | // Block 1
import graph; size(7.99cm);
real labelscalefactor = 0.5;
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
pen dotstyle = black;
real xmin = 4.087153740193288, xmax = 11.08175859031552, ymin = -4.938019122704778, ymax = 1.194137062512079;
draw(circle((7.780000000000009,-1.320000000000002), 2.0000... | [] |
486 | Let $ABCDE$ be a pentagon inscribed in a circle such that $AB = CD = 3$, $BC = DE = 10$, and $AE= 14$. The sum of the lengths of all diagonals of $ABCDE$ is equal to $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ?
$\textbf{(A) }129\qquad \textbf{(B) }247\qquad \textbf{(C) }35... | 2014 AMC 12B Problems/Problem 24 | Let $BE=a$, $AD=b$, and $AC=CE=BD=c$. Let $F$ be on $AE$ such that $CF \perp AE$.
In $\triangle CFE$ we have $\cos\theta = -\cos(\pi-\theta)=-7/c$. We use the Law of Cosines on $\triangle ABC$ to get $60\cos\theta = 109-c^2$. Eliminating $\cos\theta$ we get $c^3-109c-420=0$ which factorizes as
\[(c+7)(c+5)(c-12)=0.... | // Block 1
size(200);
defaultpen(linewidth(0.4)+fontsize(10));
pen s = linewidth(0.8)+fontsize(8);
pair O,A,B,C,D,E0,F;
O=origin;
A= dir(198);
path c = CR(O,1);
real r = 0.13535;
B = IP(c, CR(A,3*r));
C = IP(c, CR(B,10*r));
D = IP(c, CR(C,3*r));
E0 = OP(c, CR(D,10*r));
F = foot(C,A,E0);
dot("$A$", A, A-O);
dot("$B$",... | [] |
487 | Completely describe the set of all right triangles with positive integer-valued legs such that when four
copies of the triangle are arranged in square formation shown below, the incenters of the four triangles
lie on the extensions of the sides of the smaller square. (Note: the incenter of a triangle is the center
of t... | 2014 UMO Problems/Problem 3 | Let $I$ be the incenter of a triangle. Drop $I$ onto the three sides of the triangle, and let the points be $X, Y, Z$
Finally, let $a = AB, b = BC$ so that $a > b$ and let $s = BZ$.
Note that $Z$ is also a corner of the square, so $s = a - b$. But then, $AC = CX + AZ = a + b - 2(a-b) = 3b-a$. From the Pythogorean theor... | // Block 1
size(200);
path T=((0,0)--(4,0)--(4,.3)--(3.7,.3)--(3.7,0)--(4,0)--(4,3)--(0,0));
D(shift(10,0)*rotate(53)*T,black);
D(shift(15,5)*rotate(233)*T,black);
D(shift(15,0)*rotate(143)*T,black);
D(shift(10,5)*rotate(323)*T,black);
pair A, B, C, X, Y, Z, I;
A = shift(10,0)*rotate(53)*(0,0);
B = shift(10,0)*rotate(5... | [] |
488 | Find the smallest and largest possible distances between the centers of two circles of radius
$1$ such that there is an equilateral triangle of side length $1$ with two vertices on one of
the circles and the third vertex on the second circle. | 2014 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 4 | The smallest distance would be found if the two circles were externally tangent, so testing that and messing around with it yields:
Where $P$ is the point of tangency. This clearly works, so the smallest distance would be $2\cdot1=\boxed{2}$
The largest distance would be found by first finding the closest place to... | // Block 1
size(300 pt);
draw(circle((0,0),1));
draw(circle((2,0),1));
draw((1,0)--(0.5,0.86602),linewidth(.5));
draw((0,0)--(0.5,0.86602),linewidth(.5));
draw((1,0)--(1.5,0.86602),linewidth(.5));
draw((2,0)--(1.5,0.86602),linewidth(.5));
draw((0.5,0.86602)--(1.5,0.86602),linewidth(.5));
dot((0,0));
dot((2,0));
dot((1,... | [] |
489 | Triangle $ABC$ has side lengths $AB = 12$, $BC = 25$, and $CA = 17$. Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$, vertex $Q$ on $\overline{AC}$, and vertices $R$ and $S$ on $\overline{BC}$. In terms of the side length $PQ = \omega$, the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS) = \... | 2015 AIME II Problems/Problem 7 | Similar triangles can also solve the problem.
First, solve for the area of the triangle. $[ABC] = 90$. This can be done by Heron's Formula or placing an $8-15-17$ right triangle on $AC$ and solving. (The $8$ side would be collinear with line $AB$)
After finding the area, solve for the altitude to $BC$. Let $E$ be the... | // Block 1
unitsize(20);
pair A,B,C,E,F,P,Q,R,S;
A=(48/5,36/5);
B=(0,0);
C=(25,0);
E=(48/5,0);
F=(48/5,18/5);
P=(24/5,18/5);
Q=(173/10,18/5);
S=(24/5,0);
R=(173/10,0);
draw(A--B--C--cycle);
draw(P--Q);
draw(Q--R);
draw(R--S);
draw(S--P);
draw(A--E,dashed);
label("$A$",A,N);
label("$B$",B,SW);
label("$C$",C,SE);
label("... | [] |
490 | A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$. | 2015 AIME II Problems/Problem 9 | We can use the same method as in Solution 2 to find the side length of the equilateral triangle, which is $4\sqrt3$. From here, its area is
\[\dfrac{\bigl(4\sqrt3\bigr)^2\sqrt3}4=12\sqrt3.\]
The leg of the isosceles right triangle is $\dfrac{4\sqrt3}{\sqrt2}=2\sqrt6$, and the horizontal distance from the vertex to the ... | // Block 1
import olympiad;
pair V, T, B;
V = (-4, 0);
B = origin;
T = (0, 2*sqrt(2));
draw(V--B--T--cycle);
draw(rightanglemark(V, B, T));
label("Vertex", V, W);
label("Tip", T, N);
label("Base", B, SE);
label("$4$", V--B, S);
label("$2\sqrt6$", V--T, NW);
// Block 2
import olympiad; pair V, T, B; V = (-4, 0); B ... | [] |
491 | Centered at each lattice point in the coordinate plane are a circle radius $\frac{1}{10}$ and a square with sides of length $\frac{1}{5}$ whose sides are parallel to the coordinate axes. The line segment from $(0,0)$ to $(1001, 429)$ intersects $m$ of the squares and $n$ of the circles. Find $m + n$. | 2016 AIME I Problems/Problem 14 | See if you can solve the problem with the following.
