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 |
|---|---|---|---|---|---|
286 | Let $A,B,C,D$ denote four points in space and $AB$ the distance between $A$ and $B$, and so on. Show that
\[AC^2+BD^2+AD^2+BC^2\ge AB^2+CD^2.\] | 1975 USAMO Problems/Problem 2 | Solution 1
If we project points $A,B,C,D$ onto the plane parallel to $\overline{AB}$ and $\overline{CD}$, $AB$ and $CD$ stay the same but $BC, AC, AD, BD$ all decrease, making the inequality sharper. Thus, it suffices to prove the inequality when $A,B,C,D$ are coplanar:
Let $AD=a, AC=b, BC=c, BD=d, AB=m, CD=n$. We... | // Block 1
defaultpen(fontsize(8)); pair A=(2,4), B=(0,0), C=(4,0), D=(4,2); label("A",A,(0,1));label("D",D,(1,0));label("B",B,(-1,-1));label("C",C,(1,-1)); axialshade(A--C--D--cycle, lightgray, A, gray, D); draw(A--B--C--A--D--C);draw(B--D, linetype("8 8")); label("$m$",(A+B)/2,(-1,1));label("$n$",(C+D)/2,(1,0)); labe... | [] |
287 | $P$ lies between the rays $OA$ and $OB$. Find $Q$ on $OA$ and $R$ on $OB$ collinear with $P$ so that $\frac{1}{PQ} + \frac{1}{PR}$ is as large as possible. | 1979 USAMO Problems/Problem 4 | Let $r = OP, x = \angle OPR, a = \angle POR,$ and $b = \angle POQ.$ Then $\angle ORP = \pi - x - a$ and $\angle OQP = x - b.$ Using the Law of Sines on $\triangle OPR$ gives
\[PR = \sin a * \frac{r}{\sin(\pi - x - a)} = \sin a * \frac{r}{\sin(x + a)},\]
and using the Law of Sines on $\triangle OPQ$ gives
\[PQ = \sin b ... | // Block 1
pair O = (0,0), A = (14,28), Q = (20,40), B = (16,0), R = (25,0), P = (23,16);
dot(O); dot(A); dot(Q); dot(B); dot(R); dot(P);
label("O", O, S);
label("A", A, W);
label("Q", Q, W);
label("B", B, S);
label("R", R, S);
label("P", P, E);
draw(O--R--Q--O); draw(O--P);
label("r", O--P, N);
// Block 2
pair O = (0,... | [] |
288 | In triangle $ABC$, angle $A$ is twice angle $B$, angle $C$ is obtuse, and the three side lengths $a, b, c$ are integers. Determine, with proof, the minimum possible perimeter. | 1991 USAMO Problems/Problem 1 | Solution 1
(diagram by integralarefun)
After drawing the triangle, also draw the angle bisector of $\angle A$, and let it intersect $\overline{BC}$ at $D$. Notice that $\triangle ADC\sim \triangle BAC$, and let $AD=x$. Now from similarity,
\[x=\frac{bc}{a}\]
However, from the angle bisector theorem, we have
\[BD=\fr... | import olympiad; pair A, B, C, D, extensionAC; real angleABC; path braceBC; A = (0, 0); B = (2, 0); D = (1, .5); angleABC = atan(.5); //y = 4x/3 and x+2y = 2 (sides AC and BC, respectively) intersect here: C = (6/11, 8/11); braceBC = brace(C, B, .1); label("$\mathsf{A}$", A, W); label("$\mathsf{B}$", B, E); label... | [] |
289 | Let $\, D \,$ be an arbitrary point on side $\, AB \,$ of a given triangle $\, ABC, \,$ and let $\, E \,$ be the interior point where $\, CD \,$ intersects the external common tangent to the incircles of triangles $\, ACD \,$ and $\, BCD$. As $\, D \,$ assumes all positions between $\, A \,$ and $\, B \,$, prove that t... | 1991 USAMO Problems/Problem 5 | Let the incircle of $ACD$ and the incircle of $BCD$ touch line $AB$ at points $D_a,D_b$, respectively; let these circles touch $CD$ at $C_a$, $C_b$, respectively; and let them touch their common external tangent containing $E$ at $T_a,T_b$, respectively, as shown in the diagram below.
We note that
\[CE = CC_a - EC... | // Block 1
size(220);
defaultpen(1);
pair A=(0,0), B=(220,0), C=(18.7723,118.523);
pair D=(72.6,0);
pair Ia=incenter(A,D,C), Ib=incenter(B,D,C);
pair Ta=(24.9758,52.5775),Tb=(86.6196,67.4129);
pair E=IntersectionPoint((Ta--Tb),(C--D));
path Oa=circle(Ia,inradius(A,D,C));
path Ob=circle(Ib,inradius(B,D,C));
pair Da=IP(... | [] |
290 | Let $ABCD$ be a convex quadrilateral such that diagonals $AC$ and $BD$ intersect at right angles, and let $E$ be their intersection. Prove that the reflections of $E$ across $AB$, $BC$, $CD$, $DA$ are concyclic. | 1993 USAMO Problems/Problem 2 | Diagram
Work
Let $X$, $Y$, $Z$, $W$ be the foot of the altitude from point $E$ of $\triangle AEB$, $\triangle BEC$, $\triangle CED$, $\triangle DEA$.
Note that reflection of $E$ over all the points of $XYZW$ is similar to $XYZW$ with a scale of $2$ with center $E$. Thus, if $XYZW$ is cyclic, then the reflections a... | import olympiad; defaultpen(0.8pt+fontsize(12pt)); pair E; E=(0,0); label('$E$',E,N); pair A,B,C,D; A=(9,0); B=(0,13); C=(-13,0); D=(0,-11); draw(A--B--C--D--cycle,blue); label('$A$',A,E); label('$B$',B,N); label('$C$',C,W); label('$D$',D,S); pair T,R,S,Q; T=reflect(A, B)*E; R=reflect(C, B)*E; S=reflect(C, D)*E; Q=refl... | [] |
291 | A convex hexagon $ABCDEF$ is inscribed in a circle such that $AB=CD=EF$ and diagonals $AD,BE$, and $CF$ are concurrent. Let $P$ be the intersection of $AD$ and $CE$. Prove that $\frac{CP}{PE}=(\frac{AC}{CE})^2$. | 1994 USAMO Problems/Problem 3 | Let the diagonals $AD$, $BE$, $CF$ meet at $Q$.
First, let's show that the triangles $\triangle AEC$ and $\triangle QED$ are similar.
$\angle ACE=\angle ADE$ because $A$,$C$,$D$ and $E$ all lie on the circle, and $\angle ADE=\angle QDE$. $\angle AEB=\angle CED$ because $AB=CD$, and $A$,$B$,$C$,$D$ and $E$ all lie... | // Block 1
pair A,B,C,D,E,F,P,Q;
A=(-0.96,0.28);
B=(-0.352,0.936);
C=(0,1);
D=(4/5,3/5);
E=(4/5,-3/5);
F=(0,-1);
P=IntersectionPoint(A--D,C--E);
Q=IntersectionPoint(A--D,C--F);
draw(A--B);
draw(B--C);
draw(C--D);
draw(D--E,green);
draw(E--F);
draw(F--A);
draw(A--C,red);
draw(A--Q);
draw(A--E,red);
draw(B--Q);
draw(C-... | [] |
292 | Let $P$ be a point inside triangle $ABC$ such that
\[\angle APB-\angle ACB = \angle APC-\angle ABC\]
Let $D$, $E$ be the incenters of triangles $APB$, $APC$, respectively. Show that $AP$, $BD$, $CE$ meet at a point. | 1996 IMO Problems/Problem 2 | let $CF$, $BG$ be the angle bisectors of $\angle ABP$ and $\angle ACP$, respectively. Notice that they coincide with line $BD$ and $CE$. Therefore, it suffices to show $CF,BG,AP$ are concurrent.
Let $CF \cap AP = X_{1}$, and $BG\cap AP = X_{2}$. Notice that by angle bisector theorem, we have \[\frac{AX_{1}}{PX_{1}}=\... | // Block 1
import graph; size(7cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -9.124923887131423, xmax = 11.886638474419073, ymin = -8.067061524000941, ymax =... | [] |
293 | Let $ABC$ be a triangle, and $M$ an interior point such that $\angle MAB=10^\circ$, $\angle MBA=20^\circ$ , $\angle MAC= 40^\circ$ and $\angle MCA=30^\circ$. Prove that the triangle is isosceles. | 1996 USAMO Problems/Problem 5 | Solution 1
Clearly, $\angle AMB = 150^\circ$ and $\angle AMC = 110^\circ$. Now by the Law of Sines on triangles $ABM$ and $ACM$, we have \[\frac{AB}{\sin 150^\circ} = \frac{AM}{\sin 20^\circ}\] and \[\frac{AC}{\sin 110^\circ} = \frac{AM}{\sin 30^\circ}.\] Combining these equations gives us \[\frac{AB}{AC} = \frac{\sin ... | pair A,B,C,M; A=(0,0); B=(1,2); C=(2,0); M=(0.8,1.1); draw(A--B); draw(B--C); draw(C--A); draw(A--M); draw(B--M); draw(C--M); label("\(A\)",A,SW); label("\(B\)",B,N); label("\(C\)",C,SE); label("\(M\)",M,NE); | [] |
294 | Let $ABC$ be a triangle and let $I$ be the incenter. Let $N$, $M$ be the midpoints of the sides $AB$ and $CA$ respectively. The lines $BI$ and $CI$ meet $MN$ at $K$ and $L$ respectively. Prove that $AI+BI+CI>BC+KL$. | 1997 JBMO Problems/Problem 3 | First, by SAS Similarity, $\triangle ANM \sim \triangle ABC,$ so $NM \parallel BC$ and $MN = \tfrac12 BC.$ That means $\angle IBC = \angle IKN,$ and since $\angle IBN = \angle IBC,$ $\triangle NBK$ is an isosceles triangle. Similarly, $\angle MLC = \angle LCB = \angle LCM,$ making $\triangle MLC$ an isosceles as well... | // Block 1
size(9.22 cm);
pair B=(0,0), A=(50,120), C=(140,0), N=(25,60), M=(95,60), I=(60,40), K=(90,60), L=(20,60);
draw(B--A--C--B);
draw(circle(I,40));
dot(A);
label("A",(50,125));
dot(B);
label("B",B,SW);
dot(C);
label("C",C,SE);
dot(N);
label("N",(24,65));
dot(M);
label("M",M,NE);
dot(I);
label("I",I,S);
dot(K)... | [] |
295 | To clip a convex $n$-gon means to choose a pair of consecutive sides $AB, BC$ and to replace them by three segments $AM, MN,$ and $NC,$ where $M$ is the midpoint of $AB$ and $N$ is the midpoint of $BC$. In other words, one cuts off the triangle $MBN$ to obtain a convex $(n+1)$-gon. A regular hexagon $P_6$ of area $1$ i... | 1997 USAMO Problems/Problem 4 | $\textbf{Claim:}$ It is impossible to choose two non-adjacent sides and clip a whole part of it off.
$\textbf{Proof:}$ If you clip adjacent sides, you can cut off at most up to the blue lines; Clipping more is impossible due to the degrees getting larger and larger and more and more circular.
Thus, after infinitely... | // Block 1
size(200);
draw((1, 0)--(0.5, 0.866)--(-0.5, 0.866)--(-1, 0)--(-0.5, -0.866)--(0.5, -0.866)--(1, 0));
draw((1, 0)--(-0.5, 0.866)--(-0.5, -0.866)--(1, 0), blue);
draw((-1, 0)--(0.5, -0.866)--(0.5, 0.866)--(-1, 0), blue);
// Block 2
size(200); draw((1, 0)--(0.5, 0.866)--(-0.5, 0.866)--(-1, 0)--(-0.5, -0.866)--... | [] |
296 | (Răzvan Gelca)
Let $ABC$ be a triangle and $D$ the foot of the altitude from $A$. Let $E$ and $F$ be on a line through $D$ such that $AE$ is perpendicular to $BE$, $AF$ is perpendicular to $CF$, and $E$ and $F$ are different from $D$. Let $M$ and $N$ be the midpoints of the line segments $BC$ and $EF$, respectively. ... | 1998 APMO Problems/Problem 4 | We use directed angles mod $\pi$.
Since $\angle ADB$ and $\angle AEB$ are both right angles, points $ABDE$ are concyclic. It follows that
\[\angle CBA \equiv \angle DBA \equiv \angle DEA \equiv \angle FEA .\]
Similarly, the quadrilateral $ACDF$ is cyclic, so
\[\angle ACB \equiv \angle ACD \equiv \angle AFD \equiv \... | // Block 1
size(200);
defaultpen(1);
pair B=(0,0), A=(1,3), C=(4,0), X=(4,4);
pair D=foot(A,B,C);
path O1=circumcircle(A,B,D), O2=circumcircle(A,C,D);
pair E=IntersectionPoint(O1,D--X,1), F=IntersectionPoint(O2,D--X,0);
pair M=(B+C)/2;
pair N=(E+F)/2;
draw(C--F--A--B--C--A);
draw(B--E--A--D--E--F);
draw(A--N--M--A,do... | [] |
297 | Let ${\cal C}_1$ and ${\cal C}_2$ be concentric circles, with ${\cal C}_2$ in the interior of ${\cal C}_1$. From a point $A$ on ${\cal C}_1$ one draws the tangent $AB$ to ${\cal C}_2$ ($B\in {\cal C}_2$). Let $C$ be the second point of intersection of $AB$ and ${\cal C}_1$, and let $D$ be the midpoint of $AB$. A line... | 1998 USAMO Problems/Problem 2 | First, $AD=\frac{AB}{2}=\frac{AC}{4}$. Because $E$,$F$ and $B$ all lie on a circle, $AE \cdot AF=AB \cdot AB=\frac{AB}{2} \cdot 2AB=AD \cdot AC$. Therefore, $\triangle ACF \sim \triangle AED$, so $\angle ACF = \angle AED$. Thus, quadrilateral $CFED$ is cyclic, and $M$ must be the center of the circumcircle of $CFED$, w... | // Block 1
pair O,A,B,C,D,E,F,DEb,CFb,Fo,M;
O=(0,0);
A=(1.732,1);
B=(0,1);
C=(-1.732,1);
D=(0.866,1);
Fo=(-1,-0.5);
path AC,AF,DE,CF,DEbM,CFbM,C1,C2;
C1=circle(O,2);
C2=circle(O,1);
E=intersectionpoints(A--Fo,C2)[0];
F=intersectionpoints(A--Fo,C2)[1];
DEb=((D.x+E.x)/2.0,(D.y+E.y)/2.0);
CFb=((C.x+F.x)/2.0,(C.y+F.y)/2.... | [] |
298 | Let $n \geq 5$ be an integer. Find the largest integer $k$ (as a function of $n$) such that there exists a convex $n$-gon $A_{1}A_{2}\dots A_{n}$ for which exactly $k$ of the quadrilaterals $A_{i}A_{i+1}A_{i+2}A_{i+3}$ have an inscribed circle. (Here $A_{n+j} = A_{j}$.) | 1998 USAMO Problems/Problem 6 | Lemma: If quadrilaterals $A_iA_{i+1}A_{i+2}A_{i+3}$ and $A_{i+2}A_{i+3}A_{i+4}A_{i+5}$ in an equiangular $n$-gon are tangential, and $A_iA_{i+3}$ is the longest side quadrilateral $A_iA_{i+1}A_{i+2}A_{i+3}$ for all $i$, then quadrilateral $A_{i+1}A_{i+2}A_{i+3}A_{i+4}$ is not tangential.
