Found problems: 85335
A board that has some of its squares painted black is called [i]acceptable [/i] if there are no four black squares that form a $2 \times 2$ subboard. Find the largest real number $\lambda$ such that for every positive integer $n$ the following proposition holds: mercy: if an $n \times n$ board is acceptable and has fewer than $\lambda n^2$ dark squares, then an additional square black can be painted so that the board is still acceptable.
Two circles of radius 5 are externally tangent to each other and are internally tangent to a circle of radius 13 at points $A$ and $B$, as shown in the diagram. The distance $AB$ can be written in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?
[asy]
draw(circle((0,0),13));
draw(circle((5,-6.2),5));
draw(circle((-5,-6.2),5));
label("$B$", (9.5,-9.5), S);
label("$A$", (-9.5,-9.5), S);
[/asy]
$\textbf{(A) } 21 \qquad \textbf{(B) } 29 \qquad \textbf{(C) } 58 \qquad \textbf{(D) } 69 \qquad \textbf{(E) } 93 $
For any polynomial $P(x)=a_0+a_1x+\ldots+a_kx^k$ with integer coefficients, the number of odd coefficients is denoted by $o(P)$. For $i-0,1,2,\ldots$ let $Q_i(x)=(1+x)^i$. Prove that if $i_1,i_2,\ldots,i_n$ are integers satisfying $0\le i_1<i_2<\ldots<i_n$, then: \[ o(Q_{i_1}+Q_{i_2}+\ldots+Q_{i_n})\ge o(Q_{i_1}). \]
A regular $15$-gon has $L$ lines of symmetry, and the smallest positive angle for which it has rotational symmetry is $R$ degrees. What is $L+R$?
$\textbf{(A) }24\qquad\textbf{(B) }27\qquad\textbf{(C) }32\qquad\textbf{(D) }39\qquad\textbf{(E) }54$
Rectangle $ABCD$ has side lengths $AB=84$ and $AD=42$. Point $M$ is the midpoint of $\overline{AD}$, point $N$ is the trisection point of $\overline{AB}$ closer to $A$, and point $O$ is the intersection of $\overline{CM}$ and $\overline{DN}$. Point $P$ lies on the quadrilateral $BCON$, and $\overline{BP}$ bisects the area of $BCON$. Find the area of $\triangle{CDP}$.
For any positive integer $k$ denote by $S(k)$ the number of solutions $(x,y)\in \mathbb{Z}_+ \times \mathbb{Z}_+$ of the system
$$\begin{cases} \left\lceil\frac{x\cdot d}{y}\right\rceil\cdot \frac{x}{d}=\left\lceil\left(\sqrt{y}+1\right)^2\right\rceil \\ \mid x-y\mid =k , \end{cases}$$
where $d$ is the greatest common divisor of positive integers $x$ and $y.$ Determine $S(k)$ as a function of $k$. (Here $\lceil z\rceil$ denotes the smalles integer number which is bigger or equal than $z.$)
Let $ABCD$ be a cyclic quadrilateral. Let $E$ and $F$ be variable points on the sides $AB$ and $CD$, respectively, such that $AE:EB=CF:FD$. Let $P$ be the point on the segment $EF$ such that $PE:PF=AB:CD$. Prove that the ratio between the areas of triangles $APD$ and $BPC$ does not depend on the choice of $E$ and $F$.
A rectangular box with side lengths $1$, $2$, and $16$ is cut into two congruent smaller boxes with integer side lengths. Compute the square of the largest possible length of the space diagonal of one of the smaller boxes.
[i]2020 CCA Math Bonanza Lightning Round #2.2[/i]
Let $d_i(k)$ the number of divisors of $k$ greater than $i$.
Let $f(n)=\sum_{i=1}^{\lfloor \frac{n^2}{2} \rfloor}d_i(n^2-i)-2\sum_{i=1}^{\lfloor \frac{n}{2} \rfloor}d_i(n-i)$.
Find all $n \in N$ such that $f(n)$ is a perfect square.
Let $C_{1}$ and $C_{2}$ be concentric circles, with $C_{2}$ in the interior of $C_{1}$. From a point $A$ on $C_{1}$, draw the tangent $AB$ to $C_{2}$ $(B \in C_{2})$. Let $C$ be the second point of intersection of $AB$ and $C_{1}$,and let $D$ be the midpoint of $AB$. A line passing through $A$ intersects $C_{2}$ at $E$ and $F$ in such a way that the perpendicular bisectors of $DE$ and $CF$ intersect at a point $M$ on $AB$. Find, with proof, the ratio $AM/MC$.