Solution to Solution 2
This is mostly a clarification to Solution 1, but let's take the diagram for the origin to $(7,3)$. We have the origin circle and square intersected, then two squares, then the circle and square at $(7,3)$. If we take the circle and square a... | // Block 1
size(12cm);draw((0,0)--(7,3));draw(box((0,0),(7,3)),dotted);
for(int i=0;i<8;++i)for(int j=0; j<4; ++j){dot((i,j),linewidth(1));draw(box((i-.1,j-.1),(i+.1,j+.1)),linewidth(.5));draw(circle((i,j),.1),linewidth(.5));}
// Block 2
size(12cm);draw((0,0)--(7,3));draw(box((0,0),(7,3)),dotted); for(int i=0;i<8;++i)f... | [] |
492 | Equilateral $\triangle ABC$ has side length $600$. Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$, and $QA=QB=QC$, and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). Th... | 2016 AIME II Problems/Problem 14 | To make numbers more feasible, we'll scale everything down by a factor of $100$ so that $\overline{AB}=\overline{BC}=\overline{AC}=6$. We should also note that $P$ and $Q$ must lie on the line that is perpendicular to the plane of $ABC$ and also passes through the circumcenter of $ABC$ (due to $P$ and $Q$ being equidis... | pair C, D, I, P, Q, O; D=(0,0); C=(5.196152,0); P=(1.732051,7.37228); I=(1.732051,0); Q=(1.732051,-1.62772); O=(1.732051,2.87228); draw(C--Q--D--P--cycle); draw(C--D, dashed); draw(P--Q, dotted); draw(O--C, dotted); label("$C$", C, E); label("$D$", D, W); label("$I$", I, NW); label("$P$", P, N); label("$Q$", Q, S); l... | [] |
492 | Equilateral $\triangle ABC$ has side length $600$. Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$, and $QA=QB=QC$, and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). Th... | 2016 AIME II Problems/Problem 14 | Let $M$ be the midpoint of $\overline{AB}$ and $X$ the center of $\triangle ABC$. Then \[P, O, Q, M, X, C\] all lie in the same vertical plane. We can make the following observations:
The equilateral triangle has side length $600$, so $MC=300\sqrt{3}$ and $X$ divides $MC$ so that $MX=100\sqrt{3}$ and $XC=200\sqrt{3}$;... | // Block 1
unitsize(20);
pair P = (0, 12);
pair Q = (0, -3);
pair O = (P+Q)/2;
pair M = (-3, 0);
pair X = (0, 0);
pair C = (6, 0);
draw(P--O--Q);
draw(M--X--C);
draw(P--M--Q, blue);
draw(Q--C--P);
draw(circle((0, 4.5), 7.5));
label("$P$", P, N);
label("$Q$", Q, S);
label("$O$", O, E);
dot(O);
label("$M$", M, W);
label(... | [] |
493 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problems/Problem 24 | Let $AD$ intersect $OB$ at $E$ and $OC$ at $F.$
$\overarc{AB}= \overarc{BC}= \overarc{CD}=\theta$
$\angle{BAD}=\frac{1}{2} \cdot \overarc{BCD}=\theta=\angle{AOB}$
From there, $\triangle{OAB} \sim \triangle{ABE}$, thus:
$\frac{OA}{AB} = \frac{AB}{BE} = \frac{OB}{AE}$
$OA = OB$ because they are both radii of $\o... | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, E, F, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagr... | [] |
493 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problems/Problem 24 | To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by $200$ for now, then multiply it back at the end of our solution.
Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be ... | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, E, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
E=extension(B,D,O,C);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycl... | [] |
493 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problems/Problem 24 | size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
P=RADIUS*dir(270);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Dia... | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
P=RADIUS*dir(270);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//... | [] |
493 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problems/Problem 24 | Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$. Apply the law of cosines on $\Delta BOC$; let $\theta = \angle BOC$. We get ... | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
dra... | [] |
493 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problems/Problem 24 | Construct quadrilateral $ABCD$ on the circle $O$ with $AD$ being the desired side. Then, drop perpendiculars from $A$ and $D$ to the extended line of $\overline{BC}$ and let these points be $E$ and $F$, respectively. Also, let $\theta = \angle BOC$. From the Law of Cosines on $\triangle BOC$, we have $\cos \theta = \fr... | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, E, F, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
E=foot(A,B,C);
F=foot(D,B,C);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -... | [] |
493 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problems/Problem 24 | Let $s = 200$. Let $O$ be the center of the circle. Then $AC$ is twice the altitude of $\triangle OBC$ to $\overline{OB}$. Since $\triangle OBC$ is isosceles we can compute its area to be $\frac{s^2 \sqrt{7}}{4}$, hence $CA = 2 \cdot \tfrac{2 \cdot s^2\sqrt7/4}{s\sqrt2} = s\sqrt{\frac{7}{2}}$.
Now by Ptolemy's Theo... | // Block 1
pathpen = black; pointpen = black;
size(6cm);
draw(unitcircle);
pair A = D("A", dir(50), dir(50));
pair B = D("B", dir(90), dir(90));
pair C = D("C", dir(130), dir(130));
pair D = D("D", dir(170), dir(170));
pair O = D("O", (0,0), dir(-90));
draw(A--C, red);
draw(B--D, blue+dashed);
draw(A--B--C--D--cycle);
... | [] |
493 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problems/Problem 24 | Label the points as shown, and let $\angle{EOF} = \theta$. Since $\overline{OB} = \overline{OC}$, and $\triangle{OFE} \sim \triangle{OCB}$, we get that $\angle{EFO} = 90-\frac{\theta}{2}$. We assign $\alpha$ to $90-\frac{\theta}{2}$ for simplicity.
From here, by vertical angles $\angle{CFD} = \alpha$. Also, since $\tr... | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations, L is used to write alpha= statement
real RADIUS;
pair A, B, C, D, E, F, O, L;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
E=extension(A,D,O,B);
F=extension(A,D,O,C);... | [] |
493 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problems/Problem 24 | Claim: $[ABCD]$ is an isosceles trapezoid.
Proof: Notice that $[ABCD]$ is cyclic, triangle $BOC$ is isosceles, and triangle $AOB$ is congruent to $DOC$ by SSS congruence. Therefore, $\angle BAD = 180 - \angle BCD = 180-(\angle BCO + \angle DCO)=180-(\angle CBO+\angle ABO) = 180 - \angle ABC = \angle CDA$. Hence, $[ABC... | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
dra... | [] |
493 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problems/Problem 24 | Since $AB$, $BC$, and $CD$ are congruent chords, the triangles formed by connecting these sides to the circumcenter $O$ (i.e., $\triangle AOB$, $\triangle BOC$, $\triangle COD$) are congruent isosceles triangles.