Proof:
If quadrilaterals $A... | // Block 1
import geometry;
size(10cm);
pair A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U;
A = (-1,0);
B = (1,0);
draw(Circle(A,1)^^Circle(B,1));
C = (sqrt(2)/2-1,sqrt(2)/2);
D = (-sqrt(3)/2 - 1, .5);
E = (-sqrt(3)/2 - 1, -.5);
F = (-1,-1);
G = (1,-1);
H = (sqrt(3)/2 + 1, -.5);
I = (sqrt(3)/2 + 1, .5);... | [] |
299 | Let $A_1A_2A_3$ be a triangle and let $\omega_1$ be a circle in its plane passing through $A_1$ and $A_2.$ Suppose there exist circles $\omega_2, \omega_3, \dots, \omega_7$ such that for $k = 2, 3, \dots, 7,$ $\omega_k$ is externally tangent to $\omega_{k - 1}$ and passes through $A_k$ and $A_{k + 1},$ where $A_{n + 3... | 2000 USAMO Problems/Problem 5 | Solution 1
Let the circumcenter of $\triangle ABC$ be $O$, and let the center of $\omega_k$ be $O_k$. $\omega_k$ and $\omega_{k-1}$ are externally tangent at the point $A_k$, so $O_k, A_k, O_{k-1}$ are collinear.
$O$ is the intersection of the perpendicular bisectors of $\overline{A_1A_2}, \overline{A_2A_3}, \overlin... | size(300); pathpen = linewidth(0.7); pen t = linetype("2 2"); pair A = (0,0), B=3*expi(1), C=(3.5)*expi(0); /* arbitrary points */ pair O=circumcenter(A,B,C), O1 = O + 5*( ((B+C)/2) - O ), O2 = IP(O -- O + 100*( ((A+C)/2) - O ), O1 -- O1 + 10*( C - O1 )); D(MP("A_3",A,SW)--MP("A_1",B,N)--MP("A_2",C,SE)--cycle); D(MP("... | [] |
300 | $ABC$ is a triangle. $X$ lies on $BC$ and $AX$ bisects angle $A$. $Y$ lies on $CA$ and $BY$ bisects angle $B$. Angle $A$ is $60^{\circ}$. $AB + BX = AY + YB$. Find all possible values for angle $B$. | 2001 IMO Problems/Problem 5 | Let $D$ be on extension of $AB$ and $BD=BX$. Let $E$ be on $YC$ and $YE=YB$, then \[AD=AB+BD=AB+BX=AY+YB=AE\]
Since $A=60$, $\triangle{ADE}$ is equilateral. Let $\angle{ABY}=x$, then, \[\angle{YBX}=\angle{BDX}=\angle{BXD}=\angle{YEX}=x\]
We claim that $X$ must be on $BE$, i.e., $C=E$. If $X$ is not on $BE$, then $\angl... | // Block 1
import cse5;
import graph;
import olympiad;
dotfactor = 3;
unitsize(1.5inch);
pair A = (0,sqrt(3)), D= (-1, 0), E=(1,0);
pair Bb = rotate(40,E)*A;
pair B = extension(A,D,E,Bb);
pair H = foot(A,D,E);
pair X = extension(A,H,B,E);
pair Yy = bisectorpoint(A,B,E);
pair Y =extension(A,E,B,Yy);
pair C = E - (0,0.1... | [] |
301 | In triangle $ABC$, let $I$ be the incenter and $I_a$ the excenter opposite $A$. Suppose that $\overline{II_a}$ meets $\overline{BC}$ and the circumcircle of triangle $ABC$ at $A'$ and $M$, respectively. Let $N$ be the midpoint of arc $MBA$ of the circumcircle of triangle $ABC$. Let lines $NI$ and $NI_a$ intersect th... | 2001 Iran NMO (Round 3) Problems/Problem 5 | We will use directed angles mod $\pi$, and directed arcs mod $2\pi$.
Since $\widehat{MN}\equiv \widehat{NA}$, it follows that
\[\angle I_a TS \equiv \angle NTS \equiv \tfrac{1}{2} (\widehat{NA} + \widehat{AS}) \equiv \tfrac{1}{2} (\widehat{MN} + \widehat{AS}) \equiv \angle MIS \equiv \angle I_a IS .\]
It follows that ... | // Block 1
size(300);
defaultpen(1);
pair A=(1,2.7), B=(0,0), C=(3,0);
path O=circumcircle(A,B,C);
pair I=incenter(A,B,C);
pair BB=A+5(B-A);
pair II=bisectorpoint(BB,B,C);
pair Ia=IntersectionPoint(B--(B+6(II-B)),A--(A+5(I-A)));
pair Aa=IntersectionPoint(A--Ia,B--C);
pair M=IntersectionPoint(A--Ia,O,1);
pair NN=bisect... | [] |
302 | Let $ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$, respectively, such that $CD_2 = BD_1$ and $CE_2 = AE_1$, and denote by $P$ the point of intersection... | 2001 USAMO Problems/Problem 2 | Solution 1
It is well known that the excircle opposite $A$ is tangent to $\overline{BC}$ at the point $D_2$. (Proof: let the points of tangency of the excircle with the lines $BC, AB, AC$ be $D_3, F,G$ respectively. Then $AB+BD_3=AB + BF=AF = AG = AC + CG=AC + CD_3$. It follows that $2CD_3 = AB + BC - AC$, and $CD_3 = ... | pathpen = linewidth(0.7); size(300); pen d = linetype("4 4") + linewidth(0.6); pair B=(0,0), C=(10,0), A=7*expi(1),O=D(incenter(A,B,C)),D1 = D(MP("D_1",foot(O,B,C))),E1 = D(MP("E_1",foot(O,A,C),NE)),E2 = D(MP("E_2",C+A-E1,NE)); /* arbitrary points */ /* ugly construction for OA */ pair Ca = 2C-A, Cb = bisectorpoint(Ca... | [] |
303 | Given a triangle $ABC$ with $AC > BC$. On the circumcircle of triangle $ABC$ there exists point $D$, which is the midpoint of arc $AB$ that contains $C$. Let $E$ be a point on $AC$ such that $DE$ is perpendicular to $AC$. Prove that $AE = EC + CB$. | 2002 Indonesia MO Problems/Problem 4 | We use the method of phantom points.
Draw $AC$ and $BC$, and extend line $AC$ past $C$ to a point $B'$ such that $BC = B'C$. Draw point $E'$ at the midpoint of $AB'$, and $D'$ at the intersection of the perpendicular to $AC$ from $E'$ and the perpendicular bisector of $AB$.
Since $D'B = D'B', CB = CB', D'C = D'C$, we... | // Block 1
size(5cm);
draw(unitcircle);
pair A, C, B;
A = (-0.6, 0.8);
C = (0.6, 0.8);
B = (0.8, 0.6);
dot(A);
dot(C);
dot(B);
label("$A$", A, NW);
label("$C$", C, N);
label("$B$", B, SE);
draw(C--B--A);
pair Bp;
Bp = (0.88, 0.8);
dot(Bp, blue);
label("$B'$", Bp, NE, blue);
draw(A -- Bp);
pair midAB, dirAB;
midAB =... | [] |
304 | Let $\triangle{ABC}$ be an acute angled triangle. The circle with diameter $AB$ intersects the sides $AC$ and $BC$ at points $E$ and $F$ respectively. The tangents drawn to the circle through $E$ and $F$ intersect at $P$.
Show that $P$ lies on the altitude through the vertex $C$. | 2002 Pan African MO Problems/Problem 5 | Draw lines $GA$ and $BH$, where $G$ and $H$ are on $EP$ and $FP$, respectively. Because $GA$ and $GE$ are tangents as well as $HB$ and $HF$, $GA = GE$ and $HB = HF$. Additionally, because $EP$ and $FP$ are tangents, $EP = FP$.
Let $\angle GAE = a$ and $\angle HBF = b$. By the Base Angle Theorem, $\angle GEA = a$ ... | // Block 1
pair a=(-65,0),O=(0,0),b=(65,0),e=(-39,52),f=(25,60),c=(-9.286,111.429),p=(-9.286,74.286),g=(-65,32.5),h=(65,43.333);
draw(arc(O,b,a,CCW));
draw(a--b--c--a);
dot(a);
label("$A$",a,SW);
dot(b);
label("$B$",b,SE);
dot(c);
label("$C$",c,N);
dot(e);
label("$E$",e,NW);
dot(f);
label("$F$",f,NE);
dot(O);
label("$... | [] |
304 | Let $\triangle{ABC}$ be an acute angled triangle. The circle with diameter $AB$ intersects the sides $AC$ and $BC$ at points $E$ and $F$ respectively. The tangents drawn to the circle through $E$ and $F$ intersect at $P$.
Show that $P$ lies on the altitude through the vertex $C$. | 2002 Pan African MO Problems/Problem 5 | Let $D$ be the intersection of $AF$ and $BE$. Note that $\angle CED = 180^\circ - \angle AED = 180^\circ - \angle AEB = 90^\circ$, and similarly $\angle CFD = 180^\circ - \angle BFD = 180^\circ - \angle BFA = 90^\circ$. Thusly, $CEDF$ is a cyclic quadrilateral, and $CD$ is the diameter of its circumcircle.
Next, let $... | // Block 1
import graph;
pair A, B, O;
path circleAB;
A = (-5, 0);
B = (5, 0);
O = (A + B)/2;
circleAB = Circle(O, 5);
dot(A);
dot(B);
dot(O);
label("$A$", A, SW);
label("$O$", O, S);
label("$B$", B, SE);
draw(A--B);
draw(circleAB);
pair C, E, F, D;
C = (1, 8);
E = intersectionpoint(C--A, circleAB);
F = intersection... | [] |
305 | Let $ABC$ be a triangle. A circle passing through $A$ and $B$ intersects segments $AC$ and $BC$ at $D$ and $E$, respectively. Lines $AB$ and $DE$ intersect at $F$, while lines $BD$ and $CF$ intersect at $M$. Prove that $MF = MC$ if and only if $MB\cdot MD = MC^2$. | 2003 USAMO Problems/Problem 4 | Solution 1
Extend segment $DM$ through $M$ to $G$ such that $FG\parallel CD$.
Then $MF = MC$ if and only if quadrilateral $CDFG$ is a parallelogram, or, $FD\parallel CG$. Hence $MC = MF$ if and only if $\angle GCD = \angle FDA$, that is, $\angle FDA + \angle CGF = 180^\circ$.
Because quadrilateral $ABED$ is cyclic, $... | // Block 1
defaultpen(fontsize(10)+0.6); size(250); var theta=22, r=0.58; pair B=origin, A=dir(theta), C=A+(rotate(78)*0.8*A), O=IP(CR(B,r),CR(A,r)); path c=CR(O,r); pair D=IP(c,A--C), E=IP(c,B--C), F=extension(A,B,D,E), M=extension(B,D,C,F), G=extension(D,M,F,F+C-D); draw(A--B--C--A^^E--F--C^^A--F^^B--M^^E--M); draw(c... | [] |
306 | Let $M$ be a point in triangle $ABC$ such that $\angle AMC=90^{\circ}$, $\angle AMB=150^{\circ}$, $\angle BMC=120^{\circ}$. The centers of circumcircles of triangles $AMC,AMB,BMC$ are $P,Q,R$, respectively. Prove that the area of $\triangle PQR$ is greater than the area of $\triangle ABC$. | 2005 Indonesia MO Problems/Problem 4 | From the above constraints, we can let $M$ be $(0,0)$, $C$ be $(0,2c)$, $A$ be $(-2a,0)$, and $B$ be $(2b\sqrt{3},-2b)$, where $a,b,c$ are positive. Note that the centers of circumcircles are on the perpendicular bisectors of $AM$, $BM$, or $CM$.
Because $AMC$ is a right triangle, the circumcenter of $\triangle AMC... | // Block 1
import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black;
real xmin=-24.2,xmax=24.2,ymin=-24.2,ymax=24.2;
pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0);
/*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=4,gy=4... | [] |
307 | Let $ABCD$ be a convex quadrilateral. Square $AB_1A_2B$ is constructed such that the two vertices $A_2,B_1$ is located outside $ABCD$. Similarly, we construct squares $BC_1B_2C$, $CD_1C_2D$, $DA_1D_2A$. Let $K$ be the intersection of $AA_2$ and $BB_1$, $L$ be the intersection of $BB_2$ and $CC_1$, $M$ be the intersecti... | 2005 Indonesia MO Problems/Problem 7 | Let the coordinates of $A$ be $(2a,0)$, the coordinates of $B$ be $(2b_1, 2b_2)$, the coordinates of $C$ be $(-2c,0)$, and the coordinates of $D$ be $(2d_1, -2d_2)$, where all variables are rational and $a, b_2, c, d_2 \ge 0$.
Let $X$ be the midpoint of $AB$, which is point $(a+b_1,b_2)$. Additionally, mark points ... | // Block 1
import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black;
real xmin=-10.2,xmax=10.2,ymin=-10.2,ymax=10.2;
pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0);
/*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=2,gy=2... | [] |
308 | (Zuming Feng) Let $ABC$ be an acute-angled triangle, and let $P$ and $Q$ be two points on side $BC$. Construct point $C_1$ in such a way that convex quadrilateral $APBC_1$ is cyclic, $QC_1 \parallel CA$, and $C_1$ and $Q$ lie on opposite sides of line $AB$. Construct point $B_1$ in such a way that convex quadrilateral ... | 2005 USAMO Problems/Problem 3 | Solution 1
Let $B_1'$ be the second intersection of the line $C_1A$ with the circumcircle of $APC$, and let $Q'$ be the second intersection of the circumcircle of $B_1' C_1P$ and line $BC$. It is enough to show that $B_1'=B_1$ and $Q' =Q$. All our angles will be directed, and measured mod $\pi$.