This question is taken from Mathematical Olympiad Challenges , the 9-th exercise in 1.3 Power of a Point.
Let $n$ and $k$ be positive integers such that $1 \leq n \leq N+1$, $1 \leq k \leq N+1$. Show that: \[ \min_{n \neq k} |\sin n - \sin k| < \frac{2}{N}. \]
Let $ABCD$ be a parallelogram and $P$ be an arbitrary point in the plane. Let $O$ be the intersection of two diagonals $AC$ and $BD.$ The circumcircles of triangles $POB$ and $POC$ intersect the circumcircles of triangle $OAD$ at $Q$ and $R,$ respectively $(Q,R \ne O).$ Construct the parallelograms $PQAM$ and $PRDN.$
Prove that: the circumcircle of triangle $MNP$ passes through $O.$
[i]Proposed by Tran Quang Hung ([url=https://artofproblemsolving.com/community/user/68918]buratinogigle[/url])[/i]
Let $\vartriangle ABC$ be a triangle. Let $D$ be the point on $BC$ such that $DA$ is tangent to the circumcircle of $ABC$. Let $E$ be the point on the circumcircle of $ABC$ such that $DE$ is tangent to the circumcircle of $ABC$, but $E \ne A$. Let $F$ be the intersection of $AE$ and $BC$. Given that $BF/F C = 4/5$, find the maximum possible value for $\sin \angle ACB$/
The two wheels shown below are spun and the two resulting numbers are added. The probability that the sum is even is
[asy]
draw(circle((0,0),3));
draw(circle((7,0),3));
draw((0,0)--(3,0));
draw((0,-3)--(0,3));
draw((7,3)--(7,0)--(7+3*sqrt(3)/2,-3/2));
draw((7,0)--(7-3*sqrt(3)/2,-3/2));
draw((0,5)--(0,3.5)--(-0.5,4));
draw((0,3.5)--(0.5,4));
draw((7,5)--(7,3.5)--(6.5,4));
draw((7,3.5)--(7.5,4));
label("$3$",(-0.75,0),W);
label("$1$",(0.75,0.75),NE);
label("$2$",(0.75,-0.75),SE);
label("$6$",(6,0.5),NNW);
label("$5$",(7,-1),S);
label("$4$",(8,0.5),NNE);
[/asy]
$\text{(A)}\ \dfrac{1}{6} \qquad \text{(B)}\ \dfrac{1}{4} \qquad \text{(C)}\ \dfrac{1}{3} \qquad \text{(D)}\ \dfrac{5}{12} \qquad \text{(E)}\ \dfrac{4}{9}$
Points $M$ and $N$ are chosen on sides $AB$ and $BC$,respectively, in a triangle $ABC$, such that point $O$ is interserction of lines $CM$ and $AN$. Given that $AM+AN=CM+CN$. Prove that $AO+AB=CO+CB$.
A triangle $ABC$ has side lengths $AB=8$ and $BC=10.$ Given that the altitude to side $BC$ has length $4,$ what is the length of the altitude to side $AB?$
A regular hexagon $ABCDEF$ has perimeter $12$. $AB$, $CD$, and $EF$ are all extended, and the intersections of the line segments form an equilateral triangle. Compute the perimeter of the triangle.
In the figure, the chord $[CD]$ is perpendicular to the diameter $[AB]$ and intersects it at $H$. Length of $AB$ is a two-digit natural number. Changing the order of these two digits gives length of $CD$. Knowing that distance from $H$ to the center $O$ is a positive rational number, calculate $AB$.
[img]https://cdn.artofproblemsolving.com/attachments/5/f/eb9c61579a38118b4f753bbc19a9a50e0732dc.png[/img]
There are $6$ different bus lines in a city, each stopping at exactly $5$ stations and running in both directions. Nevertheless, for every two different stations there is always a bus line connecting these two stations. Determine the maximum number of stations in this city.
[i](Karl Czakler)[/i]
Consider two circles of radius one, and let $O$ and $O'$ denote their centers. Point $M$ is selected on either circle. If $OO' = 2014$, what is the largest possible area of triangle $OMO'$?