Let $\alpha = \angle OCB$. In $\triangle BOC$, the side length $BC=200$ and $OC = OB = 200\sqrt{2}$. By dro... | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, E, F, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagr... | [] |
494 | Circles with centers $P, Q$ and $R$, having radii $1, 2$ and $3$, respectively, lie on the same side of line $l$ and are tangent to $l$ at $P', Q'$ and $R'$, respectively, with $Q'$ between $P'$ and $R'$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $PQ... | 2016 AMC 12A Problems/Problem 15 | Notice that we can find $[P'PQRR']$ in two different ways: $[P'PQQ']+[Q'QRR']$ and $[PQR]+[P'PRR'].$ Since we want $[PQR],$ we use the latter method, so we have $[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR'].$
$\break$
$P'Q'=\sqrt{PQ^2-(QQ'-PP')^2}=\sqrt{3^2-1^2}=\sqrt{8}=2\sqrt{2}$. Additionally, $Q'R'=\sqrt{QR^2-(RR'-QQ')^2}=\sq... | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
pair P,Q,R,Pp,Qp,Rp;
pair A,B;
//Variable Definitions
A=(-5, 0);
B=(8, 0);
P=(-2.828,1);
Q=(0,2);
R=(4.899,3);
Pp=foot(P,A,B);
Qp=foot(Q,A,B);
Rp=foot(R,A,B);
path PQR = P--Q--R--cycle;
//Initial Diagram
dot(P);
dot(Q);
dot(R);
dot(Pp);
dot(Qp);... | [] |
494 | Circles with centers $P, Q$ and $R$, having radii $1, 2$ and $3$, respectively, lie on the same side of line $l$ and are tangent to $l$ at $P', Q'$ and $R'$, respectively, with $Q'$ between $P'$ and $R'$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $PQ... | 2016 AMC 12A Problems/Problem 15 | The above diagram can be achieved relatively simply using basic knowledge of the Pythagorean theorem and the fact that the radius from the center to the point of tangency is perpendicular to the tangent line. From there, observe that $[PQRY]$ can be calculated in two ways: $[\triangle PQX] + [QZYX] + [\triangle QRZ]$ a... | // Block 1
// Initial Pen Sizing
size(250);
defaultpen(linewidth(0.4));
defaultpen(fontsize(10pt));
// Variable Declarations
pair P,Q,R,Pp,Qp,Rp,X,Y,Z,A,B;
// Variable Definitions
A=(-5, 0);
B=(8, 0);
P=(-2.828,1);
Q=(0,2);
R=(4.899,3);
X=(0,1);
Y=(4.899,1);
Z=(4.899,2);
Pp=foot(P,A,B);
Qp=foot(Q,A,B);
Rp=foot(R,A,B)... | [] |
495 | Circle $C_0$ has radius $1$, and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r<1$ and is internally tangent to $C_0$ at point $A_0$. Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\circ}$ counterclockwise from $A_0$ on $C_1$. Circle $C_2$ has radius $r^2$ and is internally tangent... | 2017 AIME II Problems/Problem 12 | Using the invariance again as in Solution 3, assume $B$ is $d$ away from the origin. The locus of possible points is a circle with radius $d$. Consider the following diagram.
Let the distance from $B$ to $(1,0)$ be $x$. As $B$ is invariant, $x = r(BB' + x) \implies x = r\frac{d\sqrt{2}}{1-r}$. Then by Power of a Poin... | // Block 1
size(7cm);
draw(circle((0,0), 49/61));
draw((0,0)--(0.790110185, 0.144853534));
draw((0,0)--(-0.144853534, 0.790110185));
draw((-0.144853534, 0.790110185)--(1,0));
draw((0,0)--(1,0));
draw(rightanglemark((-0.144853534, 0.790110185), (0,0), (0.790110185, 0.144853534), 3));
label("$O$", (0,0), SW);
label("$(1... | [] |
496 | The diameter $AB$ of a circle of radius $2$ is extended to a point $D$ outside the circle so that $BD=3$. Point $E$ is chosen so that $ED=5$ and line $ED$ is perpendicular to line $AD$. Segment $AE$ intersects the circle at a point $C$ between $A$ and $E$. What is the area of $\triangle ABC$?
$\textbf{(A)}\ \frac{120... | 2017 AMC 12B Problems/Problem 18 | Let $O$ be the center of the circle. Note that $EC + CA = EA = \sqrt{AD^2 + DE^2} = \sqrt{(2+2+3)^2 + 5^2} = \sqrt{74}$. However, by Power of a Point, $(EC)(EC + CA) = EO^2 - R^2 = (2+3)^2 + 5^2 - 2^2 = 25 + 25 - 4 = 46 \implies EC = \frac{46}{\sqrt{74}}$, so $AC = \sqrt{74} - \frac{46}{\sqrt{74}} = \frac{28}{\sqrt{74}... | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(8.865514650638614cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen ... | [] |
497 | $ABCD$ is a parallelogram. $g$ is a line passing $A$. Prove that the distance from $C$ to $g$ is either the sum or the difference of the distance from $B$ to $g$, and the distance from $D$ to $g$. | 2017 Indonesia MO Problems/Problem 1 | In order to prove that the distance from $C$ to $g$ is either the sum or the difference of the distance from $B$ to $g$, and the distance from $D$ to $g$, we will use casework to show that the statement is true for all scenarios.