Since points $C_1,... | size(300); defaultpen(1); pair A=(2,5), B=(-1,0), C=(5,0); pair C1=(.5,5.7); path O1=circumcircle(A,B,C1); pair P=IntersectionPoint(O1,B--C,1); path O2=circumcircle(A,P,C); pair B1=IntersectionPoint(O2,C1--5A-4C1,0); path O=circumcircle(B1,C1,P); pair Q=IntersectionPoint(O,B--C,1); draw(C1--P--A--B--C--A); draw(P--B1... | [] |
309 | In triangle $ABC$, $M$ is the midpoint of side $BC$ and $G$ is the centroid of triangle $ABC$. A line $l$ passes through $G$, intersecting line $AB$ at $P$ and line $AC$ at $Q$, where $P\ne B$ and $Q\ne C$. If $[XYZ]$ denotes the area of triangle $XYZ$, show that $\frac{[BGM]}{[PAG]}+\frac{[CMG]}{[QGA]}=\frac32$. | 2006 Indonesia MO Problems/Problem 5 | First, since $G$ is the centroid of the triangle and $BM = MC$, $[BMG] = [CMG] = \frac16 [ABC]$. Also, note that $[BGA] = \frac13 [ABC]$, so $[APG] = \frac{AP}{AB} \cdot \frac13 [ABC]$. Similarly, $[AQG] = \frac{AQ}{AC} \cdot \frac13 [ABC]$. Substituting the areas results in
\begin{align*} \frac{[BGM]}{[PAG]}+\frac{... | // Block 1
pair B=(0,0),A=(20,90),C=(100,0),M=(50,0),G=(40,30);
pair b=(20/3,30),c=(220/3,30);
draw(A--B--C--A);
draw(A--M);
draw(b--c);
dot(G);
label("$A$",A,N);
label("$B$",B,SW);
label("$C$",C,SE);
label("$M$",M,S);
label("$B'$",b,W);
label("$C'$",c,E);
label("$G$",G,NW);
draw((4,18)--(65.714,38.571));
label("$P$",(... | [] |
310 | Consider five points $A,B,C,D$, and $E$ such that $ABCD$ is a parallelogram and $BCED$ is a cyclic quadrilateral.
Let $\ell$ be a line passing through $A$. Suppose that $\ell$ intersects the interior of the segment $DC$ at $F$ and intersects
line $BC$ at $G$. Suppose also that $EF=EG=EC$. Prove that $\ell$ is the bise... | 2007 IMO Problems/Problem 2 | Since $\angle{DAF}=\angle{CGF}$, $\angle{BAF}=\angle{CFG}$, it suffices to prove $CF=CG$.
Let $\angle{FCE}=\alpha$, $\angle{GCE}=\beta$, $\angle{CDE}=\gamma$. We have:
\[CF=2CE\cos{\alpha}, CG=2CE\cos{\beta}\]
so,
\[\dfrac{CF}{CG}=\dfrac{\cos{\alpha}}{\cos{\beta}}\]
Meantime, using Law of Sines on $\triangle{DEC}$, w... | // Block 1
import cse5;
import graph;
import olympiad;
dotfactor = 3;
unitsize(1.5inch);
path circle = Circle(origin, 1);
draw(circle);
pair D = (-sqrt(3)/2, -0.5), C = (sqrt(3)/2, -0.5);
//G = bisectorpoint(C, B, D);
pair Ee = rotate(38,C)*D;
pair E = IP(C--Ee, circle,1);
pair Gg = rotate(76,C)*D;
path circle2 = Cir... | [] |
311 | Points $A,B,C,D$ are on circle $S$, such that $AB$ is the diameter of $S$, but $CD$ is not the diameter. Given also that $C$ and $D$ are on different sides of $AB$. The tangents of $S$ at $C$ and $D$ intersect at $P$. Points $Q$ and $R$ are the intersections of line $AC$ with line $BD$ and line $AD$ with line $BC$, res... | 2007 Indonesia MO Problems/Problem 7 | Let $O$ be the center of the circle. Let $\angle CBA = b$ and $\angle DBA = a$. To prove that $P$ is on line $QR$, we can show that $\angle PRD = \angle QRD$.
Because $AB$ is a diameter, we know that $\angle ACB = 90^\circ$ and $\angle ADB = 90^\circ$. By the Vertical Angle Theorem, $\angle RAC = \angle QAD$, so ... | // Block 1
size(225);
pair a=(-50,0),b=(50,0),c=(-48,14),d=(-30,-40),o=(0,0);
draw(circle(o,50));
draw(a--c--b--d--a);
dot(a);
label("$A$",a,NE);
dot(b);
label("$B$",b,E);
dot(c);
label("$C$",c,NE);
dot(d);
label("$D$",d,S);
dot(o);
label("$O$",o,SE);
pair r=(-57.692,15.385),q=(-57.692,-53.846);
draw (r--b--q);
dot(r)... | [] |
312 | (Kiran Kedlaya, Sungyoon Kim) Let $ABC$ be an acute triangle with $\omega$, $\Omega$, and $R$ being its incircle, circumcircle, and circumradius, respectively. The circle $\omega_A$ is tangent internally to $\Omega$ at $A$ and externally tangent to $\omega$. Circle $\Omega_A$ is internally tangent to $\Omega$ at $A$ ... | 2007 USAMO Problems/Problem 6 | Solution 1
Lemma.
\[P_{A}Q_{A}=\frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}}\]
Proof. Note $P_{A}$ and $Q_{A}$ lie on $AO$ since for a pair of tangent circles, the point of tangency and the two centers are collinear.
Let $\omega$ touch $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. Note $AE=AF=s-a$. Cons... | size(400); defaultpen(fontsize(8)); pair A=(2,8), B=(0,0), C=(13,0), I=incenter(A,B,C), O=circumcenter(A,B,C), p_a, q_a, X, Y, X1, Y1, D, E, F; real r=abs(I-foot(I,A,B)), R=abs(A-O), a=abs(B-C), b=abs(A-C), c=abs(A-B), x=(((b+c-a)/2)^2)/(r^2+4*r*R+((b+c-a)/2)^2), y=((b+c-a)/2)^2/(r^2+((b+c-a)/2)^2); p_a=x*(O-A)+A; q_a=... | [] |
313 | Given a convex, $n$-sided polygon $P$, form a $2n$-sided polygon $\text{clip}(P)$ by cutting off each corner of $P$ at the edges’ trisection points. In other words, $\text{clip}(P)$ is the polygon whose vertices are the $2n$ edge trisection points of $P$, connected in order around the boundary of $P$. Let $P_1$ be an i... | 2008 iTest Problems/Problem 99 | Let $D_n$ be the difference in the areas between $P_n$ and $P_{n+1}$. Let our trapezoid be $P_1 = ABCD$ (and $[ABCD] = \frac{12(3+13)}{2} = 96$); then without loss of generality construct diagonal $BD$.
Let $A_1, A_2$ be the trisection points on $\overline{AB},\overline{AD}$, respectively, that are closest to $A$. Th... | // Block 1
pathpen = linewidth(0.7);pen d = linetype("4 4")+linewidth(0.7); pair A=(0,0),B=(5,12),C=(8,12),D=(13,0),A1=B/3,A2=D/3; D(MP("A",A)--MP("B",B,N)--MP("C",C,N)--MP("D",D)--cycle); D(B--D); D(MP("A_1",A1,NW) -- MP("A_2",A2)); D(D+2(C-D)/3 -- B+2(C-B)/3); D(4B/9 -- (A+D)/3+2*((A+B)/3 - (A+D)/3)/3, d); D(4D/9 --... | [] |
314 | (Zuming Feng) Let $ABC$ be an acute, scalene triangle, and let $M$, $N$, and $P$ be the midpoints of $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively. Let the perpendicular bisectors of $\overline{AB}$ and $\overline{AC}$ intersect ray $AM$ in points $D$ and $E$ respectively, and let lines $BD$ and $... | 2008 USAMO Problems/Problem 2 | Solution 1 (synthetic)
Without Loss of Generality, assume $AB >AC$. It is sufficient to prove that $\angle OFA = 90^{\circ}$, where $O$ is the circumcenter of $\triangle{ABC},$ as this would immediately prove that $A,P,O,F,N$ are concyclic.
By applying the Menelaus' Theorem in the Triangle $\triangle BFC$ for the trans... | // Block 1
/* setup and variables */ size(280); pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8); pair B=(0,0),C=(5,0),A=(1,4); /* A.x > C.x/2 */ /* construction and drawing */ pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F... | ["https://artofproblemsolving.com/wiki/images/0/05/2008usamo2-sol8.png", "https://artofproblemsolving.com/wiki/images/d/df/2008usamo2-sol9.png"] |
315 | Trapezoid $ABCD$, with $\overline{AB}||\overline{CD}$, is inscribed in circle $\omega$ and point $G$ lies inside triangle $BCD$. Rays $AG$ and $BG$ meet $\omega$ again at points $P$ and $Q$, respectively. Let the line through $G$ parallel to $\overline{AB}$ intersect $\overline{BD}$ and $\overline{BC}$ at points $R$ ... | 2009 USAMO Problems/Problem 5 | We will use directed angles in this solution. Extend $QR$ to $T$ as follows:
If:
Note that \begin{align*}\measuredangle GBT+\measuredangle TRG&=\frac{m\widehat{TQ}}{2}+\measuredangle TRB+\measuredangle BRG\\ &=\frac{m\widehat{TQ}+m\widehat{DQ}+m\widehat{CB}+m\widehat{BT}}{2}.\\ \end{align*}
Thus, $BTRG$ is cyclic.
... | // Block 1
import cse5;
import graph;
import olympiad;
dotfactor = 3;
unitsize(1.5inch);
path circle = Circle(origin, 1);
draw(circle);
pair A = (-.6, .8), B = (.6, .8), C = (.9, -sqrt(.19)), D = (-.9, -sqrt(.19)), G = bisectorpoint(C, B, D);
draw(A--B--C--D--cycle); draw(B--D);
dot("$A$", A, NW); dot("$B$", B, NE); ... | [] |
315 | Trapezoid $ABCD$, with $\overline{AB}||\overline{CD}$, is inscribed in circle $\omega$ and point $G$ lies inside triangle $BCD$. Rays $AG$ and $BG$ meet $\omega$ again at points $P$ and $Q$, respectively. Let the line through $G$ parallel to $\overline{AB}$ intersect $\overline{BD}$ and $\overline{BC}$ at points $R$ ... | 2009 USAMO Problems/Problem 5 | Extend $QR$ to $T$, and let line $l \parallel AB$ intersect $\omega$ at $K$ and another point $V$, as shown:
If:
Suppose that $VP \cap CB = S'$, and $AC \cap QV = R'$. Pascal's theorem on the tuple $(V, P, A, C, B, Q)$ implies that the points $S'$, $R'$, and $G = PA \cap BQ$ are collinear. However, $AC$ and $BD$ are... | // Block 1
import cse5;
import graph;
import olympiad;
dotfactor = 3;
unitsize(1.5inch);
path circle = Circle(origin, 1);
draw(circle);
pair A = (-.6, .8), B = (.6, .8), C = (.9, -sqrt(.19)), D = (-.9, -sqrt(.19)), G = bisectorpoint(C, B, D);
draw(A--B--C--D--cycle); draw(B--D);
dot("$A$", A, NW); dot("$B$", B, NE); ... | [] |
316 | Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$). The lines $AP$, $BP$ and $CP$ meet again its circumcircle $\Gamma$ at $K$, $L$, respectively $M$. The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$. Show that from $SC = SP$ follows $MK = ML$. | 2010 IMO Problems/Problem 4 | Solution 1
Without loss of generality, suppose that $AS > BS$. By Power of a Point, $SP^2 = SC^2 = SB \cdot SA$, so $\overline{SP}$ is tangent to the circumcircle of $\triangle ABP$. Thus, $\angle KPS = 180 - \angle SPA = \widehat{AP}/2 = \angle ABP$. It follows that after some angle-chasing,
$\begin{aligned} \widehat... | // Block 1
import graph; size(10.71cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(9); defaultpen(dps); pen ds=black; real xmin=-2.22,xmax=4.48,ymin=-1.99,ymax=3; pen ccqqqq=rgb(0.8,0,0), qqzzqq=rgb(0,0.6,0), evevff=rgb(0.9,0.9,1); filldraw(arc((0.08,1.23),0.37,-17.48,12.32)--(0.08,1.23)--cycle,evevff,blue); filld... | [] |
317 | Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter
$AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto
lines $AX, BX, AZ, BZ$, respectively. Prove that the acute angle
formed by lines $PQ$ and $RS$ is half the size of $\angle XOZ$, where
$O$ is the midpoint of segment $AB$. | 2010 USAMO Problems/Problem 1 | Let $\alpha = \angle BAZ$, $\beta = \angle ABX$.
Since $XY$ is a chord of the circle with diameter $AB$,
$\angle XAY = \angle XBY = \gamma$. From the chord $YZ$,
we conclude $\angle YAZ = \angle YBZ = \delta$.
Triangles $BQY$ and $APY$ are both right-triangles, and share the
angle $\gamma$, therefore they are simi... | // Block 1
import olympiad;
// Scale
unitsize(1inch);
real r = 1.75;
// Semi-circle: centre O, radius r, diameter A--B.
pair O = (0,0); dot(O); label("$O$", O, plain.S);
pair A = r * plain.W; dot(A); label("$A$", A, unit(A));
pair B = r * plain.E; dot(B); label("$B$", B, unit(B));
draw(arc(O, r, 0, 180)--cycle);
// ... | [] |
317 | Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter
$AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto
lines $AX, BX, AZ, BZ$, respectively. Prove that the acute angle
formed by lines $PQ$ and $RS$ is half the size of $\angle XOZ$, where
$O$ is the midpoint of segment $AB$. | 2010 USAMO Problems/Problem 1 | Let $T$ be the foot of the perpendicular from $Y$ to $\overline{AB}$, let $O$ be the center of the semi-circle.
Since we have a semi-circle, if we were to reflect it over $\overline {AB}$, we would have a full circle, with $\triangle{AXB}$ and $\triangle{AZB}$ inscribed in it. Now, notice that $Y$ is a point on that f... | // Block 1
currentpicture=new picture;
size(12cm);
pair O, A, B, X, Y, Z, P, Q, R, SS, T;
O=(0, 0);
A=(-1, 0);
B=(1, 0);
X=(Cos(144), Sin(144));
Y=(Cos(105), Sin(105));
Z=(Cos(27), Sin(27));
P=foot(Y, A, X);
Q=foot(Y, B, X);
R=foot(Y, A, Z);
SS=foot(Y, B, Z);
T=foot(Y, A, B);
dot(O); dot(A); dot(B); dot(X); dot(Y); dot... | [] |
318 | Let $ABC$ be a triangle with $\angle A = 90^{\circ}$. Points $D$
and $E$ lie on sides $AC$ and $AB$, respectively, such that $\angle ABD = \angle DBC$ and $\angle ACE = \angle ECB$. Segments $BD$ and
$CE$ meet at $I$. Determine whether or not it is possible for
segments $AB, AC, BI, ID, CI, IE$ to all have integer len... | 2010 USAMO Problems/Problem 4 | We know that angle $BIC = 135^{\circ}$, as the other two angles in triangle $BIC$ add to $45^{\circ}$. Assume that only $AB, AC, BI$, and $CI$ are integers. Using the Law of Cosines on triangle BIC,
$BC^2 = BI^2 + CI^2 - 2BI\cdot CI \cdot \cos 135^{\circ}$. Observing that $BC^2 = AB^2 + AC^2$ is an integer and tha... | // Block 1
import olympiad;
// Scale
unitsize(1inch);
// Shape
real h = 1.75;
real w = 2.5;
// Points
void ldot(pair p, string l, pair dir=p) { dot(p); label(l, p, unit(dir)); }
pair A = origin; ldot(A, "$A$", plain.SW);
pair B = w * plain.E; ldot(B, "$B$", plain.SE);
pair C = h * plain.N; ldot(C, "$C$", plain.NW);
... | [] |
319 | Given a triangle $ABC$, let $P$ and $Q$ be points on segments $\overline{AB}$ and $\overline{AC}$, respectively, such that $AP = AQ$. Let $S$ and $R$ be distinct points on segment $\overline{BC}$ such that $S$ lies between $B$ and $R$, $\angle{BPS} = \angle{PRS}$, and $\angle{CQR} = \angle{QSR}$. Prove that $P$, $Q$,... | 2012 USAJMO Problems/Problem 1 | Since $\angle BPS = \angle PRS$, the circumcircle of triangle $PRS$ is tangent to $AB$ at $P$. Similarly, since $\angle CQR = \angle QSR$, the circumcircle of triangle $QRS$ is tangent to $AC$ at $Q$.