[i]Proposed by Evan Chen[/i]
Given is a triangle $ABC$.On the extensions of the side $AB$ we consider points $A_1,B_1$ such that $AB_1=BA_1$ (with $A_1$ lying closer to $B$).On the extensions of the side $BC$ we consider points $B_4,C_4$ such that $CB_4=BC_4$ (with $B_4$ lying closer to $C$).On the extensions of the side $AC$ we consider points $C_1,A_4$ such that $AC_1=CA_4$ (with $C_1$ lying closer to $A$).On the segment $A_1A_4$ we consider points $A_2,A_3$ such that $A_1A_2=A_3A_4=mA_1A_4$ where $0<m<\frac{1}{2}$.Points $B_2,B_3$ and $C_2,C_3$ are defined similarly,on the segments $B_1B_4,C_1C_4$ respectively.If $D\equiv BB_2\cap CC_2 \ , \ E\equiv AA_3\cap CC_2 \ , \ F\equiv AA_3\cap BB_3$, $\ G\equiv BB_3\cap CC_3 \ , \ H\equiv AA_2\cap CC_3$ and $I\equiv AA_2\cap BB_2$,prove that the diagonals $DG,EH,FI$ of the hexagon $DEFGHI$ are concurrent.
[hide=Diagram][asy]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; /* point style */
real xmin = -7.984603447540051, xmax = 21.28710511372557, ymin = -6.555010307713199, ymax = 10.006614273002825; /* image dimensions */
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draw((1.1583842866003107,4.638449718549554)--(0.,0.)--(7.,0.)--cycle, aqaqaq);
/* draw figures */
draw((1.1583842866003107,4.638449718549554)--(0.,0.), uququq);
draw((0.,0.)--(7.,0.), uququq);
draw((7.,0.)--(1.1583842866003107,4.638449718549554), uququq);
draw((1.1583842866003107,4.638449718549554)--(1.623345080409327,6.500264738079558));
draw((0.,0.)--(-0.46496079380901606,-1.8618150195300045));
draw((-3.0803965232149757,0.)--(0.,0.));
draw((7.,0.)--(10.080396523214976,0.));
draw((1.1583842866003107,4.638449718549554)--(0.007284204967787214,5.552463941947242));
draw((7.,0.)--(8.151100081632526,-0.9140142233976905));
draw((-0.46496079380901606,-1.8618150195300045)--(8.151100081632526,-0.9140142233976905));
draw((-3.0803965232149757,0.)--(0.007284204967787214,5.552463941947242));
draw((10.080396523214976,0.)--(1.623345080409327,6.500264738079558));
draw((0.,0.)--(3.7376079411107392,4.8751985535596685));
draw((-0.7646359770779035,4.164347956460432)--(7.,0.));
draw((1.1583842866003107,4.638449718549554)--(5.997084862772141,-1.150964422430769));
draw((0.,0.)--(7.966133662513563,1.6250661845198895));
draw((-2.308476341169285,1.3881159854868106)--(7.,0.));
draw((1.1583842866003107,4.638449718549554)--(1.6890544250513695,-1.624864820496926));
draw((2.0395968109217,2.660375186246903)--(2.9561195753832448,0.6030390855677443), linetype("2 2"));
draw((3.4388364046369224,1.909931693481981)--(1.4816619768719694,0.8229159040072803), linetype("2 2"));
draw((1.3969966570225139,1.8221911417546572)--(4.301698851378541,0.8775330211014288), linetype("2 2"));
/* dots and labels */
dot((1.1583842866003107,4.638449718549554),linewidth(3.pt) + dotstyle);
label("$A$", (0.6263408942608304,4.2), NE * labelscalefactor);
dot((0.,0.),linewidth(3.pt) + dotstyle);
label("$B$", (-0.44658827292841696,0.04763072114368767), NE * labelscalefactor);
dot((7.,0.),linewidth(3.pt) + dotstyle);
label("$C$", (7.008893888822507,0.18518574257820614), NE * labelscalefactor);
dot((1.623345080409327,6.500264738079558),linewidth(3.pt) + dotstyle);
label("$B_1$", (1.7267810657369815,6.6777827542874775), NE * labelscalefactor);
dot((-0.46496079380901606,-1.8618150195300045),linewidth(3.pt) + dotstyle);
label("$A_1$", (-1.1068523758141076,-1.6305405403574376), NE * labelscalefactor);
dot((10.080396523214976,0.),linewidth(3.pt) + dotstyle);
label("$B_4$", (10.062615364668826,-0.612633381742001), NE * labelscalefactor);
dot((-3.0803965232149757,0.),linewidth(3.pt) + dotstyle);