Case 1: $g$ passes through $D$ or $B$
If $g$ passes through $D$, then the distance from ... | // Block 1
pair A=(0,0),B=(50,0),C=(60,40),D=(10,40);
draw(A--B--C--D--A);
dot(A);
label("A",A,SW);
dot(B);
label("B",B,SE);
dot(C);
label("C",C,NE);
dot(D);
label("D",D,NW);
pair e=(48,40),XA=(55.082,45.902),XB=(29.508,24.59),XC=(25.574,21.311);
draw(A--XA--C);
draw(D--XC);
draw(B--XB);
dot(e);
label("E",e,N);
dot(XA... | [] |
498 | Suppose n points on the circumference of a circle
are joined by straight line segments in all possible ways
and that no point that is not one of the original n points
is contained in more than two of the segments. How
many triangles are formed by the segments? Count all
triangles whose sides lie along the segments, inc... | 2017 UNCO Math Contest II Problems/Problem 9 | Since any $3$ pairwise intersecting lines make a triangle, so we can split the possibilities into $4$ cases:
Case 1: All vertices of the triangle are points on the circle
Any $3$ points will work, so it's ${n}\choose{3}$
Case 2: 2 vertices of the triangle are points on the circle
There will be $4$ points to cho... | // Block 1
pair A=dir(0),B=dir(80),C=dir(160);
draw(unitcircle);
draw(A--B--C--A);
// Block 2
pair A=dir(0),B=dir(80),C=dir(160),D=dir(310);
draw(unitcircle);
draw(A--C--B--D);
// Block 3
pair A=dir(0),B=dir(80),C=dir(160),D=dir(210),E=dir(300);
draw(unitcircle);
draw(A--C--E);
draw(B--D);
// Block 4
pair A=dir(0),B=di... | [] |
499 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortes... | 2018 AMC 10A Problems/Problem 23 | Note that the hypotenuse of the field is $5,$ and the area of the field is $6.$ Let $x$ be the side-length of square $S.$
We partition the field into a red triangle, a yellow triangle, and a green triangle, as shown below:
Let the brackets denote areas. By area addition, we set up an equation for $x:$
\begin{align*} ... | // Block 1
/* Edited by MRENTHUSIASM */
size(180);
pair A, B, C, D, F;
A = origin;
B = (4,0);
C = (0,3);
D = (2/7,2/7);
F = foot(D,B,C);
fill(A--D--C--cycle, red);
fill(A--D--B--cycle, yellow);
fill(B--D--C--cycle, green);
draw(A--B--C--cycle);
label("$5$", midpoint(B--C), NE);
label("$4$", midpoint(A--B), S);
label("$... | [] |
499 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortes... | 2018 AMC 10A Problems/Problem 23 | Let the square have side length $x$. Connect the upper-right vertex of square $S$ with the two vertices of the triangle's hypotenuse. This divides the triangle in several regions whose areas must add up to the area of the whole triangle, which is $6$.
Square $S$ has area $x^2$, and the two thin triangle regions have a... | // Block 1
/* Edited by MRENTHUSIASM */
size(180);
pair A, B, C, D, F;
A = origin;
B = (4,0);
C = (0,3);
D = (2/7,2/7);
F = foot(D,B,C);
fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray);
draw(A--B--C--cycle);
label("$5$", midpoint(B--C), NE);
label("$4$", midpoint(A--B), S);
label("$3$", midpoint(A--C), W);
label("$2$", ... | [] |
499 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortes... | 2018 AMC 10A Problems/Problem 23 | On the diagram above, find two smaller triangles similar to the large one with side lengths $3$, $4$, and $5$; consequently, the segments with length $\frac{5}{2}$ and $\frac{10}{3}$.
With $\ell$ being the side length of the square, we need to find an expression for $\ell$. Using the hypotenuse, we can see that $\frac... | // Block 1
/* Edited by MRENTHUSIASM */
size(180);
pair A, B, C, D, F;
A = origin;
B = (4,0);
C = (0,3);
D = (2/7,2/7);
F = foot(D,B,C);
fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray);
draw(A--B--C--cycle);
label("$4$", midpoint(A--B), S);
label("$3$", midpoint(A--C), W);
label("$2$", midpoint(D--F), SE);
label("$S$", ... | [] |
499 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortes... | 2018 AMC 10A Problems/Problem 23 | Let $AD=x$. Note that $\triangle DEF$ is a $3{-}4{-}5$ triangle, so $EF=\frac{5}{4}x$ and $FD=\frac{3}{4}x$. $BF=BD+FD=4-x+\frac{3}{4}x=4-\frac{1}{4}x$. We know that $GE$ is $2$ from the problem so $GF=2+\frac{5}{4}x$. $\triangle FGB$ is also a $3{-}4{-}5$ triangle with $GF:BF=3:5$. We now have $3\left(4-\frac{1}{4}x\r... | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(180);
real labelscalefactor = 1.5; /* changes label-to-point distance */
// pen dps = linewidth(0.5) + fontsize(10);
// defaultpen(dps); /* default pen style */
// pen dotstyl... | [] |
499 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortes... | 2018 AMC 10A Problems/Problem 23 | Denote $A,B,C$ to be the three vertices of the triangular field. Also denote $A,M,D,N$ to be the vertices of the square $S$. Let $X$ be on $BC$ such that $AC\parallel DX$ and $Y$ be on $BC$ such that $AB\parallel DY$. Let $P$ and $Q$ be the foot of the altitudes from $X$ to $AC$ and from $Y$ to $AB$ respectively.
Note... | // Block 1
size(240);
pair A, B, C, D, F, X, Y, P, Q, M, N;
A = origin; label(A, "$A$", SW);
B = (4,0); label(B, "$B$", S);
C = (0,3); label(C, "$C$", W);
D = (2/7,2/7); label(D, "$D$", NE);
F = foot(D,B,C); label(F, "$F$", NE);
X = (2/7,39/14); label(X, "$X$", NE, red);
Y = (76/21,2/7); label(Y, "$Y$", NE, red);
P = f... | [] |
500 | Line segment $\overline{AB}$ is a diameter of a circle with $AB = 24$. Point $C$, not equal to $A$ or $B$, lies on the circle. As point $C$ moves around the circle, the centroid (center of mass) of $\triangle ABC$ traces out a closed curve missing two points. To the nearest positive integer, what is the area of the reg... | 2018 AMC 10B Problems/Problem 12 | By the Inscribed Angle Theorem, $\triangle ABC$ is a right triangle with $\angle C=90^{\circ}.$ So, its circumcenter is the midpoint of $\overline{AB},$ and its median from $C$ is half as long as $\overline{AB}.$ For each $\triangle ABC,$ let $O$ and $G$ be its circumcenter and centroid, respectively. It follows that $... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
pair O, A, B, C1, C2, G1, G2, M1, M2;
O = (0,0);
A = (-12,0);
B = (12,0);
C1 = (36/5,48/5);
C2 = (-96/17,-180/17);
G1 = O + 1/3 * C1;
G2 = O + 1/3 * C2;
M1 = (4,0);
M2 = (-4,0);
draw(Circle(O,12));
draw(Circle(O,4),red);
dot("$O$", O, (3/5,-4/5), linewidth(4.5));
dot("... | [] |
501 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A... | 2018 AMC 10B Problems/Problem 24 | The desired area (hexagon $MPNQOR$) consists of an equilateral triangle ($\triangle MNO$) and three right triangles ($\triangle MPN,\triangle NQO,$ and $\triangle ORM$).
Notice that $\overline {AD}$ (not shown) and $\overline {BC}$ are parallel. $\overline {XY}$ divides transversals $\overline {AB}$ and $\overline {CD... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
draw(polygon(6));
pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R;
A = dir(120);
B = dir(60);
C = dir(0);
D = dir(300);
E = dir(240);
F = dir(180);
X = midpoint(A--B);
Y = midpoint(C--D);
Z = midpoint(E--F);
M = intersectionpoint(A--E,X--Z);
N = intersectionpoint(A--C,X... | [] |
501 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A... | 2018 AMC 10B Problems/Problem 24 | Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of $3$ isosceles trapezoids (namely $AXZF,CYXB,$ and $EZYD$) and $3$ right triangles (namely $\triangle XPN,\triangle YQO,$ and $\triangle ZRM$).