For the sake of contradiction, suppose that the circumcircles of triangles $PRS$ and $QRS$ are not the same circle.... | // Block 1
import markers;
unitsize(0.5 cm);
pair A, B, C, O, P, Q, R, S;
A = (2,12);
B = (0,0);
C = (14,0);
P = intersectionpoint(A--B,Circle(A,8));
Q = intersectionpoint(A--C,Circle(A,8));
O = extension(P, P + rotate(90)*(A - P), Q, Q + rotate(90)*(A - Q));
S = intersectionpoint(B--C,arc(O, abs(O - P), 180, 270));... | [] |
320 | Let $P$ be a point in the plane of triangle $ABC$, and $\gamma$ a line passing through $P$. Let $A'$, $B'$, $C'$ be the points where the reflections of lines $PA$, $PB$, $PC$ with respect to $\gamma$ intersect lines $BC$, $AC$, $AB$, respectively. Prove that $A'$, $B'$, $C'$ are collinear. | 2012 USAMO Problems/Problem 5 | By the sine law on triangle $AB'P$,
\[\frac{AB'}{\sin \angle APB'} = \frac{AP}{\sin \angle AB'P},\]
so
\[AB' = AP \cdot \frac{\sin \angle APB'}{\sin \angle AB'P}.\]
Similarly,
\begin{align*} B'C &= CP \cdot \frac{\sin \angle CPB'}{\sin \angle CB'P}, \\ CA' &= CP \cdot \frac{\sin \angle CPA'}{\sin \angle CA'P}, \\ A'... | // Block 1
import graph;
import geometry;
unitsize(0.5 cm);
pair[] A, B, C;
pair P, R;
A[0] = (2,12);
B[0] = (0,0);
C[0] = (14,0);
P = (4,5);
R = 5*dir(70);
A[1] = extension(B[0],C[0],P,reflect(P + R,P - R)*(A[0]));
B[1] = extension(C[0],A[0],P,reflect(P + R,P - R)*(B[0]));
C[1] = extension(A[0],B[0],P,reflect(P + R... | [] |
321 | Let the excircle of triangle $ABC$ opposite the vertex $A$ be tangent to the side $BC$ at the point $A_1$. Define the points $B_1$ on $CA$ and $C_1$ on $AB$ analogously, using the excircles opposite $B$ and $C$, respectively. Suppose that the circumcentre of triangle $A_1B_1C_1$ lies on the circumcircle of triangle $AB... | 2013 IMO Problems/Problem 3 | Let the excenters opposite $A,B,C$ be $I_a,I_b,I_c$. Let the midpoint of $\overline{I_bI_c}$ be $M_a$, which lies on $(ABC)$, the nine-point circle of $\triangle I_aI_bI_c$; analogously define $M_b,M_c$.
$M_aB=M_aC$ and $BC_1=s-a=B_1C$, so $\triangle M_aBC_1\cong\triangle M_aCB_1$ (SAS), thus $M_a$ is equidistant from... | // Block 1
unitsize(2.5cm);
void b() {
pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10); pen dd=linetype("4 8");
/* Define the excenter */
pair excenter(pair A=(0,0), pair B=(0,0), pair C=(0,0)) {
return extension(A,bisectorpoint(C,A,B),B,rotate(90,B)*bisectorpoint(A,B,C));
}
/* Draw points */
pair ... | [] |
322 | Convex quadrilateral $ABCD$ has $\angle{ABC}=\angle{CDA}=90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside $\triangle{SCT}$ and
\[\angle{CHS}-\angle{CSB}=90^{\circ},\quad \angle{THC}-\angle{DTC} = 90^{\circ}.\... | 2014 IMO Problems/Problem 3 | Denote $\angle{HSB}=v$, $\angle{HTD}=u$, $\angle{HSC}=s$, $\angle{HTC}=t$, $\angle{HCS}=x$, $\angle{HCT}=y$, $\angle{AHS}=z$, $\angle{AHT}=w$. Since $\angle{CHS}-\angle{CSB} =90$ and $\angle{CHT}-\angle{CTD}=90$, we have $\angle{CSA}=x+90$, $\angle{CTA}=y+90$.
Since $\angle{CHS} = \angle{CSB}+90$, the tangent of the... | // Block 1
import cse5;
import graph;
import olympiad;
dotfactor = 3;
unitsize(1.5inch);
path circle = Circle(origin, 1);
// draw(circle);
pair A = (0,1), C=(0,-1);
pair Oo = (0,-0.05);
pair Bb = rotate(-8,Oo)*(2,-0.05), Dd =rotate(-8, Oo)*(-2,-0.05);
pair B = IP(Dd--Bb, circle, 1);
pair D = IP(Dd--Bb, circle, 0);
p... | [] |
323 | Points $P$ and $Q$ lie on side $BC$ of acute-angled $\triangle{ABC}$ so that $\angle{PAB}=\angle{BCA}$ and $\angle{CAQ}=\angle{ABC}$. Points $M$ and $N$ lie on lines $AP$ and $AQ$, respectively, such that $P$ is the midpoint of $AM$, and $Q$ is the midpoint of $AN$. Prove that lines $BM$ and $CN$ intersect on the circu... | 2014 IMO Problems/Problem 4 | Solution 1
We are trying to prove that the intersection of $BM$ and $CN$, call it point $D$, is on the circumcircle of triangle $ABC$. In other words, we are trying to prove $\angle {BDC} + \angle {BAC} = 180$.
Let the intersection of $BM$ and $AN$ be point $E$, and the intersection of $AM$ and $CN$ be point $F$.
L... | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10.60000000000002cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen ... | ["https://artofproblemsolving.com/wiki/images/thumb/2/25/IMO2014_P4.png/600px-IMO2014_P4.png"] |
324 | Let $ABCD$ be a cyclic quadrilateral with $AB = 7$ and $CD = 8$. Points $P$ and $Q$ are selected on segment $AB$ such that $AP = BQ = 3$. Points $R$ and $S$ are selected on segment $CD$ such that $CR = DS = 2$. Prove that $PQRS$ is a cyclic quadrilateral. | 2024 USAJMO Problems/Problem 1 | First, let $E$ and $F$ be the midpoints of $AB$ and $CD$, respectively. It is clear that $AE=BE=3.5$, $PE=QE=0.5$, $DF=CF=4$, and $SF=RF=2$. Also, let $O$ be the circumcenter of $ABCD$.
By properties of cyclic quadrilaterals, we know that the circumcenter of a cyclic quadrilateral is the intersection of its sides' ... | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/wiki/User:Azjps/geogebra */
import graph; size(12cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* p... | [] |
325 | In $\triangle ABC$, $BD$ is a median. $CF$ intersects $BD$ at $E$ so that $\overline{BE}=\overline{ED}$. Point $F$ is on $AB$. Then, if $\overline{BF}=5$,
$\overline{BA}$ equals:
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ \text{none of these}$ | 1959 AHSME Problems/Problem 40 | Draw $\overline{DG} \parallel \overline{FC}$ with $G$ on $\overline{AB}$. We know that $GF = BF = 5$, since $\triangle BFE \sim \triangle BGD$.
Likewise, since $\triangle ADG \sim \triangle ACF$, we know that $AG=5$.
Thus, $AF=AG+GF+FB=5+5+5=15$, which is answer $\fbox{\textbf{(C)}}$. | // Block 1
import geometry;
point A = (0,0);
point B = (5,8);
point C = (16,0);
point D = midpoint(A--C);
point E = midpoint(B--D);
point F, G;
triangle ABC = triangle(A,B,C);
// Triangle ABC
draw(ABC);
dot(A);
label("A",A,SW);
dot(B);
label("B",B,N);
dot(C);
label("C",C,SE);
// Segment BD
draw(B--D);
dot(D);
label(... | [] |
326 | The base of a triangle is of length $b$, and the altitude is of length $h$.
A rectangle of height $x$ is inscribed in the triangle with the base of the rectangle in the base of the triangle. The area of the rectangle is:
$\textbf{(A)}\ \frac{bx}{h}(h-x)\qquad \textbf{(B)}\ \frac{hx}{b}(b-x)\qquad \textbf{(C)}\ \frac{... | 1960 AHSME Problems/Problem 37 | Let $AB=b$, $DE=h$, and $WX = YZ = x$.
Since $CD$ is perpendicular to $AB$, $ND = WX$. That means $CN = h-x$.
The sides of the rectangle are parallel, so $XY \parallel WZ$. That means by AA Similarity, $\triangle CXY \sim \triangle CAB$. Letting $n$ be the length of the base of the rectangle, that means
\[\frac{h-x... | // Block 1
pair A=(0,0),B=(56,0),C=(20,48),D=(20,0),W=(10,0),X=(10,24),Y=(38,24),Z=(38,0);
draw(A--B--C--A);
draw((10,0)--(10,24)--(38,24)--(38,0));
draw(C--D);
dot(A);
dot(B);
dot(C);
dot(D);
dot(W);
dot(X);
dot(Y);
dot(Z);
dot((20,24));
label("$A$",A,S);
label("$B$",B,S);
label("$C$",C,N);
label("$D$",D,S);
label("$W... | [] |
327 | In $\triangle ABC$ the ratio $AC:CB$ is $3:4$. The bisector of the exterior angle at $C$ intersects $BA$ extended at $P$
($A$ is between $P$ and $B$). The ratio $PA:AB$ is:
$\textbf{(A)}\ 1:3 \qquad \textbf{(B)}\ 3:4 \qquad \textbf{(C)}\ 4:3 \qquad \textbf{(D)}\ 3:1 \qquad \textbf{(E)}\ 7:1$ | 1961 AHSME Problems/Problem 31 | Let $AC = 3n$ and $BC = 4n$. Draw $X$, where $X$ is on $BC$ and $AC \parallel PX$. By AA Similarity, $\triangle ABC \sim \triangle PBX$, so $PX = 3an$, $BX = 4an$, and $CX = 4an - 4n$.
Also, let $\angle ABC = a$ and $\angle BAC = b$. Since the angles of a triangle add up to $180^{\circ}$, $\angle BCA = 180-a-b$. ... | // Block 1
draw((0,0)--(40,0)--(16,18)--(0,0));
draw((40,0)--(64,72)--(16,18));
draw((40,0)--(160,0)--(64,72),dotted);
dot((0,0));
label("B",(0,0),SW);
dot((16,18));
label("A",(16,18),NW);
dot((40,0));
label("C",(40,0),S);
dot((64,72));
label("P",(64,72),N);
dot((160,0));
label("X",(160,0),SE);
label("$4n$",(20,0),S);... | [] |
328 | In $\triangle ABC$ the median from $A$ is given perpendicular to the median from $B$. If $BC=7$ and $AC=6$, find the length of $AB$.
$\textbf{(A)}\ 4\qquad \textbf{(B)}\ \sqrt{17} \qquad \textbf{(C)}\ 4.25\qquad \textbf{(D)}\ 2\sqrt{5} \qquad \textbf{(E)}\ 4.5$ | 1961 AHSME Problems/Problem 36 | By SAS Similarity, $\triangle ABC \sim \triangle MNC$, so $AB \parallel MN$. Thus, by AA Similarity, $\triangle AGB \sim \triangle NGM$.
Let $a = GN$ and $b = GM$, so $AG = 2a$ and $BG = 2b$. By the Pythagorean Theorem,
\[4a^2 + b^2 = 9\]
\[a^2 + 4b^2 = \frac{49}{4}\]
Adding the two equations yields $5a^2 + 5b^2 = \... | // Block 1
draw((-16,0)--(8,0));
draw((-16,0)--(16,-24));
draw((16,-24)--(0,24)--(0,-12));
draw((-16,0)--(0,24));
draw((0,2)--(2,2)--(2,0));
draw((0,-12)--(8,0),dotted);
dot((16,-24));
label("C",(16,-24),SE);
dot((-16,0));
label("A",(-16,0),W);
dot((0,24));
label("B",(0,24),N);
label("3",(8,-18),SW);
label("3",(-8,-6... | [] |
329 | $\triangle ABC$ is inscribed in a semicircle of radius $r$ so that its base $AB$ coincides with diameter $AB$.
Point $C$ does not coincide with either $A$ or $B$. Let $s=AC+BC$. Then, for all permissible positions of $C$:
$\textbf{(A)}\ s^2\le8r^2\qquad \textbf{(B)}\ s^2=8r^2 \qquad \textbf{(C)}\ s^2 \ge 8r^2 \qquad\... | 1961 AHSME Problems/Problem 38 | Since $s=AC+BC$, $s^2 = AC^2 + 2 \cdot AC \cdot BC + BC^2$. Since $\triangle ABC$ is inscribed and $AB$ is the diameter, $\triangle ABC$ is a right triangle, and by the Pythagorean Theorem, $AC^2 + BC^2 = AC^2 = (2r)^2$. Thus, $s^2 = 4r^2 + 2 \cdot AC \cdot BC$.
The area of $\triangle ABC$ is $\frac{AC \cdot BC}{... | // Block 1
draw((-50,0)--(-30,40)--(50,0)--(-50,0));
draw(Arc((0,0),50,0,180));
draw(rightanglemark((-50,0),(-30,40),(50,0),200));
dot((-50,0));
label("A",(-50,0),SW);
dot((-30,40));
label("C",(-30,40),NW);
dot((50,0));
label("B",(50,0),SE);
// Block 2
draw((-50,0)--(-30,40)--(50,0)--(-50,0)); draw(Arc((0,0),50,0,180))... | [] |
330 | Acute-angled $\triangle ABC$ is inscribed in a circle with center at $O$; $\stackrel \frown {AB} = 120^\circ$ and $\stackrel \frown {BC} = 72^\circ$.