label("$C_4$", (-3.3077327187664096,-0.612633381742001), NE * labelscalefactor);
dot((0.007284204967787214,5.552463941947242),linewidth(3.pt) + dotstyle);
label("$C_1$", (0.1036318128096586,5.714897604245849), NE * labelscalefactor);
dot((8.151100081632526,-0.9140142233976905),linewidth(3.pt) + dotstyle);
label("$A_4$", (8.521999124602214,-1.1903644717669786), NE * labelscalefactor);
dot((-2.308476341169285,1.3881159854868106),linewidth(3.pt) + dotstyle);
label("$C_3$", (-2.9776006673235647,1.7808239912186203), NE * labelscalefactor);
dot((-0.7646359770779035,4.164347956460432),linewidth(3.pt) + dotstyle);
label("$C_2$", (-1.1618743843879151,4.504413415622086), NE * labelscalefactor);
dot((1.6890544250513695,-1.624864820496926),linewidth(3.pt) + dotstyle);
label("$A_2$", (1.6167370485893664,-2.125738617521704), NE * labelscalefactor);
dot((5.997084862772141,-1.150964422430769),linewidth(3.pt) + dotstyle);
label("$A_3$", (6.211074764502297,-1.603029536070534), NE * labelscalefactor);
dot((7.966133662513563,1.6250661845198895),linewidth(3.pt) + dotstyle);
label("$B_3$", (8.081823056011753,1.7808239912186203), NE * labelscalefactor);
dot((3.7376079411107392,4.8751985535596685),linewidth(3.pt) + dotstyle);
label("$B_2$", (3.8451283958285725,5.027122497073257), NE * labelscalefactor);
dot((2.0395968109217,2.660375186246903),linewidth(3.pt) + dotstyle);
label("$D$", (1.7542920700238853,2.991308179842383), NE * labelscalefactor);
dot((3.4388364046369224,1.909931693481981),linewidth(3.pt) + dotstyle);
label("$E$", (3.542507348672631,2.083445038374561), NE * labelscalefactor);
dot((4.301698851378541,0.8775330211014288),linewidth(3.pt) + dotstyle);
label("$F$", (4.22,0.93), NE * labelscalefactor);
dot((2.9561195753832448,0.6030390855677443),linewidth(3.pt) + dotstyle);
label("$G$", (2.909754250073844,0.10265272971749505), NE * labelscalefactor);
dot((1.4816619768719694,0.8229159040072803),linewidth(3.pt) + dotstyle);
label("$H$", (0.9839839499905795,0.43278478116033936), NE * labelscalefactor);
dot((1.3969966570225139,1.8221911417546572),linewidth(3.pt) + dotstyle);
label("$I$", (0.9839839499905795,1.8908680083662353), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy][/hide]
Let $a_1, a_2, \ldots$ be a strictly increasing sequence of positive integers, such that for any positive integer $n$, $a_n$ is not representable in the for $\sum_{i=1}^{n-1}c_ia_i$ for $c_i \in \{0, 1\}$. For every positive integer $m$, let $f(m)$ denote the number of $a_i$ that are at most $m$. Show that for any positive integers $m, k$, we have that $$f(m) \leq a_k+\frac{m} {k+1}.$$
In triangle $ABC$ ($AB\ne AC$), where all angles are greater than $45^\circ$, the altitude $AD$ is drawn. Let $\omega_1$ and $\omega_2$ be-- circles with diameters $AC$ and $AB$, respectively. The angle bisector of $\angle ADB$ secondarily intersects $\omega_1$ at point $P$, and the angle bisector of $\angle ADC$ secondarily intersects $\omega_2$ at point $Q$. The line $AP$ intersects $\omega_2$ at the point $R$. Prove that the circumcenter of triangle $PQR$ lies on line $BC$.
A circle that passes through the vertex $A$ of a rectangle $ABCD$ intersects the side $AB$ at a second point $E$ different from $B.$ A line passing through $B$ is tangent to this circle at a point $T,$ and the circle with center $B$ and passing through $T$ intersects the side $BC$ at the point $F.$ Show that if $\angle CDF= \angle BFE,$ then $\angle EDF=\angle CDF.$
If $a$ and $b$ are arbitrary positive real numbers and $m$ an integer, prove that
\[\Bigr( 1+\frac ab \Bigl)^m +\Bigr( 1+\frac ba \Bigl)^m \geq 2^{m+1}.\]