Finding the trapezoid's area, we know that one base of each tr... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
draw(polygon(6));
pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R;
A = dir(120);
B = dir(60);
C = dir(0);
D = dir(300);
E = dir(240);
F = dir(180);
X = midpoint(A--B);
Y = midpoint(C--D);
Z = midpoint(E--F);
M = intersectionpoint(A--E,X--Z);
N = intersectionpoint(A--C,X... | [] |
501 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A... | 2018 AMC 10B Problems/Problem 24 | We could also subtract $\triangle APM,\triangle CQN,$ and $\triangle ERO$ from $\triangle ACE.$
Since $\angle BAF = 120^{\circ}$ and $\angle BAC = \angle FAE = 30^{\circ},$ we have $\angle CAE = \angle BAF-\angle BAC-\angle FAE=60^{\circ}.$
Since $AX=BX$ and $FZ=EZ,$ we have $AF \parallel XZ,$ from which $\angle AMX=... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
draw(polygon(6));
pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R;
A = dir(120);
B = dir(60);
C = dir(0);
D = dir(300);
E = dir(240);
F = dir(180);
X = midpoint(A--B);
Y = midpoint(C--D);
Z = midpoint(E--F);
M = intersectionpoint(A--E,X--Z);
N = intersectionpoint(A--C,X... | [] |
501 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A... | 2018 AMC 10B Problems/Problem 24 | We partition hexagon $ABCDEF$ into $48$ congruent $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangles, as shown below:
Let the brackets denote areas. Note that the desired region contains $15$ of the $48$ small triangles, so the answer is \[\frac{15}{48}[ABCDEF]=\frac{15}{48}\cdot\frac{3\sqrt3}{2}=\boxed{\textbf{(C)}... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R;
A = dir(120);
B = dir(60);
C = dir(0);
D = dir(300);
E = dir(240);
F = dir(180);
X = midpoint(A--B);
Y = midpoint(C--D);
Z = midpoint(E--F);
M = intersectionpoint(A--E,X--Z);
N = intersectionpoint(A--C,X--Y);
O = intersec... | [] |
501 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A... | 2018 AMC 10B Problems/Problem 24 | If we try to coordinate bash this problem, it's going to look very ugly with a lot of radicals. However, we can alter and skew the diagram in such a way that all ratios of lengths and areas stay the same while making it a lot easier to work with. Then, we can find the ratio of the area of the wanted region to the area ... | // Block 1
unitsize(1cm);
draw((0,0)--(4,0)--(6,3.464)--(2,3.464)--(0,0));
draw((2,0)--(1,1.732));
draw((5,1.732)--(4,3.464));
draw((1.5, 0.866)--(3, 3.464)--(4.5, 0.866)--cycle);
draw((2,0)--(2,3.464)--(5,1.732)--cycle);
// Block 2
unitsize(1cm);
fill((1,4)--(1,3.5)--(2,3)--cycle,red);
fill((1,1)--(1.5,1)--(1,2)--cycl... | [] |
501 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A... | 2018 AMC 10B Problems/Problem 24 | (~MRENTHUSIASM's diagram + modifications)
Lemma 1: $MP$ is midsegment of $\triangle ANE$.
As $ZX$ is a midsegment of trapezoid $FEBA$, we know that $FA||ZX||EB$. Furthermore, by subdividing trapezoid $FEBA$, $ZM,MP,PX$ are midsegments of $\triangle EFA, \triangle ANE, \triangle, ANB$ respectively.
Lemma 2: $[ACE] = \... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
draw(polygon(6));
pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R, G;
A = dir(120);
B = dir(60);
C = dir(0);
D = dir(300);
E = dir(240);
F = dir(180);
X = midpoint(A--B);
Y = midpoint(C--D);
Z = midpoint(E--F);
G = midpoint(A--D);
M = intersectionpoint(A--E,X--Z);
N =... | [] |
502 | Square $ABCD$ has side length $30$. Point $P$ lies inside the square so that $AP = 12$ and $BP = 26$. The centroids of $\triangle{ABP}$, $\triangle{BCP}$, $\triangle{CDP}$, and $\triangle{DAP}$ are the vertices of a convex quadrilateral. What is the area of that quadrilateral?
$\textbf{(A) }100\sqrt{2}\qquad\textbf... | 2018 AMC 12B Problems/Problem 13 | As shown below, let $M_1,M_2,M_3,M_4$ be the midpoints of $\overline{AB},\overline{BC},\overline{CD},\overline{DA},$ respectively, and $G_1,G_2,G_3,G_4$ be the centroids of $\triangle{ABP},\triangle{BCP},\triangle{CDP},\triangle{DAP},$ respectively.
By SAS, we conclude that $\triangle G_1G_2P\sim\triangle M_1M_2P, \tr... | // Block 1
/* Made by MRENTHUSIASM */
unitsize(210);
pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3);
pair M1 = midpoint(A--B);
pair M2 = midpoint(B--C);
pair M3 = midpoint(C--D);
pair M4 = midpoint(D--A);
pair G1 = centroid(A,B,P);
pair G2 = centroid(B,C,P);
pair G3 = centroid(C,D,P);
pair G4 = cen... | [] |
503 | The solutions to the equation $(z+6)^8=81$ are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled $A,B,$ and $C$. What is the least possible area of $\triangle ABC?$
$\textbf{(A) } \frac{1}{6}\sqrt{6} \qquad \textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2} \qquad \textbf{... | 2018 AMC 12B Problems/Problem 16 | Recall that translations preserve the shapes and the sizes for all objects. We translate the solutions to the given equation $6$ units right, so they become the solutions to the equation $z^8=81.$
We rewrite $z$ to the polar form \[z=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta,\] where $r$ is the magnitude of ... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
int xMin = -2;
int xMax = 2;
int yMin = -2;
int yMax = 2;
int numRays = 24;
//Draws a polar grid that goes out to a number of circles
//equal to big, with numRays specifying the number of rays:
void polarGrid(int big, int numRays)
{
for (int i = 1; i < big+1; ++i... | [] |
504 | Ajay is standing at point $A$ near Pontianak, Indonesia, $0^\circ$ latitude and $110^\circ \text{ E}$ longitude. Billy is standing at point $B$ near Big Baldy Mountain, Idaho, USA, $45^\circ \text{ N}$ latitude and $115^\circ \text{ W}$ longitude. Assume that Earth is a perfect sphere with center $C.$ What is the degre... | 2018 AMC 12B Problems/Problem 23 | This solution refers to the Diagram section.