A point $E$ is taken in minor arc $AC$ such that $OE$ is perpendicular to $AC$. Then the ratio of the magnitudes of $\angle OBE$ and $\angle BAC$ is:
$\textbf{(A)}\ \fr... | 1963 AHSME Problems/Problem 22 | Because $\stackrel \frown {AB} = 120^\circ$ and $\stackrel \frown {BC} = 72^\circ$, $\stackrel \frown {AC} = 168^\circ$. Also, $OA = OC$ and $OE \perp AC$, so $\angle AOE = \angle COE = 84^\circ$. Since $\angle BOC = 72^\circ$, $\angle BOE = 156^\circ$. Finally, $\triangle BOE$ is an isosceles triangle, so $\angle O... | // Block 1
draw(circle((0,0),1));
dot((-1,0));
pair A=(-1,0),B=(0.5,0.866),C=(0.978,-0.208),O=(0,0),E=(-0.105,-0.995);
label("A",(-1,0),W);
dot((0.5,0.866));
label("B",(0.5,0.866),NE);
dot((0.978,-0.208));
label("C",(0.978,-0.208),SE);
dot((0,0));
label("O",(0,0),NE);
dot(E);
label("E",E,S);
draw(A--B--C--A);
draw(E--O... | [] |
331 | In $\triangle ABC$ lines $CE$ and $AD$ are drawn so that
$\dfrac{CD}{DB}=\dfrac{3}{1}$ and $\dfrac{AE}{EB}=\dfrac{3}{2}$. Let $r=\dfrac{CP}{PE}$
where $P$ is the intersection point of $CE$ and $AD$. Then $r$ equals:
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \dfrac{3}{2}\qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qq... | 1963 AHSME Problems/Problem 39 | Draw line $PB$, and let $[PEB] = 2b$, $[PDB] = a$, and $[CAP] = c$, so $[CPD] = 3a$ and $[APE] = 3b$. Because $\triangle CAE$ and $\triangle CEB$ share an altitude,
\[c + 3b = \tfrac{3}{2} (3a+a+2b)\]
\[c + 3b = 6a + 3b\]
\[c = 6a\]
Because $\triangle ACD$ and $\triangle ABD$ share an altitude,
\[6a+3a = 3(a+2b+3b)\]... | // Block 1
size(8cm);
pair A = (0, 0), B = (9, 0), C = (3, 6);
pair D = (7.5, 1.5), E = (6.5, 0);
pair P = intersectionpoints(A--D, C--E)[0];
draw(A--B--C--cycle);
draw(A--D);
draw(C--E);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("$D$", D, NE);
label("$E$", E, S);
label("$P$", P, S);
draw(P--B... | [] |
332 | The length of a rectangle is $5$ inches and its width is less than $4$ inches. The rectangle is folded so that two
diagonally opposite vertices coincide. If the length of the crease is $\sqrt {6}$, then the width is:
$\textbf{(A)}\ \sqrt {2} \qquad \textbf{(B) }\ \sqrt {3} \qquad \textbf{(C) }\ 2 \qquad \textbf{(... | 1965 AHSME Problems/Problem 35 | Let the rectangle be $ABCD$ with $AB=CD=w$ and $AD=BC=5$, as in the diagram. We desire a line such that reflecting point $C$ across that line yields point $A$. For this to happen, the line must be perpendicular to the diagonal $\overline{AC}$, and it must go through the midpoint of $\overline{AC}$ (let it be point $M$)... | // Block 1
import geometry;
point M;
segment l;
// Rectangle ABCD
draw((0,sqrt(5))--(0,0)--(5,0)--(5,sqrt(5))--(0,sqrt(5)));
dot((0,sqrt(5)));
label("A", (0,sqrt(5)), NW);
dot((0,0));
label("B", (0,0), SW);
dot((5,0));
label("C", (5,0), SE);
dot((5,sqrt(5)));
label("D", (5, sqrt(5)), NE);
// Segment AC and point M
M... | [] |
333 | Given distinct straight lines $OA$ and $OB$. From a point in $OA$ a perpendicular is drawn to $OB$;
from the foot of this perpendicular a line is drawn perpendicular to $OA$.
From the foot of this second perpendicular a line is drawn perpendicular to $OB$;
and so on indefinitely. The lengths of the first and second ... | 1965 AHSME Problems/Problem 36 | For simplicity, let the first perpendicular from $\overleftrightarrow{OA}$ to $\overleftrightarrow{OB}$ be $\overline{AB}$, and let the second perpendicular have foot $C$ on $\overleftrightarrow{OA}$. Further, let the perpendicular from $C$ to $\overleftrightarrow{OB}$ have foot $D$ and length $c$, as in the diagram. A... | // Block 1
import geometry;
point O=(0,0);
point A=(10,5);
point B=(10,0);
point C;
point D;
line OA=line(O,A);
line OB=line(O,B);
// Lines OA and OB
draw(OA);
draw(OB);
// Points O, A, and B
dot(O);
label("O",O,S);
dot(A);
label("A",A,NW);
dot(B);
label("B",B,S);
// Segments AB, BC, and CD
draw(A--B);
pair[] x=inte... | [] |
334 | Point $E$ is selected on side $AB$ of $\triangle{ABC}$ in such a way that $AE: EB = 1: 3$ and point $D$ is selected on side $BC$
such that $CD: DB = 1: 2$. The point of intersection of $AD$ and $CE$ is $F$. Then $\frac {EF}{FC} + \frac {AF}{FD}$ is:
$\textbf{(A)}\ \frac {4}{5} \qquad \textbf{(B) }\ \frac {5}{4} \qq... | 1965 AHSME Problems/Problem 37 | We use mass points for this problem. Let $\text{m} A$ denote the mass of point $A$.
Rewrite the expression we are finding as
\[\frac{EF}{FC} + \frac{AF}{FD} = \frac{FE}{FC} + \frac{FA}{FD} = \frac{\text{m} C}{\text{m} E} + \frac{\text{m} D}{\text{m} A}\]
Now, let $\text{m} C = 2$. We then have $2 \cdot 1 = \text{m} B ... | // Block 1
import geometry;
point A = (0,0);
point B = (16,0);
point C = (3, 10);
point D, E, F;
real d;
// Triangle ABC
draw(A--B--C--A);
dot(A);
label("A", A, SW);
dot(B);
label("B", B, SE);
dot(C);
label("C", C, NW);
// Segments AD and CE
D = 2/3*C+1/3*B;
dot(D);
label("D", D, NE);
draw(A--D);
E = midpoint(A--mid... | [] |
335 | A foreman noticed an inspector checking a $3$"-hole with a $2$"-plug and a $1$"-plug and suggested that two more gauges
be inserted to be sure that the fit was snug. If the new gauges are alike, then the diameter, $d$, of each, to the nearest hundredth of an inch, is:
$\textbf{(A)}\ .87 \qquad \textbf{(B) }\ .86 \q... | 1965 AHSME Problems/Problem 39 | Let the center of the $3$" circle be $O$, that of the $2$" circle be $A$, that of the $1$" circle be $B$, and those of the circles of unknown radius (let their radii have length $r$) be $C$ and $D$, as in the diagram. Also, in this problem, no two circles share a center, so let the circles be named by their correspondi... | // Block 1
import geometry;
point O = (0,0);
point A = (-1/2,0);
point B = (1,0);
point C, D, E;
// 1", 2", and 3" circles
draw(circle(O,3/2));
dot(O);
label("O", O, S);
draw(circle(A,1));
dot(A);
label("A", A, SW);
draw(circle(B,1/2));
dot(B);
label("B", B, SE);
// Other two circles
C=(15/14*3/5,15/14*4/5);
D=(15/1... | [] |
336 | In quadrilateral $ABCD$ with diagonals $AC$ and $BD$, intersecting at $O$, $BO=4$, $OD = 6$, $AO=8$, $OC=3$, and $AB=6$. The length of $AD$ is:
$\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}$ | 1967 AHSME Problems/Problem 32 | (Diagram not to scale)
Since $AO \cdot OC = BO \cdot OD$, $ABCD$ is cyclic through power of a point. From the given information, we see that $\triangle{AOB}\sim \triangle{DOC}$ and $\triangle{BOC} \sim \triangle{AOD}$. Hence, we can find $CD=\frac{9}{2}$ and $AD=2 \cdot BC$. Letting $BC$ be $x$, we can use Ptolemy's t... | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(5cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black;... | [] |
337 | Located inside equilateral triangle $ABC$ is a point $P$ such that $PA=8$, $PB=6$, and $PC=10$. To the nearest integer the area of triangle $ABC$ is:
$\textbf{(A)}\ 159\qquad \textbf{(B)}\ 131\qquad \textbf{(C)}\ 95\qquad \textbf{(D)}\ 79\qquad \textbf{(E)}\ 50$ | 1967 AHSME Problems/Problem 40 | Notice that $6^2+8^2=10^2.$ That makes us want to construct a right triangle.
Rotate $\triangle APC$ $60^{\circ}$ about A. Note that $\triangle PAC \cong \triangle P'AB$, so
\[\angle P'AP = \angle PAB + \angle P'AB = \angle PAB + \angle PAC = 60^{\circ}.\]
Therefore, $\triangle APP'$ is equilateral, so $P'P=8$, whic... | // Block 1
draw((0,10)--(8.66,-5)--(-8.66,-5)--cycle);
label("$A$",(0,10),N);
label("$B$",(-9.5,-5.2),N);
label("$C$",(9.5,-5.2),N);
dot((-3,0));
label("$P$",(-3,-2),N);
draw((-3,0)--(0,10));
draw((-3,0)--(-8.66,-5));
draw((-3,0)--(8.66,-5));
dot((-9,7.5));
label("$P'$",(-9.2,7.5),N);
draw((-9,7.5)--(0,10));
draw((-9... | [] |
337 | Located inside equilateral triangle $ABC$ is a point $P$ such that $PA=8$, $PB=6$, and $PC=10$. To the nearest integer the area of triangle $ABC$ is:
$\textbf{(A)}\ 159\qquad \textbf{(B)}\ 131\qquad \textbf{(C)}\ 95\qquad \textbf{(D)}\ 79\qquad \textbf{(E)}\ 50$ | 1967 AHSME Problems/Problem 40 | Rotate $P$ and $B$ $60^{\circ}$ CCW around $A$, becoming $X$ and $C$. Rotate $P$ and $C$ $60^{\circ}$ CCW around $B$, becoming $Y$ and $A$. Rotate $P$ and $A$ $60^{\circ}$ CCW around $C$, becoming $Z$ and $B$:
Notice that since $\triangle AXC\cong\triangle APB$, $\triangle BYA\cong\triangle BPC$, and $\triangle CZ... | // Block 1
import graph;
import geometry;
size(12cm);
pair A, B, C, P, X, Y, Z;
// Define the equilateral triangle ABC
real a = sqrt(100+48*sqrt(3));
A = (0, 0);
B = rotate(60)*A + (a, 0);
C = rotate(120)*B + (a, 0);
// Define point P using given distances
pair[] P_candidates = intersectionpoints(Circle(A,8), Circle... | [] |
338 | Let $n$ be the number of points $P$ interior to the region bounded by a circle with radius $1$, such that the sum of squares of the distances from $P$ to the endpoints of a given diameter is $3$. Then $n$ is:
$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 4\quad \text{(E) } \infty$ | 1969 AHSME Problems/Problem 28 | Let $A$ and $B$ be points on diameter. Extend $AP$, and mark intersection with circle as point $C$.
Because $AB$ is a diameter, $\angle ACB = 90^\circ$. Also, by Exterior Angle Theorem, $\angle ACB + \angle CBP = \angle APB$, so $\angle APB > \angle ACB$, making $\angle APB$ an obtuse angle.
By the Law of Cosines, ... | // Block 1
draw(circle((0,0),50));
draw((-50,0)--(-10,20)--(50,0)--(-50,0));
draw((-10,20)--(30,40)--(50,0),dotted);
dot((-50,0));
label("$A$",(-50,0),W);
dot((50,0));
label("$B$",(50,0),E);
dot((-10,20));
label("$P$",(-10,20),S);
dot((30,40));
label("$C$",(30,40),NE);
// Block 2
draw(circle((0,0),50)); draw((-50,0)--(... | [] |
339 | Each circle in an infinite sequence with decreasing radii is tangent externally to the one following it and to both sides of a given right angle.
The ratio of the area of the first circle to the sum of areas of all other circles in the sequence, is
$\textbf{(A) }(4+3\sqrt{2}):4\qquad \textbf{(B) }9\sqrt{2}:2\qquad \t... | 1971 AHSME Problems/Problem 35 | WLOG, let the radius of the largest circle be $1$ (we can do this because scaling the diagram preserves the ratio of areas). Let the largest circle be $\omega_1$ and the second largest circle be $\omega_2$. As in the diagram above, let $\omega_1$ have center $A$ with points of tangency at $B$ and $D$. Let the tangents ... | // Block 1
import geometry;
point A = origin;
point B = dir(135);
point C = (0,sqrt(2));
point D = dir(45);
point X = 1/(1+sqrt(2)/2)*(B-C)+C;
point Y = 1/(1+sqrt(2)/2)*(D-C)+C;
circle i = incircle(triangle(C,X,Y));
point R;
// Defining R
pair[] r = intersectionpoints(i,triangle(C,X,Y));
R = r[1];
// Circles
dra... | [] |
340 | A sector with acute central angle $\theta$ is cut from a circle of radius 6. The radius of the circle circumscribed about the sector is
$\textbf{(A)}\ 3\cos\theta \qquad \textbf{(B)}\ 3\sec\theta \qquad \textbf{(C)}\ 3 \cos \frac12 \theta \qquad \textbf{(D)}\ 3 \sec \frac12 \theta \qquad \textbf{(E)}\ 3$ | 1973 AHSME Problems/Problem 15 | Let $O$ be the center of the circle and $A,B$ be two points on the circle such that $\angle AOB = \theta$. If the circle circumscribes the sector, then the circle must circumscribe $\triangle AOB$.
Draw the perpendicular bisectors of $OA$ and $OB$ and mark the intersection as point $C$, and draw a line from $C$ to ... | // Block 1
draw((-120,-160)--(0,0)--(120,-160));
draw((-60,-80)--(0,-125)--(60,-80),dotted);
draw((0,0)--(0,-125));
draw(arc((0,0),200,233.13,306.87));
dot((0,0));
label("O",(0,0),N);
dot((-120,-160));
label("A",(-120,-160),SW);
dot((120,-160));
label("B",(120,-160),SE);
// Block 2
draw((-120,-160)--(0,0)--(120,-160));... | [] |
341 | In the unit circle shown in the figure, chords $PQ$ and $MN$ are parallel to the unit radius $OR$ of the circle with center at $O$. Chords $MP$, $PQ$, and $NR$ are each $s$ units long and chord $MN$ is $d$ units long.
Of the three equations
\[\textbf{I.}\ d-s=1, \qquad \textbf{II.}\ ds=1, \qquad \textbf{III.}\ d^2... | 1973 AHSME Problems/Problem 35 | First, let $S$ be on circle $O$ so $RS$ is a diameter. In order to prove that the three statements are true (or false), we first show that $SM = QN = s,$ and then we examine each statement one by one.
Lemma 1: $SM = QN = s$
Since $OM = ON,$ by the Base Angle Theorem, $\angle OMN = \angle ONM.$ By the Alternate In... | // Block 1
pair O=(0,0), R=(10,0), M=(-8.5,5.3), N=(8.5,5.3), P=(-3,9.5), Q=(3,9.5);
draw(Circle(O,10));
draw(O--R--N--M--P--Q);
dot(O);
dot(R);
dot(N);
dot(M);
dot(P);
dot(Q);
label("1", (5,0), S);
label("s", (10.3,2.6));
label("s", (-5,7));
label("s", (0,8.5));
label("O", O, S);
label("R", R, E);
label("M", M, W);
... | [] |
342 | The edges of a regular tetrahedron with vertices $A ,~ B,~ C$, and $D$ each have length one.
Find the least possible distance between a pair of points $P$ and $Q$, where $P$ is on edge $AB$ and $Q$ is on edge $CD$.
$\textbf{(A) }\frac{1}{2}\qquad \textbf{(B) }\frac{3}{4}\qquad \textbf{(C) }\frac{\sqrt{2}}{2}\qq... | 1979 AHSME Problems/Problem 23 | Note that the distance $PQ$ will be minimized when $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $CD$.
To find this distance, consider triangle $\triangle PCQ$. $Q$ is the midpoint of $CD$, so $CQ=\frac{1}{2}$. Additionally, since $CP$ is the altitude of equilateral $\triangle ABC$, $CP=\frac{\sqrt{3}}{2}$.
... | // Block 1
size(150);
import patterns;
import olympiad;
pair D=(0,0),C=(1,-1),B=(2.5,-0.2),A=(1,2),AA,BB,CC,DD,P,Q,aux;
add("hatch",hatch());
//AA=new A and etc.
draw(rotate(100,D)*(A--B--C--D--cycle));
AA=rotate(100,D)*A;
BB=rotate(100,D)*D;
CC=rotate(100,D)*C;
DD=rotate(100,D)*B;
draw(BB--DD);
P=midpoint(AA--BB);
Q=m... | [] |
343 | Sides $AB,~ BC$, and $CD$ of (simple*) quadrilateral $ABCD$ have lengths $4,~ 5$, and $20$, respectively.
If vertex angles $B$ and $C$ are obtuse and $\sin C = - \cos B =\frac{3}{5}$, then side $AD$ has length
A polygon is called “simple” if it is not self intersecting.
$\textbf{(A) }24\qquad \textbf{(B) }24.5\qquad ... | 1979 AHSME Problems/Problem 24 | We know that $\sin(C)=-\cos(B)=\frac{3}{5}$. Since $B$ and $C$ are obtuse, we have $\sin(180-C)=\cos(180-B)=\frac{3}{5}$. It is known that $\sin(x)=\cos(90-x)$, so $180-C=90-(180-C)=180-B$. We simplify this as follows:
\[-90+C=180-B\]
\[B+C=270^{\circ}\]
Since $B+C=270^{\circ}$, we know that $A+D=360-(B+C)=90^{\circ... | // Block 1
size(10cm);
label("A",(-1,0));
dot((0,0));
label("B",(-1,4));
dot((0,4));
label("E",(-1,7));
dot((0,7));
label("C",(4,8));
dot((4,7));
label("D",(24,8));
dot((24,7));
draw((0,0)--(0,4));
draw((0,4)--(4,7));
draw((4,7)--(24,7));
draw((24,7)--(0,0));
draw((0,4)--(0,7), dashed);
draw((0,7)--(4,7), dashed);
//di... | [] |
344 | In a narrow alley of width $w$ a ladder of length $a$ is placed with its foot at point $P$ between the walls.
Resting against one wall at $Q$, the distance $k$ above the ground makes a $45^\circ$ angle with the ground.
Resting against the other wall at $R$, a distance $h$ above the ground, the ladder makes a $75^\cir... | 1982 AHSME Problems/Problem 22 | We know that $m\angle QPL=45^{\circ}$ and $m\angle RPT=75^{\circ}.$ Therefore, $m\angle QPR=60^{\circ}.$ \[\qquad\]
Because the two ladders are the same length, we know that
\[RP=PQ=a.\]
Since $\triangle QPR$ is isosceles with vertex angle $60^{\circ},$ we can conclude that it must be equilateral.\[\qquad\]
Now, sinc... | // Block 1
import olympiad;
size(200);
pair T,P,Q,M,L,R;
T=(0,0);
P=(2.5,0);
Q=(10,7.5);
M=(0,7.5);
L=(10,0);
R=(0,10);
draw(R--T--L--Q);
draw(P--Q--R--cycle);
draw(Q--M);
label("$P$", P, S);
dot(P);
label("$Q$", Q, E);
dot(Q);
label("$R$", R, W);
dot(R);
label("$S$", M, W);
dot(M);
... | [] |
345 | The solid shown has a square base of side length $s$. The upper edge is parallel to the base and has length $2s$. All other edges have length $s$. Given that $s=6\sqrt{2}$, what is the volume of the solid? | 1983 AIME Problems/Problem 11 | Solution 1
First, we find the height of the solid by dropping a perpendicular from the midpoint of $AD$ to $EF$. The hypotenuse of the triangle formed is the median of equilateral triangle $ADE$, and one of the legs is $3\sqrt{2}$. We apply the Pythagorean Theorem to deduce that the height is $6$.
Next, we complete t... | // Block 1
size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5); currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s... | [] |
346 | The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$. Suppose that the radius of the circle is $5$, that $BC=6$, and that $AD$ is bisected by $BC$. Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$. It follows that the sine of the central angle o... | 1983 AIME Problems/Problem 15 | Solution 1
As with some of the other solutions, we analyze this with a locus—but a different one. We'll consider: given a point $P$ and a line $\ell,$ what is the set of points $X$ such that the midpoint of $PX$ lies on line $\ell$? The answer to this question is: a line $m$ parallel to $\ell$, such that $m$ and $P$ ... | // Block 1
size(170); pair P = (0,0), L1 = (-3, 0), L2 = (1.5, 1.5), M1 = (-3, 1), M2 = (1.5, 2.5); pair X = .6*M1 + .4*M2, M = (P+X)/2; draw(L1--L2); draw(M1--M2); draw(P--X, dotted); dot(M); dot("$P$", P, S); dot("$X$", X, N); label("$\ell$", L2, E); label("$m$", M2, E);
// Block 2
size(170); pair O = (0,0), D = (0, ... | ["https://artofproblemsolving.com/wiki/images/thumb/b/b7/Aime1983p15s2.png/500px-Aime1983p15s2.png", "https://artofproblemsolving.com/wiki/images/thumb/3/3c/Dgram.png/800px-Dgram.png"] |
347 | A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. | 1983 AIME Problems/Problem 4 | Because we are given a right angle, we look for ways to apply the Pythagorean Theorem. Let the foot of the perpendicular from $O$ to $AB$ be $D$ and let the foot of the perpendicular from $O$ to the line $BC$ be $E$. Let $OE=x$ and $OD=y$. We're trying to find $x^2+y^2$.
Applying the Pythagorean Theorem, $OA^2 = OD^... | // Block 1
size(150); defaultpen(linewidth(0.6)+fontsize(11));
real r=10;
pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r);
pair D=(A.x,0),F=(0,B.y);
path P=circle(O,r);
pair C=intersectionpoint(B--(B.x+r,B.y),P);
draw(P);
draw(C--B--O--A--B);
draw(D--O--F--B,dashed);
dot(O); dot(A); dot(B); dot(C);
label("$O$",O,SW);
label("$A$... | [] |
347 | A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. | 1983 AIME Problems/Problem 4 | We'll use the law of cosines. Let $O$ be the center of the circle; we wish to find $OB$. We know how long $OA$ and $AB$ are, so if we can find $\cos \angle OAB$, we'll be in good shape.
We can find $\cos \angle OAB$ using angles $OAC$ and $BAC$. First we note that by Pythagoras,
\[AC = \sqrt{AB^2 + BC^2} = \sqrt{... | // Block 1
size(150); defaultpen(linewidth(0.6)+fontsize(11));
real r=10;
pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r);
pair D=(A.x,0),F=(0,B.y);
path P=circle(O,r);
pair C=intersectionpoint(B--(B.x+r,B.y),P);
draw(P);
draw(C--B--O--A--B);
draw(O--B); draw(A--C);
dot(O); dot(A); dot(B); dot(C);
label("$O$",O,SW);
label("$A$"... | [] |
347 | A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. | 1983 AIME Problems/Problem 4 | Mark the midpoint $M$ of $AC$. Then, drop perpendiculars from $O$ to $AB$ (with foot $T_1$), $M$ to $OT_1$ (with foot $T_2$), and $M$ to $AB$ (with foot $T_3$).
First notice that by computation, $OAC$ is a $\sqrt {50} - \sqrt {40} - \sqrt {50}$ isosceles triangle, so $AC = MO$.
Then, notice that $\angle MOT_2 = \angl... | // Block 1
size(200);
pair dl(string name, pair loc, pair offset) {
dot(loc);
label(name,loc,offset);
return loc;
};
pair a[] = {(0,0),(0,5),(1,5),(1,7),(-2,6),(-5,5),(-2,5),(-2,6),(0,6)};
string n[] = {"O","$T_1$","B","C","M","A","$T_3$","M","$T_2$"};
for(int i=0;i<a.length;++i) {
dl(n[i],a[i],dir(degrees(a[i],fal... | [] |
347 | A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. | 1983 AIME Problems/Problem 4 | I will use the law of cosines in triangle $\triangle OAC$ and $\triangle OBC$.
$AC = \sqrt{AB^2 + BC^2} = \sqrt{6^2 + 2^2} = 2 \sqrt{10}$
$\cos \angle ACB = \frac{2}{2\sqrt{10}} = \frac{1}{\sqrt{10}}$
$\cos \angle ACO = \frac{AC^2+OC^2-OA^2}{2 \cdot AC \cdot OC} = \frac{(2\sqrt{10})^2+(\sqrt{50})^2-(\sqrt{50})^2}{2 ... | // Block 1
size(150); defaultpen(linewidth(0.6)+fontsize(11));
real r=10;
pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r);
pair D=(A.x,0),F=(0,B.y);
path P=circle(O,r);
pair C=intersectionpoint(B--(B.x+r,B.y),P);
draw(P);
draw(C--B--O--A--B);
draw(O--B); draw(A--C); draw(O--C);
dot(O); dot(A); dot(B); dot(C);
label("$O$",O,SW);... | [] |
348 | In tetrahedron $ABCD$, edge $AB$ has length 3 cm. The area of face $ABC$ is $15\mbox{cm}^2$ and the area of face $ABD$ is $12 \mbox { cm}^2$. These two faces meet each other at a $30^\circ$ angle. Find the volume of the tetrahedron in $\mbox{cm}^3$. | 1984 AIME Problems/Problem 9 | Position face $ABC$ on the bottom. Since $[\triangle ABD] = 12 = \frac{1}{2} \cdot AB \cdot h_{ABD}$, we find that $h_{ABD} = 8$. Because the problem does not specify, we may assume both $ABC$ and $ABD$ to be isosceles triangles. Thus, the height of $ABD$ forms a $30-60-90$ with the height of the tetrahedron. So, $h = ... | // Block 1
/* modified version of olympiad modules */
import three;
real markscalefactor = 0.03;
path3 rightanglemark(triple A, triple B, triple C, real s=8)
{
triple P,Q,R;
P=s*markscalefactor*unit(A-B)+B;
R=s*markscalefactor*unit(C-B)+B;
Q=P+R-B;
return P--Q--R;
}
path3 anglemark(triple A, triple B, triple C, re... | [] |
349 | An ellipse has foci at $(9,20)$ and $(49,55)$ in the $xy$-plane and is tangent to the $x$-axis. What is the length of its major axis? | 1985 AIME Problems/Problem 11 | An ellipse is defined to be the locus of points $P$ such that the sum of the distances between $P$ and the two foci is constant. Let $F_1 = (9, 20)$, $F_2 = (49, 55)$ and $X = (x, 0)$ be the point of tangency of the ellipse with the $x$-axis. Then $X$ must be the point on the axis such that the sum $F_1X + F_2X$ is m... | // Block 1
size(200);
pointpen=black;pathpen=black+linewidth(0.6);pen f = fontsize(10);
pair F1=(9,20),F2=(49,55);
D(shift((F1+F2)/2)*rotate(41.186)*scale(85/2,10*11^.5)*unitcircle);
D((-20,0)--(80,0)--(0,0)--(0,80)--(0,-60));
path p = F1--(49,-55);
pair X = IP(p,(0,0)--(80,0));
D(p,dashed);D(F1--X--F2);D(F1);D(F2);D((... | [] |
350 | In a circle, parallel chords of lengths 2, 3, and 4 determine central angles of $\alpha$, $\beta$, and $\alpha + \beta$ radians, respectively, where $\alpha + \beta < \pi$. If $\cos \alpha$, which is a positive rational number, is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator... | 1985 AIME Problems/Problem 9 | All chords of a given length in a given circle subtend the same arc and therefore the same central angle. Thus, by the given, we can re-arrange our chords into a triangle with the circle as its circumcircle.
This triangle has semiperimeter $\frac{2 + 3 + 4}{2}$ so by Heron's formula it has area $K = \sqrt{\frac92 ... | // Block 1
size(200); pointpen = black; pathpen = black + linewidth(0.8); real r = 8/15^0.5, a = 57.91, b = 93.135; pair O = (0,0), A = r*expi(pi/3); D(CR(O,r)); D(O--rotate(a/2)*A--rotate(-a/2)*A--cycle); D(O--rotate(b/2)*A--rotate(-b/2)*A--cycle); D(O--rotate((a+b)/2)*A--rotate(-(a+b)/2)*A--cycle); MP("2",(rotate(a/... | [] |
350 | In a circle, parallel chords of lengths 2, 3, and 4 determine central angles of $\alpha$, $\beta$, and $\alpha + \beta$ radians, respectively, where $\alpha + \beta < \pi$. If $\cos \alpha$, which is a positive rational number, is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator... | 1985 AIME Problems/Problem 9 | It’s easy to see in triangle which lengths 2, 3, and 4, that the angle opposite the side 2 is $\frac{\alpha}{2}$, and using the Law of Cosines, we get:
\[2^2 = 3^2 + 4^2 - 2\cdot3\cdot4\cos\frac{\alpha}{2}\]
Which, rearranges to:
\[21 = 24\cos\frac{\alpha}{2}\]
And, that gets us:
\[\cos\frac{\alpha}{2} = 7/8\]
U... | // Block 1
size(200);
pointpen = black; pathpen = black + linewidth(0.8);
real r = 8/15^0.5, a = 57.91, b = 93.135;
pair O = (0,0), A = r*expi(pi/3), A1 = rotate(a/2)*A, A2 = rotate(-a/2)*A, A3 = rotate(-a/2-b)*A;
D(CR(O,r));
D(O--A1--A2--cycle);
D(O--A2--A3--cycle);
D(O--A1--A3--cycle);
MP("2",(A1+A2)/2,NE);
MP("3",(... | [] |
351 | $ABCDE$ is a regular pentagon. $AP, AQ$ and $AR$ are the perpendiculars dropped from $A$ onto $CD, CB$ extended and $DE$ extended,
respectively. Let $O$ be the center of the pentagon. If $OP = 1$, then $AO + AQ + AR$ equals
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 1 + \sqrt{5}\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ ... | 1986 AHSME Problems/Problem 28 | To solve the problem, we compute the area of regular pentagon $ABCDE$ in two different ways. First, we can divide regular pentagon $ABCDE$ into five congruent triangles.