Let $D$ be the orthogonal projection of $B$ onto the equator. Note that $\angle BDA = \angle BDC = 90^\circ$ and $\angle BCD = 45^\circ.$ Recall that $115^\circ \text{ W}$ longitude is the same as $245^\circ \text{ E}$ longitude, so $\angle ACD=135^\circ.$
We obtain the fo... | // Block 1
/* Made by MRENTHUSIASM */
size(250);
import graph3;
import solids;
currentprojection=orthographic((0.2,-0.5,0.2));
triple A, B, C, D;
A = (1,0,0);
B = (-1/2,1/2,sqrt(2)/2);
C = (0,0,0);
D = (-1/2,1/2,0);
draw(unitsphere,white,light=White);
draw(surface(A--B--C--cycle),yellow);
dot(A^^B^^C^^D,linewidth(4.5)... | [] |
504 | Ajay is standing at point $A$ near Pontianak, Indonesia, $0^\circ$ latitude and $110^\circ \text{ E}$ longitude. Billy is standing at point $B$ near Big Baldy Mountain, Idaho, USA, $45^\circ \text{ N}$ latitude and $115^\circ \text{ W}$ longitude. Assume that Earth is a perfect sphere with center $C.$ What is the degre... | 2018 AMC 12B Problems/Problem 23 | This solution refers to the Diagram section.
Let $D$ be the orthogonal projection of $B$ onto the equator. Note that $\angle BDA = \angle BDC = 90^\circ$ and $\angle BCD = 45^\circ.$ Recall that $115^\circ \text{ W}$ longitude is the same as $245^\circ \text{ E}$ longitude, so $\angle ACD=135^\circ.$
Without loss of ... | // Block 1
/* Made by MRENTHUSIASM */
size(300);
import graph3;
import solids;
currentprojection=orthographic((0.2,-0.5,0.2));
triple A, B, C, D;
A = (1,0,0);
B = (-1/2,1/2,sqrt(2)/2);
C = (0,0,0);
D = (-1/2,1/2,0);
draw(unitsphere,white,light=White);
dot(A^^B^^C^^D,linewidth(4.5));
draw(Circle(C,1,(0,0,1))^^B--C--D--... | [] |
505 | Circles $\omega_1$, $\omega_2$, and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$, $P_2$, and $P_3$ lie on $\omega_1$, $\omega_2$, and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\om... | 2018 AMC 12B Problems/Problem 25 | First, note that because the $\angle P_1=\angle P_2=\angle P_3=\pi/3$, the arcs inside the shaded equilateral triangle are each $2\pi/3$. Also, the distances between the centers of any two of the $3$ given circles are each $8$.
Draw the circle $\Gamma$ concentric with $\omega_1$ with radius $2$. Because the arc of $\om... | // Block 1
unitsize(12);
pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A;
real theta = 41.5;
pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1;
filldraw(P1--P2--P3--cycle, gray(0.9));
draw(Circle(A, 4));
draw(Circle(B, 4));
draw(Circle(C, 4));
dot(P1);
dot(P2);
dot(P3... | [] |
505 | Circles $\omega_1$, $\omega_2$, and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$, $P_2$, and $P_3$ lie on $\omega_1$, $\omega_2$, and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\om... | 2018 AMC 12B Problems/Problem 25 | Suppose $P_1P_2 = x$, noticed that $OMP_1H$ is rectangle.
\[OO_1 = \frac{O_1O_2}{\sqrt{3}} = \frac83\sqrt{3}\]
\[OH = P_1M = \frac{1}{2}P_1P_2 = \frac12 x\]
\[O_1H = O_1P_1 - HP_1 = O_1P_1 - OM = O_1P_1 - \frac{P_1P_2}{2\sqrt{3}} = 4 - \frac{\sqrt{3}}{6}x\]
$\triangle O_1HO$ is right triangle, we can use Pythagorean th... | // Block 1
import geometry;
unitsize(12);
pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A;
real theta = 41.5;
pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1;
filldraw(P1--P2--P3--cycle, gray(0.9));
draw(Circle(A, 4));
draw(Circle(B, 4));
draw(Circle(C, 4));
dot(P1);... | [] |
506 | In $\triangle ABC$, the sides have integer lengths and $AB=AC$. Circle $\omega$ has its center at the incenter of $\triangle ABC$. An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that ... | 2019 AIME I Problems/Problem 11 | First, assume $BC=2$ and $AB=AC=x$. The triangle can be scaled later if necessary. Let $I$ be the incenter and let $r$ be the inradius. Let the points at which the incircle intersects $AB$, $BC$, and $CA$ be denoted $M$, $N$, and $O$, respectively.
Next, we calculate $r$ in terms of $x$. Note the right triangle form... | // Block 1
unitsize(1cm);
var x = 9;
pair A = (0,sqrt(x^2-1));
pair B = (-1,0);
pair C = (1,0);
dot(Label("$A$",A,NE),A);
dot(Label("$B$",B,SW),B);
dot(Label("$C$",C,SE),C);
draw(A--B--C--cycle);
var r = sqrt((x-1)/(x+1));
pair I = (0,r);
dot(Label("$I$",I,SE),I);
draw(circle(I,r));
draw(Label("$r$"),I--I+r*SSW... | [] |
506 | In $\triangle ABC$, the sides have integer lengths and $AB=AC$. Circle $\omega$ has its center at the incenter of $\triangle ABC$. An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that ... | 2019 AIME I Problems/Problem 11 | Before we start thinking about the problem, let’s draw it out;
For the sake of space, I've drawn only 2 of the 3 excircles because the third one looks the same as the second large one because the triangle is isosceles. By the incenter-excenter lemma, $AII_A$ and $BII_B$ are collinear, $E$ is the tangent of circle $I... | // Block 1
unitsize(1cm);
var x = 9;
pair A = (0,sqrt(x^2-1));
pair B = (-1,0);
pair C = (1,0);
dot(Label("$A$",A,NE),A);
dot(Label("$B$",B,SW),B);
dot(Label("$C$",C,SE),C);
draw(A--B--C--cycle);
var r = sqrt((x-1)/(x+1));
pair I = (0,r);
dot(Label("$I$",I,SE),I);
draw(circle(I,r));
pair G = intersectionpoint... | [] |
507 | Triangle $ABC$ has side lengths $AB=4$, $BC=5$, and $CA=6$. Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$. The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$. Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$, where $a$, $b$... | 2019 AIME I Problems/Problem 13 | Notice that \[\angle DFE=\angle CFE-\angle CFD=\angle CBE-\angle CAD=180-B-A=C.\]By the Law of Cosines, \[\cos C=\frac{AC^2+BC^2-AB^2}{2\cdot AC\cdot BC}=\frac34.\]Then, \[DE^2=DF^2+EF^2-2\cdot DF\cdot EF\cos C=32\implies DE=4\sqrt2.\]Let $X=\overline{AB}\cap\overline{CF}$, $a=XB$, and $b=XD$. Then, \[XA\cdot XD=XC\cdo... | // Block 1
unitsize(20);
pair A, B, C, D, E, F, X, O1, O2;
A = (0, 0); B = (4, 0);
C = intersectionpoints(circle(A, 6), circle(B, 5))[0];
D = B + (5/4 * (1 + sqrt(2)), 0); E = D + (4 * sqrt(2), 0);
F = intersectionpoints(circle(D, 2), circle(E, 7))[1];
X = extension(A, E, C, F);
O1 = circumcenter(C, A, D);
O2 = circumc... | [] |
508 | Let $\overline{AB}$ be a chord of a circle $\omega$, and let $P$ be a point on the chord $\overline{AB}$. Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$. Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$. Circles $\omega_1$ and $\omega_2$ intersect at... | 2019 AIME I Problems/Problem 15 | Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$, respectively. There is a homothety at $A$ sending $\omega$ to $\omega_1$ that sends $B$ to $P$ and $O$ to $O_1$, so $\overline{OO_2}\parallel\overline{O_1P}$. Similarly, $\overline{OO_1}\parallel\overline{O_2P}$, so $OO_1PO_2$ is a parallelogram. Moreover... | // Block 1
size(8cm);
pair O, A, B, P, O1, O2, Q, X, Y;
O=(0, 0);
A=dir(140); B=dir(40);
P=(3A+5B)/8;
O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);
O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O);
Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1];
X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1... | [] |
508 | Let $\overline{AB}$ be a chord of a circle $\omega$, and let $P$ be a point on the chord $\overline{AB}$. Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$. Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$. Circles $\omega_1$ and $\omega_2$ intersect at... | 2019 AIME I Problems/Problem 15 | $PX \cdot PY=AP \cdot PB=5 \cdot 3=15$ by power of a point. Also, $PX+PY=XY=11$, so $PX$ and $PY$ are solutions to the quadratic $x^2-11x+15=0$ so $PX$ and $PY$ is $\frac{11\pm\sqrt{61}}{2}$ in some order. Now, because we want $PQ^2$ and it is known to be rational, we can guess that $PQ$ is irrational or the problem wo... | // Block 1
size(8cm);
pair O, A, B, P, O1, O2, Q, X, Y;
O=(0, 0);
A=dir(140); B=dir(40);
P=(3A+5B)/8;
O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);
O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O);
Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1];
X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1... | [] |
509 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problems/Problem 6 | NOTE: this solution is wrong. The equation is correct due to similar triangles as described in solution 8, not PoP.
Because $\angle KLN = \angle KMN = 90^{\circ}$, $KLMN$ is a cyclic quadrilateral. (THE FOLLOWING SENTENCE IS WRONG) Hence, by Power of Point, \[KO\cdot KM = KL^2 \implies KM=\dfrac{28^2}{8}=98 \implies... | // Block 1
size(250);
real h = sqrt(98^2+65^2);
real l = sqrt(h^2-28^2);
pair K = (0,0);
pair N = (h, 0);
pair M = ((98^2)/h, (98*65)/h);
pair L = ((28^2)/h, (28*l)/h);
pair P = ((28^2)/h, 0);
pair O = ((28^2)/h, (8*65)/h);
draw(K--L--N);
draw(K--M--N--cycle);
draw(L--M);
label("K", K, SW);
label("L", L, NW);
label("M"... | [] |
509 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problems/Problem 6 | First, let $P$ be the intersection of $LO$ and $KN$ as shown above. Note that $m\angle KPL = 90^{\circ}$ as given in the problem. Since $\angle KPL \cong \angle KLN$ and $\angle PKL \cong \angle LKN$, $\triangle PKL \sim \triangle LKN$ by AA similarity. Similarly, $\triangle KMN \sim \triangle KPO$. Using these similar... | // Block 1
size(250);
real h = sqrt(98^2+65^2);
real l = sqrt(h^2-28^2);
pair K = (0,0);
pair N = (h, 0);
pair M = ((98^2)/h, (98*65)/h);
pair L = ((28^2)/h, (28*l)/h);
pair P = ((28^2)/h, 0);
pair O = ((28^2)/h, (8*65)/h);
draw(K--L--N);
draw(K--M--N--cycle);
draw(L--M);
label("K", K, SW);
label("L", L, NW);
label("M"... | [] |
509 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problems/Problem 6 | Notice that $KLMN$ is inscribed in the circle with diameter $\overline{KN}$ and $XOMN$ is inscribed in the circle with diameter $\overline{ON}$. Furthermore, $(XLN)$ is tangent to $\overline{KL}$. Then, \[KO\cdot KM=KX\cdot KN=KL^2\implies KM=\frac{28^2}{8}=98,\]and $MO=KM-KO=\boxed{090}$.