If $s$ is the side length of the regular pentagon, then each of the triangles $AOB$, $BOC$, $COD$, $DOE$, and $EOA$ has base $s$ and height 1, so th... | // Block 1
unitsize(2 cm);
pair A, B, C, D, E, O, P, Q, R;
A = dir(90);
B = dir(90 - 360/5);
C = dir(90 - 2*360/5);
D = dir(90 - 3*360/5);
E = dir(90 - 4*360/5);
O = (0,0);
P = (C + D)/2;
Q = (A + reflect(B,C)*(A))/2;
R = (A + reflect(D,E)*(A))/2;
draw((2*R - E)--D--C--(2*Q - B));
draw(A--P);
draw(A--Q);
draw(A--R);... | [] |
352 | Let triangle $ABC$ be a right triangle in the xy-plane with a right angle at $C_{}$. Given that the length of the hypotenuse $AB$ is $60$, and that the medians through $A$ and $B$ lie along the lines $y=x+3$ and $y=2x+4$ respectively, find the area of triangle $ABC$. | 1986 AIME Problems/Problem 15 | Let $\theta_1$ be the angle that the median through $A$ makes with the positive $y$-axis, and let $\theta_2$ be the angle that the median through $B$ makes with the positive $x$-axis. The tangents of these two angles are the slopes of the respective medians; in other words, $\tan \theta_1 = 1$, and $\tan \theta_2 =2$.... | // Block 1
size(170);
pair A = (0,0), B = (3, 2), C = (3, 0);
pair M = (B+C)/2, NN = (A+C)/2, G = (A+B+C)/3;
draw(A--B--C--cycle);
draw(A--M);
draw(B--NN);
label("$A$", A, S);
label("$C$", C, S);
label("$B$", B, N);
label("$M$", M, NW);
label("$N$", NN, NW);
label("$a$", M, E);
label("$b$", NN, S);
label("$G$", G, NW);... | [] |
352 | Let triangle $ABC$ be a right triangle in the xy-plane with a right angle at $C_{}$. Given that the length of the hypotenuse $AB$ is $60$, and that the medians through $A$ and $B$ lie along the lines $y=x+3$ and $y=2x+4$ respectively, find the area of triangle $ABC$. | 1986 AIME Problems/Problem 15 | We first seek to find the angle between the lines $y = x + 3$ and $y = 2x + 4$.
Let the acute angle the red line makes with the $x-$ axis be $\alpha$ and the acute angle the blue line makes with the $x-$ axis be $\beta$. Then, we know that $\tan \alpha = 1$ and $\tan \beta = 2$. Note that the acute angle between the r... | // Block 1
import graph;
size(150);
Label f;
f.p=fontsize(6);
xaxis(-8,8,Ticks(f, 2.0));
yaxis(-8,8,Ticks(f, 2.0));
real f(real x)
{
return (x + 3);
}
real g( real x){
return (2x + 4);
}
draw(graph(f,-8,5),red+linewidth(1));
draw(graph(g,-6,2),blue+linewidth(1));
// Block 2
import graph; size(4cm);
real label... | [] |
353 | In $\triangle ABC$, $AB= 425$, $BC=450$, and $AC=510$. An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle. If these three segments are of an equal length $d$, find $d$. | 1986 AIME Problems/Problem 9 | Solution 1
Let the points at which the segments hit the triangle be called $D, D', E, E', F, F'$ as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are similar ($\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF$). The remaining three se... | // Block 1
size(200); pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10); pair C=(0,0),A=(510,0),B=IP(circle(C,450),circle(A,425)); /* construct remaining points */ pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C); pair D=IP(Ea--(Ea+A-C),A--B),F=IP(Da--(Da+C-B),A--C... | [] |
354 | There are two natural ways to inscribe a square in a given isosceles right triangle.
If it is done as in Figure 1 below, then one finds that the area of the square is $441 \text{cm}^2$.
What is the area (in $\text{cm}^2$) of the square inscribed in the same $\triangle ABC$ as shown in Figure 2 below?
$\textbf{(A)}... | 1987 AHSME Problems/Problem 21 | We are given that the area of the inscribed square is $441$, so the side length of that square is $21$. Since the square divides the $45-45-90$ larger triangle into 2 smaller congruent $45-45-90$, then the legs of the larger isosceles right triangle ($BC$ and $AB$) are equal to $42$.
We now have that $3S=42\sqrt{2}$... | // Block 1
draw((0,0)--(10,0)--(0,10)--cycle);
draw((6.5,3.25)--(3.25,0)--(0,3.25)--(3.25,6.5));
label("A", (0,10), W);
label("B", (0,0), W);
label("C", (10,0), E);
label("S", (25/3,11/6), E);
label("S", (11/6,25/3), E);
label("S", (5,5), NE);
// Block 2
draw((0,0)--(10,0)--(0,10)--cycle); draw((6.5,3.25)--(3.25,0)--(0... | [] |
355 | A square flag has a red cross of uniform width with a blue square in the center on a white background as shown. (The cross is symmetric with respect to each of the diagonals of the square.) If the entire cross (both the red arms and the blue center) takes up 36% of the area of the flag, what percent of the area of the ... | 1989 AHSME Problems/Problem 21 | The diagram can be quartered as shown:
and reassembled into two smaller squares of side $k$, each of which looks like this:
The border in this figure is the former cross, which still occupies 36% of the area. Therefore the inner square occupies 64% of the area, from which we deduce that it is $0.8k \times 0.8k$, and ... | // Block 1
draw((0,0)--(0,5)--(5,5)--(5,0)--(0,0));
draw((0,1)--(4,5));
draw((1,0)--(5,4));
draw((0,4)--(4,0));
draw((1,5)--(5,1));
draw((0,0)--(5,5),dotted);
draw((0,5)--(5,0),dotted);
// Block 2
draw((0,0)--(0,5)--(5,5)--(5,0)--(0,0));
draw((0,1)--(4,1)--(4,5));
draw((1,0)--(1,4)--(5,4));
label("blue",(0.5,0.5));
lab... | [] |
355 | A square flag has a red cross of uniform width with a blue square in the center on a white background as shown. (The cross is symmetric with respect to each of the diagonals of the square.) If the entire cross (both the red arms and the blue center) takes up 36% of the area of the flag, what percent of the area of the ... | 1989 AHSME Problems/Problem 21 | Let the side length of the flag be 10, and let each red segment on the border of the flag be $x$. Then the four white triangles can be placed together to form a white square of area $100 - 36 = 64$ and side length $10 - 2x$.
Since $(10 - 2x)^2 = 64$, we have $x = 1$. Then the blue area is $(x \sqrt{2})^2 = 2$, which... | // Block 1
unitsize(4 cm);
pair[] A, B, C;
real t = 0.2;
A[1] = (0,0);
A[2] = (1,0);
A[3] = (1,1);
A[4] = (0,1);
B[1] = (t,0);
B[2] = (1 - t,0);
B[3] = (1,t);
B[4] = (1,1 - t);
B[5] = (1 - t,1);
B[6] = (t,1);
B[7] = (0,1 - t);
B[8] = (0,t);
C[1] = extension(B[1],B[4],B[7],B[2]);
C[2] = extension(B[3],B[6],B[1],B[4]);... | [] |
356 | Let $a$, $b$, $c$ be the three sides of a triangle, and let $\alpha$, $\beta$, $\gamma$, be the angles opposite them. If $a^2+b^2=1989c^2$, find
$\frac{\cot \gamma}{\cot \alpha+\cot \beta}$
Contents
1 Problem
2 Solution
2.1 Solution 1
2.2 Solution 2
2.3 Solution 3
2.4 Solution 4
2.5 Solution 5
2.6 Solution 6
2.7 Sol... | 1989 AIME Problems/Problem 10 | Solution 1
We draw the altitude $h$ to $c$, to get two right triangles.
Then $\cot{\alpha}+\cot{\beta}=\frac{c}{h}$, from the definition of the cotangent.
Let $K$ be the area of $\triangle ABC.$ Then $h=\frac{2K}{c}$, so $\cot{\alpha}+\cot{\beta}=\frac{c^2}{2K}$.
By identical logic, we can find similar expressions ... | // Block 1
size(170); pair A = (0,0), B = (3, 0), C = (1, 4); pair P = .5*(C + reflect(A,B)*C); draw(A--B--C--cycle); draw(C--P, dotted); draw(rightanglemark(C,P, B, 4)); label("$A$", A, S); label("$B$", B, S); label("$C$", C, N); label("$a$", (B+C)/2, NE); label("$b$", (A+C)/2, NW); label("$c$", (A+B)/2, S); label("$h... | [] |
357 | Two skaters, Allie and Billie, are at points $A$ and $B$, respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$. At the same time Allie leaves $A$, Billie leaves $... | 1989 AIME Problems/Problem 6 | Label the point of intersection as $C$. Since $d = rt$, $AC = 8t$ and $BC = 7t$. According to the law of cosines,
\begin{align*}(7t)^2 &= (8t)^2 + 100^2 - 2 \cdot 8t \cdot 100 \cdot \cos 60^\circ\\ 0 &= 15t^2 - 800t + 10000 = 3t^2 - 160t + 2000\\ t &= \frac{160 \pm \sqrt{160^2 - 4\cdot 3 \cdot 2000}}{6} = 20, \frac{1... | // Block 1
pointpen=black; pathpen=black+linewidth(0.7);
pair A=(0,0),B=(10,0),C=16*expi(pi/3);
D(B--A); D(A--C); D(B--C,dashed); MP("A",A,SW);MP("B",B,SE);MP("C",C,N);MP("60^{\circ}",A+(0.3,0),NE);MP("100",(A+B)/2);MP("8t",(A+C)/2,NW);MP("7t",(B+C)/2,NE);
// Block 2
pointpen=black; pathpen=black+linewidth(0.7); pair... | [] |
357 | Two skaters, Allie and Billie, are at points $A$ and $B$, respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$. At the same time Allie leaves $A$, Billie leaves $... | 1989 AIME Problems/Problem 6 | Let $P$ be the point of intersection between the skaters, Allie and Billie. We can draw a line that goes through $P$ and is parallel to $\overline{AB}$. Letting this line be the $x$-axis, we can reflect $B$ over the $x$-axis to get $B'$. As reflections preserve length, $B'X = XB$.
We then draw lines $BB'$ and $PB'$. W... | // Block 1
draw((0,0)--(11,0)--(7,14)--cycle);
draw((7,14)--(11,28));
draw((11,28)--(11,0));
label("$A$",(-1,-1),N);
label("$B$",(12,-1),N);
label("$P$",(6,15),N);
label("$B'$",(12,29),N);
draw((-10,14)--(20,14));
label("$X$",(12.5,15),N);
draw((7,0)--(7,14),dashed);
label("$Y$",(7,-4),N);
draw((6,0)--(6,1));
draw((6,1... | [] |
357 | Two skaters, Allie and Billie, are at points $A$ and $B$, respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$. At the same time Allie leaves $A$, Billie leaves $... | 1989 AIME Problems/Problem 6 | We can define $x$ to be the time elapsed since both Allie and Billie moved away from points $A$ and $B$ respectfully. Also, set the point of intersection to be $M$. Then we can produce the following diagram:
Now, if we drop an altitude from point$M$, we get :
We know this from the $30-60-90$ triangle that is form... | // Block 1
draw((0,0)--(100,0)--(80,139)--cycle);
label("8x",(0,0)--(80,139),NW);
label("7x",(100,0)--(80,139),NE);
label("100",(0,0)--(100,0),S);
dot((0,0));
label("A",(0,0),S);
dot((100,0));
label("B",(100,0),S);
dot((80,139));
label("M",(80,139),N);
// Block 2
size(300); draw((0,0)--(100,0)--(80,139)--cycle);
label(... | [] |
357 | Two skaters, Allie and Billie, are at points $A$ and $B$, respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$. At the same time Allie leaves $A$, Billie leaves $... | 1989 AIME Problems/Problem 6 | Drop the altitude from $M$ to $AB$, and call it $P$. $\triangle AMP$ is a $30-60-90$ triangle, so $AP = 4t$ and $MP = 4\sqrt{3}t$, and by the Pythagorean theorem on $\triangle MPB$, $PB = t$. $AP + BP = AB$, so $t=20$. Therefore, $8t = \boxed{160}$.
~~Disphenoid_lover | // Block 1
draw((0,0)--(100,0)--(80,139)--cycle);
label("8x",(0,0)--(80,139),NW);
label("7x",(100,0)--(80,139),NE);
label("100",(0,0)--(100,0),S);
dot((0,0));
label("A",(0,0),S);
dot((100,0));
label("B",(100,0),S);
dot((80,139));
label("M",(80,139),N);
draw((80,139)--(80,0),dashed); // Altitude MP
label("$P$",(80,0),S)... | [] |
358 | In the figure $ABCD$ is a quadrilateral with right angles at $A$ and $C$. Points $E$ and $F$ are on $\overline{AC}$, and $\overline{DE}$ and $\overline{BF}$ are perpendicual to $\overline{AC}$. If $AE=3, DE=5,$ and $CE=7$, then $BF=$
$\text{(A) } 3.6\quad \text{(B) } 4\quad \text{(C) } 4.2\quad \text{(D) } 4.5\quad \t... | 1990 AHSME Problems/Problem 20 | Label the angles as shown in the diagram. Since $\angle DEC$ forms a linear pair with $\angle DEA$, $\angle DEA$ is a right angle.
Let $\angle DAE = \alpha$ and $\angle ADE = \beta$.
Since $\alpha + \beta = 90^\circ$, and $\alpha + \angle BAF = 90^\circ$, then $\beta = \angle BAF$. By the same logic, $\angle ABF = ... | // Block 1
pair A = (0,0), B = (7,4.2), C = (10, 0), D = (3, -5), E = (3, 0), F = (7,0);
draw(A--B--C--D--cycle,dot);
draw(A--E--F--C,dot);
draw(D--E--F--B,dot);
markscalefactor = 0.075;
draw(rightanglemark(B, A, D));
draw(rightanglemark(D, E, A));
draw(rightanglemark(B, F, A));
draw(rightanglemark(D, C, B));
draw(ri... | [] |
359 | Consider a pyramid $P-ABCD$ whose base $ABCD$ is square and whose vertex $P$ is equidistant from $A,B,C$ and $D$. If $AB=1$ and $\angle{APB}=2\theta$, then the volume of the pyramid is
$\text{(A) } \frac{\sin(\theta)}{6}\quad \text{(B) } \frac{\cot(\theta)}{6}\quad \text{(C) } \frac{1}{6\sin(\theta)}\quad \text{(D) } ... | 1990 AHSME Problems/Problem 21 | As the base has area $1$, the volume will be one third of the height. Drop a line from $P$ to $AB$, bisecting it at $Q$.
Then $\angle QPB=\theta$, so $\cot\theta=\frac{PQ}{BQ}=2PQ$. Therefore $PQ=\tfrac12\cot\theta$.
Now turning to the dotted triangle, by Pythagoras, the square of the pyramid's height is \[PQ^2-(\tfr... | // Block 1
import three;unitsize(1cm);size(200);real h = 0.7;
//currentprojection=perspective(1/3,-1,1/2);
triple P = (.5,.5,h);
draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle);
draw((0,0,0)--P--(1,1,0)^^(1,0,0)--P--(0,1,0));
draw((.5,.5,0)--P--(.5,0,0)--cycle,dotted);
dot((0,0,0));dot((1,0,0));
label("P",P,N);label("Q... | [] |
360 | Nine congruent spheres are packed inside a unit cube in such a way that one of them has its center at the center of the cube and each of the others is tangent to the center sphere and to three faces of the cube. What is the radius of each sphere?
$\text{(A) } 1-\frac{\sqrt{3}}{2}\quad \text{(B) } \frac{2\sqrt{3}-3}{2}... | 1990 AHSME Problems/Problem 25 | Let $r$ be the radius, let $C$ be the center of the cube, and let $P$ be the center of one of the eight outer spheres, noting that $PC=2r$.