(Solution by TheUltimate123)... | // Block 1
size(8cm);
pair K, L, M, NN, X, O;
K=(-sqrt(98^2+65^2)/2, 0);
NN=(sqrt(98^2+65^2)/2, 0);
L=sqrt(98^2+65^2)/2*dir(180-2*aSin(28/sqrt(98^2+65^2)));
M=sqrt(98^2+65^2)/2*dir(2*aSin(65/sqrt(98^2+65^2)));
X=foot(L, K, NN);
O=extension(L, X, K, M);
draw(K -- L -- M -- NN -- K -- M); draw(L -- NN); draw(arc((K+NN)/2... | [] |
509 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problems/Problem 6 | (Diagram by vedadehhc)
Call the base of the altitude from $L$ to $NK$ point $P$. Let $PO=x$. Now, we have that $KP=\sqrt{64-x^2}$ by the Pythagorean Theorem. Once again by Pythagorean, $LO=\sqrt{720+x^2}-x$. Using Power of a Point, we have
\[(KO)(OM)=(LO)(OQ)\] ($Q$ is the intersection of $OL$ with the circle $\neq ... | // Block 1
size(250);
real h = sqrt(98^2+65^2);
real l = sqrt(h^2-28^2);
pair K = (0,0);
pair N = (h, 0);
pair M = ((98^2)/h, (98*65)/h);
pair L = ((28^2)/h, (28*l)/h);
pair P = ((28^2)/h, 0);
pair O = ((28^2)/h, (8*65)/h);
draw(K--L--N);
draw(K--M--N--cycle);
draw(L--M);
label("K", K, SW);
label("L", L, NW);
label("M"... | [] |
509 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problems/Problem 6 | Note that since $\angle KLN = \angle KMN$, quadrilateral $KLMN$ is cyclic. Therefore, we have \[\angle LMK = \angle LNK = 90^{\circ} - \angle LKN = \angle KLP,\]so $\triangle KLO \sim \triangle KML$, giving \[\frac{KM}{28} = \frac{28}{8} \implies KM = 98.\] Therefore, $OM = 98-8 = \boxed{90}$. | // Block 1
size(250);
real h = sqrt(98^2+65^2);
real l = sqrt(h^2-28^2);
pair K = (0,0);
pair N = (h, 0);
pair M = ((98^2)/h, (98*65)/h);
pair L = ((28^2)/h, (28*l)/h);
pair P = ((28^2)/h, 0);
pair O = ((28^2)/h, (8*65)/h);
draw(K--L--N);
draw(K--M--N--cycle);
draw(L--M);
label("K", K, SW);
label("L", L, NW);
label("M"... | [] |
509 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problems/Problem 6 | By Pythagorean Theorem, $KM^2+65^2 = KN^2 = 28^2 + LN^2$. Thus, $LN^2 = KM^2 + 65^2 - 28^2$.
By Pythagorean Theorem, $KP^2 + LP^2 = 28^2$, and $PN^2 + LP^2 = LN^2$.
\[PN^2 = (KN - KP)^2 = (\sqrt{KM^2 + 65^2} - KP)^2\]
It follows that
\[(\sqrt{KM^2 + 65^2} - KP)^2 + LP^2 = KM^2 + 65^2 - 28^2\]
\[KM^2 + 65^2 - 2\sqrt... | // Block 1
size(250);
real h = sqrt(98^2+65^2);
real l = sqrt(h^2-28^2);
pair K = (0,0);
pair N = (h, 0);
pair M = ((98^2)/h, (98*65)/h);
pair L = ((28^2)/h, (28*l)/h);
pair P = ((28^2)/h, 0);
pair O = ((28^2)/h, (8*65)/h);
draw(K--L--N);
draw(K--M--N--cycle);
draw(L--M);
label("K", K, SW);
label("L", L, NW);
label("M"... | [] |
510 | Two different points, $C$ and $D$, lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$, $BC=AD=10$, and $CA=DB=17$. The intersection of these two triangular regions has area $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2019 AIME II Problems/Problem 1 | - Diagram by Brendanb4321
Extend $AB$ to form a right triangle with legs $6$ and $8$ such that $AD$ is the hypotenuse and connect the points $CD$ so
that you have a rectangle. (We know that $\triangle ADE$ is a $6-8-10$, since $\triangle DEB$ is an $8-15-17$.) The base $CD$ of the rectangle will be $9+6+6=21$. Now,... | // Block 1
unitsize(10);
pair A = (0,0);
pair B = (9,0);
pair C = (15,8);
pair D = (-6,8);
pair E = (-6,0);
draw(A--B--C--cycle);
draw(B--D--A);
label("$A$",A,dir(-120));
label("$B$",B,dir(-60));
label("$C$",C,dir(60));
label("$D$",D,dir(120));
label("$E$",E,dir(-135));
label("$9$",(A+B)/2,dir(-90));
label("$10$",(D+A)... | [] |
510 | Two different points, $C$ and $D$, lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$, $BC=AD=10$, and $CA=DB=17$. The intersection of these two triangular regions has area $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2019 AIME II Problems/Problem 1 | - Diagram by Brendanb4321 extended by Duoquinquagintillion
Begin with the first step of solution 1, seeing $AD$ is the hypotenuse of a $6-8-10$ triangle and calling the intersection of $DB$ and $AC$ point $E$. Next, notice $DB$ is the hypotenuse of an $8-15-17$ triangle. Drop an altitude from $E$ with length $h$, so t... | // Block 1
unitsize(10);
pair A = (0,0);
pair B = (9,0);
pair C = (15,8);
pair D = (-6,8);
draw(A--B--C--cycle);
draw(B--D--A);
label("$A$",A,dir(-120));
label("$B$",B,dir(-60));
label("$C$",C,dir(60));
label("$D$",D,dir(120));
label("$9$",(A+B)/2,dir(-90));
label("$10$",(D+A)/2,dir(-150));
label("$10$",(C+B)/2,dir(-30... | [] |
510 | Two different points, $C$ and $D$, lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$, $BC=AD=10$, and $CA=DB=17$. The intersection of these two triangular regions has area $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2019 AIME II Problems/Problem 1 | First, let's define H as the intersection of CB and DA, and define G as the midpoint of AB. Next, let E and F be the feet of the perpendicular lines from C and D to line AB respectively. We get a pleasing line of symmetry HG where A maps to B, C maps to D, and E maps to F. We notice that 8-15-17 and 8-6-10 are both pyt... | //Made by Afly. I used some resources. //Took me 10 min to get everything right. import olympiad; unitsize(18); pair A = (0,0); pair B = (0,8); pair C = (6,0); pair D = (15,0); pair E = (21,0); pair F = (21,8); pair G = (21/2,0); pair H = intersectionpoints(B--D,C--F)[0]; pen dash1 = linetype(new real [] {9,9})+linewid... | [] |
511 | Triangle $ABC$ has side lengths $AB=7, BC=8,$ and $CA=9.$ Circle $\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\omega_1$ and $\omega_2$ not equal to $A.$ Then $AK=\tfrac mn,$ where $m$ an... | 2019 AIME II Problems/Problem 11 | -Diagram by Brendanb4321
Note that from the tangency condition that the supplement of $\angle CAB$ with respects to lines $AB$ and $AC$ are equal to $\angle AKB$ and $\angle AKC$, respectively, so from tangent-chord, \[\angle AKC=\angle AKB=180^{\circ}-\angle BAC\] Also note that $\angle ABK=\angle KAC$$^{(*)}$, so ... | // Block 1
unitsize(20);
pair B = (0,0);
pair A = (2,sqrt(45));
pair C = (8,0);
draw(circumcircle(A,B,(-17/8,0)),rgb(.7,.7,.7));
draw(circumcircle(A,C,(49/8,0)),rgb(.7,.7,.7));
draw(B--A--C--cycle);
label("$A$",A,dir(105));
label("$B$",B,dir(-135));
label("$C$",C,dir(-75));
dot((2.68,2.25));
label("$K$",(2.68,2.25),2*d... | [] |
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