Back, in the corner of the unit cube, a smaller cube whose inner corner coincides with $P$. The cube is of dimensions $1\times 1\times 1$ and its space diagonal is of length $\sqr... | // Block 1
draw((0,0)--(0,1)--(1,1)--(1,0)--cycle);
for(int i=0;i<4;++i)draw(rotate(i*90,(.5,.5))*circle((0.25,0.25),0.225));
//draw(circle((0.75,0.25),0.225));
draw(circle((0.5,0.5),0.19),dashed);
draw((0,0)--(.5,.5),dotted);
dot((.25,.25));dot((.5,.5));
label("P",(.25,.25),S);label("C",(.5,.5),N);
// Block 2
draw((0,... | [] |
361 | The rectangle $ABCD^{}_{}$ below has dimensions $AB^{}_{} = 12 \sqrt{3}$ and $BC^{}_{} = 13 \sqrt{3}$. Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at $P^{}_{}$. If triangle $ABP^{}_{}$ is cut out and removed, edges $\overline{AP}$ and $\overline{BP}$ are joined, and the figure is then creased along segmen... | 1990 AIME Problems/Problem 14 | Solution 1(Synthetic)
Our triangular pyramid has base $12\sqrt{3} - 13\sqrt{3} - 13\sqrt{3} \triangle$. The area of this isosceles triangle is easy to find by $[ACD] = \frac{1}{2}bh$, where we can find $h_{ACD}$ to be $\sqrt{399}$ by the Pythagorean Theorem. Thus $A = \frac 12(12\sqrt{3})\sqrt{399} = 18\sqrt{133}$.
... | // Block 1
import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9); currentprojection = perspective(20,-20,12); triple O=(0,0,0); triple A=(0, 399^(0.5), 0); triple D=(108^(0.5), 0, 0); triple C=(-108^(0.5), 0, 0); triple Pa; pair Ci=circumcenter((A.x,A.y),(C.x,C.y),(D.x,D.y)); triple ... | [] |
362 | A triangle has vertices $P_{}^{}=(-8,5)$, $Q_{}^{}=(-15,-19)$, and $R_{}^{}=(1,-7)$. The equation of the bisector of $\angle P$ can be written in the form $ax+2y+c=0_{}^{}$. Find $a+c_{}^{}$.
Contents
1 Problem
2 Solution
2.1 Solution 1
2.2 Solution 2
2.3 Solution 3
2.4 Solution 4
2.5 Solution 5 (Trigonometry)
3 ... | 1990 AIME Problems/Problem 7 | Use the distance formula to determine the lengths of each of the sides of the triangle. We find that it has lengths of side $15,\ 20,\ 25$, indicating that it is a $3-4-5$ right triangle. At this point, we just need to find another point that lies on the bisector of $\angle P$.
Solution 1
Use the angle bisector theor... | // Block 1
import graph; pointpen=black;pathpen=black+linewidth(0.7);pen f = fontsize(10); pair P=(-8,5),Q=(-15,-19),R=(1,-7),S=(7,-15),T=(-4,-17),U=IP(P--T,Q--R); MP("P",P,N,f);MP("Q",Q,W,f);MP("R",R,E,f);MP("P'",U,SE,f); D(P--Q--R--cycle);D(U);D(P--U); D((-17,0)--(4,0),Arrows(2mm));D((0,-21)--(0,7),Arrows(2mm));
// B... | [] |
363 | Rhombus $PQRS^{}_{}$ is inscribed in rectangle $ABCD^{}_{}$ so that vertices $P^{}_{}$, $Q^{}_{}$, $R^{}_{}$, and $S^{}_{}$ are interior points on sides $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, and $\overline{DA}$, respectively. It is given that $PB^{}_{}=15$, $BQ^{}_{}=20$, $PR^{}_{}=30$, and $QS^{}_{}=40$. ... | 1991 AIME Problems/Problem 12 | Note that this is a modified version of the original diagram. By the Pythagorean theorem, the side length of the rhombus is $25$. Since a rhombus is a parallelogram, its diagonals bisect each other. Thus, $\triangle PBQ \cong \triangle RDS$ and $\triangle QCR \cong \triangle SAP$ by symmetry.
Let $RC=x$ and $QC=y$. Ap... | // Block 1
defaultpen(fontsize(12)+linewidth(1.3));
pair A=(0,28.8), B=(38.4,28.8), C=(38.4,0), D=(0,0), O, P=(23.4,28.8), Q=(38.4,8.8), R=(15,0), S=(0,20);
pair F=(0, Q.y);
O=intersectionpoint(A--C,B--D);
draw(A--B--C--D--cycle);
draw(P--R..Q--S);
draw(P--Q--R--S--cycle);
draw(Q--F, dashed);
draw(rightanglemark(Q... | [] |
364 | A hexagon is inscribed in a circle. Five of the sides have length $81$ and the sixth, denoted by $\overline{AB}$, has length $31$. Find the sum of the lengths of the three diagonals that can be drawn from $A_{}^{}$. | 1991 AIME Problems/Problem 14 | Let $x=AC=BF$, $y=AD=BE$, and $z=AE=BD$.
Ptolemy's Theorem on $ABCD$ gives $81y+31\cdot 81=xz$, and Ptolemy on $ACDF$ gives $x\cdot z+81^2=y^2$.
Subtracting these equations give $y^2-81y-112\cdot 81=0$, and from this $y=144$. Ptolemy on $ADEF$ gives $81y+81^2=z^2$, and from this $z=135$. Finally, plugging back into th... | // Block 1
defaultpen(fontsize(9));
pair A=expi(-pi/2-acos(475/486)), B=expi(-pi/2+acos(475/486)), C=expi(-pi/2+acos(475/486)+acos(7/18)), D=expi(-pi/2+acos(475/486)+2*acos(7/18)), E=expi(-pi/2+acos(475/486)+3*acos(7/18)), F=expi(-pi/2-acos(475/486)-acos(7/18));
draw(unitcircle);draw(A--B--C--D--E--F--A);draw(A--C..A--... | [] |
365 | Rectangle $ABCD_{}^{}$ has sides $\overline {AB}$ of length 4 and $\overline {CB}$ of length 3. Divide $\overline {AB}$ into 168 congruent segments with points $A_{}^{}=P_0, P_1, \ldots, P_{168}=B$, and divide $\overline {CB}$ into 168 congruent segments with points $C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B$. For $1_{}^{} ... | 1991 AIME Problems/Problem 2 | The length of the diagonal is $\sqrt{3^2 + 4^2} = 5$ (a 3-4-5 right triangle). For each $k$, $\overline{P_kQ_k}$ is the hypotenuse of a $3-4-5$ right triangle with sides of $3 \cdot \frac{168-k}{168}, 4 \cdot \frac{168-k}{168}$. Thus, its length is $5 \cdot \frac{168-k}{168}$. Let $a_k=\frac{5(168-k)}{168}$. We want to... | // Block 1
real r = 0.35; size(220);
pointpen=black;pathpen=black+linewidth(0.65);pen f = fontsize(8);
pair A=(0,0),B=(4,0),C=(4,3),D=(0,3);
D(A--B--C--D--cycle);
pair P1=A+(r,0),P2=A+(2r,0),P3=B-(r,0),P4=B-(2r,0);
pair Q1=C-(0,r),Q2=C-(0,2r),Q3=B+(0,r),Q4=B+(0,2r);
D(A--C);D(P1--Q1);D(P2--Q2);D(P3--Q3);D(P4--Q4);
MP("... | [] |
366 | In $\triangle{ABC}$, $\angle ABC=120^\circ,AB=3$ and $BC=4$. If perpendiculars constructed to $\overline{AB}$ at $A$ and to $\overline{BC}$ at $C$ meet at $D$, then $CD=$
$\text{(A) } 3\quad \text{(B) } \frac{8}{\sqrt{3}}\quad \text{(C) } 5\quad \text{(D) } \frac{11}{2}\quad \text{(E) } \frac{10}{\sqrt{3}}$ | 1992 AHSME Problems/Problem 25 | We begin by drawing a diagram.
We extend $CB$ and $DA$ to meet at $E.$ This gives us a couple right triangles in $CED$ and $BEA.$
We see that $\angle E = 30^\circ$. Hence, $\triangle BEA$ and $\triangle DEC$ are 30-60-90 triangles.
Using the side ratios of 30-60-90 triangles, we have $BE=2BA=6$. This tells us that $... | // Block 1
import olympiad; import cse5; import geometry; size(150);
defaultpen(fontsize(10pt));
defaultpen(0.8);
dotfactor = 4;
pair A = origin;
pair C = A+dir(55);
pair D = A+dir(0);
pair B = extension(A,A+dir(90),C,C+dir(-155));
label("$A$",A,S);
label("$C$",C,NE);
label("$D$",D,SE);
label("$B$",B,NW);
label("$4$",B... | [] |
367 | A circle of radius $r$ has chords $\overline{AB}$ of length $10$ and $\overline{CD}$ of length 7. When $\overline{AB}$ and $\overline{CD}$ are extended through $B$ and $C$, respectively, they intersect at $P$, which is outside of the circle. If $\angle{APD}=60^\circ$ and $BP=8$, then $r^2=$
$\text{(A) } 70\quad \text{... | 1992 AHSME Problems/Problem 27 | Applying Power of a Point on $P$, we find that $PC=9$ and thus $PD=16$. Observing that $PD=2BP$ and that $\angle BPD=60^{\circ}$, we conclude that $BPD$ is a $30-60-90$ right triangle with right angle at $B$. Thus, $BD=8\sqrt{3}$ and triangle $ABD$ is also right. Using that fact that the circumcircle of a right triangl... | // Block 1
import olympiad;
import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; path circ = Circle(origin, 1); pair A = dir(degrees(7pi/12)); pair D = dir(degrees(-5pi/12)); pair B = dir(degrees(2pi/12)); pair C = dir(degrees(-2pi/12)); pair P = extension(A, B, C, D); d... | [] |
368 | The sides of $\triangle ABC$ have lengths $6,8,$ and $10$. A circle with center $P$ and radius $1$ rolls around the inside of $\triangle ABC$, always remaining tangent to at least one side of the triangle. When $P$ first returns to its original position, through what distance has $P$ traveled?
$\text{(A) } 10\quad \te... | 1993 AHSME Problems/Problem 27 | Start by considering the triangle traced by $P$ as the circle moves around the triangle. It turns out this triangle is similar to the $6-8-10$ triangle (Proof: Realize that the slope of the line made while the circle is on $AC$ is the same as line $AC$ and that it makes a right angle when the circle switches from being... | // Block 1
draw(circle((4,1),1),black+linewidth(.75));
draw((0,0)--(8,0)--(8,6)--cycle,black+linewidth(.75));
draw((3,1)--(7,1)--(7,4)--cycle,black+linewidth(.75));
draw((3,1)--(3,0),black+linewidth(.75));
draw((3,1)--(2.4,1.8),black+linewidth(.75));
draw((7,1)--(8,1),black+linewidth(.75));
draw((7,1)--(7,0),black+line... | [] |
369 | Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and K... | 1993 AIME Problems/Problem 13 | Solution 1
Consider the unit cicle of radius 50. Assume that they start at points $(-50,100)$ and $(-50,-100).$ Then at time $t$, they end up at points $(-50+t,100)$ and $(-50+3t,-100).$
The equation of the line connecting these points and the equation of the circle are \begin{align}y&=-\frac{100}{t}x+200-\frac{5000}{t... | // Block 1
size(8cm); defaultpen(linewidth(0.7)); pair A,B,C,D,P,Q,O,X; A=(0,0); B=(0,160); C=(200,0); D=(200,53.333); P=(100,0); Q=(123.529,94.118); O=(100,50); X=(300,0); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(O); dot(X); draw(A--B--X--cycle); draw(C--D); draw(P--O--Q); draw(O--X); draw(Circle(O,5... | [] |
369 | Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and K... | 1993 AIME Problems/Problem 13 | Consider our diagram here, where Jenny goes from $A$ to $D$ and Kenny goes from $B$ to $C$. Really what we are asking is the length of $AD$, knowing that $AB$ and $CD$ are tangents and $AB$ is perpendicular to the parallel lines. Draw lines $EF$ and $GH$ as shown such that they are parallel to $BC$ and $AD$ and are tan... | // Block 1
import math; import geometry; import olympiad;
point A,B,C,D,F,G,H,I,M,O,P; A=(0,0); B=(0,200); C=(160,200); D=(160/3,0); F=(0,150); G=(400/3,150); H=(80,50); I=(0,50); M=(0,100); O=(320/3,100); P=(80,150);
draw(I--H--D--A--B--C--G--F); draw(M--O--H); draw(circle((50,100),50)); draw(G--H--P);
label("A",A,SW)... | [] |
370 | Let $\overline{CH}$ be an altitude of $\triangle ABC$. Let $R\,$ and $S\,$ be the points where the circles inscribed in the triangles $ACH\,$ and $BCH^{}_{}$ are tangent to $\overline{CH}$. If $AB = 1995\,$, $AC = 1994\,$, and $BC = 1993\,$, then $RS\,$ can be expressed as $m/n\,$, where $m\,$ and $n\,$ are relatively ... | 1993 AIME Problems/Problem 15 | From the Pythagorean Theorem, $AH^2+CH^2=1994^2$, and $(1995-AH)^2+CH^2=1993^2$.
Subtracting those two equations yields $AH^2-(1995-AH)^2=3987$.
After simplification, we see that $2*1995AH-1995^2=3987$, or $AH=\frac{1995}{2}+\frac{3987}{2*1995}$.
Note that $AH+BH=1995$.
Therefore we have that $BH=\frac{1995}{2}-... | // Block 1
unitsize(48);
pair A,B,C,H;
A=(8,0); B=origin; C=(3,4); H=(3,0); draw(A--B--C--cycle); draw(C--H);
label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,N); label("$H$",H,NE);
draw(circle((2,1),1));
pair [] x=intersectionpoints(C--H,circle((2,1),1));
dot(x[0]); label("$S$",x[0],SW);
draw(circle((4.29843788128,... | [] |
371 | Given a point $P^{}_{}$ on a triangular piece of paper $ABC,\,$ consider the creases that are formed in the paper when $A, B,\,$ and $C\,$ are folded onto $P.\,$ Let us call $P_{}^{}$ a fold point of $\triangle ABC\,$ if these creases, which number three unless $P^{}_{}$ is one of the vertices, do not intersect. Supp... | 1994 AIME Problems/Problem 15 | Let $O_{AB}$ be the intersection of the perpendicular bisectors (in other words, the intersections of the creases) of $\overline{PA}$ and $\overline{PB}$, and so forth. Then $O_{AB}, O_{BC}, O_{CA}$ are, respectively, the circumcenters of $\triangle PAB, PBC, PCA$. According to the problem statement, the circumcenters ... | // Block 1
pair project(pair X, pair Y, real r){return X+r*(Y-X);}
path endptproject(pair X, pair Y, real a, real b){return project(X,Y,a)--project(X,Y,b);}
pathpen = linewidth(1); size(250); pen dots = linetype("2 3") + linewidth(0.7), dashes = linetype("8 6")+linewidth(0.7)+blue, bluedots = linetype("1 4") + linewid... | [] |
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