A baker uses $6\tfrac{2}{3}$ cups of flour when she prepares $\tfrac{5}{3}$ recipes of rolls. She will use $9\tfrac{3}{4}$ cups of flour when she prepares $\tfrac{m}{n}$ recipes of rolls where m and n are relatively prime positive integers. Find $m + n.$

In one of the adjoining figures a square of side $2$ is dissected into four pieces so that $E$ and $F$ are the midpoints of opposite sides and $AG$ is perpendicular to $BF$. These four pieces can then be reassembled into a rectangle as shown in the second figure. The ratio of height to base, $XY$ / $YZ$, in this rectangle is [asy] size(180); defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,1), B=(0,-1), C=(2,-1), D=(2,1), E=(1,-1), F=(1,1), G=(.8,.6); pair X=(4,sqrt(5)), Y=(4,-sqrt(5)), Z=(4+2/sqrt(5),-sqrt(5)), W=(4+2/sqrt(5),sqrt(5)), T=(4,0), U=(4+2/sqrt(5),-4/sqrt(5)), V=(4+2/sqrt(5),1/sqrt(5)); draw(A--B--C--D--A^^B--F^^E--D^^A--G^^rightanglemark(A,G,F)); draw(X--Y--Z--W--X^^T--V--X^^Y--U); label("A", A, NW); label("B", B, SW); label("C", C, SE); label("D", D, NE); label("E", E, S); label("F", F, N); label("G", G, E); label("X", X, NW); label("Y", Y, SW); label("Z", Z, SE); label("W", W, NE); [/asy] $ \textbf{(A)}\ 4\qquad\textbf{(B)}\ 1+2\sqrt{3}\qquad\textbf{(C)}\ 2\sqrt{5}\qquad\textbf{(D)}\ \frac{8+4\sqrt{3}}{3}\qquad\textbf{(E)}\ 5 $

Let $a$ and $b$ be distinct real numbers for which \[\dfrac ab+\dfrac{a+10b}{b+10a}=2.\] Find $\dfrac ab$. $\textbf{(A) }0.6\qquad\textbf{(B) }0.7\qquad\textbf{(C) }0.8\qquad\textbf{(D) }0.9\qquad\textbf{(E) }1$

Let $ ABC$ be an acute triangle, let $ M,N$ be the midpoints of minor arcs $ \widehat{CA},\widehat{AB}$ of the circumcircle of triangle $ ABC,$ point $ D$ is the midpoint of segment $ MN,$ point $ G$ lies on minor arc $ \widehat{BC}.$ Denote by $ I,I_{1},I_{2}$ the incenters of triangle $ ABC,ABG,ACG$ respectively.Let $ P$ be the second intersection of the circumcircle of triangle $ GI_{1}I_{2}$ with the circumcircle of triangle $ ABC.$ Prove that three points $ D,I,P$ are collinear.

A point $P$ is chosen in the interior of $\triangle ABC$ so that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$, the resulting smaller triangles, $t_1$, $t_2$, and $t_3$ in the figure, have areas 4, 9, and 49, respectively. Find the area of $\triangle ABC$. [asy] size(200); pathpen=black+linewidth(0.65);pointpen=black; pair A=(0,0),B=(12,0),C=(4,5); D(A--B--C--cycle); D(A+(B-A)*3/4--A+(C-A)*3/4); D(B+(C-B)*5/6--B+(A-B)*5/6);D(C+(B-C)*5/12--C+(A-C)*5/12); MP("A",C,N);MP("B",A,SW);MP("C",B,SE); /* sorry mixed up points according to resources diagram. */ MP("t_3",(A+B+(B-A)*3/4+(A-B)*5/6)/2+(-1,0.8),N); MP("t_2",(B+C+(B-C)*5/12+(C-B)*5/6)/2+(-0.3,0.1),WSW); MP("t_1",(A+C+(C-A)*3/4+(A-C)*5/12)/2+(0,0.15),ESE);[/asy]

Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a whole number of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the ratio $3: 2: 1$, what is the least possible total for the number of bananas?

Let A be be a point outside the circle C, and AB and AC be the two tangents from A to this circle C. Let L be an arbitrary tangent to C that cuts AB and AC in P and Q. A line through P parallel to AC cuts BC in R. Prove that while L varies, QR passes through a fixed point. :)

Find an upper bound for the ratio $$\frac{x_1x_2+2x_2x_3+x_3x_4}{x_1^2+x_2^2+x_3^2+x_4^2}$$ over all quadruples of real numbers $(x_1,x_2,x_3,x_4)\neq (0,0,0,0)$. [i]Note.[/i] The smaller the bound, the better the solution.

The ratio of the sides of a parallelogram is $\lambda>1$. Given $\lambda$, determine the maximum of the acute angle subtended by the diagonals of the parallelogram.

In a triangle $ABC$, the external bisector of $\angle BAC$ intersects the ray $BC$ at $D$. The feet of the perpendiculars from $B$ and $C$ to line $AD$ are $E$ and $F$, respectively and the foot of the perpendicular from $D$ to $AC$ is $G$. Show that $\angle DGE + \angle DGF = 180^{\circ}$.

An equilateral triangle is originally painted black. Each time the triangle is changed, the middle fourth of each black triangle turns white. After five changes, what fractional part of the original area of the black triangle remains black? [asy] unitsize(36); fill((0,0)--(2,0)--(1,sqrt(3))--cycle,gray); draw((0,0)--(2,0)--(1,sqrt(3))--cycle,linewidth(1)); fill((4,0)--(6,0)--(5,sqrt(3))--cycle,gray); fill((5,0)--(9/2,sqrt(3)/2)--(11/2,sqrt(3)/2)--cycle,white); draw((5,sqrt(3))--(4,0)--(5,0)--(9/2,sqrt(3)/2)--(11/2,sqrt(3)/2)--(5,0)--(6,0)--cycle,linewidth(1)); fill((8,0)--(10,0)--(9,sqrt(3))--cycle,gray); fill((9,0)--(17/2,sqrt(3)/2)--(19/2,sqrt(3)/2)--cycle,white); fill((17/2,0)--(33/4,sqrt(3)/4)--(35/4,sqrt(3)/4)--cycle,white); fill((9,sqrt(3)/2)--(35/4,3*sqrt(3)/4)--(37/4,3*sqrt(3)/4)--cycle,white); fill((19/2,0)--(37/4,sqrt(3)/4)--(39/4,sqrt(3)/4)--cycle,white); draw((9,sqrt(3))--(35/4,3*sqrt(3)/4)--(37/4,3*sqrt(3)/4)--(9,sqrt(3)/2)--(35/4,3*sqrt(3)/4)--(33/4,sqrt(3)/4)--(35/4,sqrt(3)/4)--(17/2,0)--(33/4,sqrt(3)/4)--(8,0)--(9,0)--(17/2,sqrt(3)/2)--(19/2,sqrt(3)/2)--(9,0)--(19/2,0)--(37/4,sqrt(3)/4)--(39/4,sqrt(3)/4)--(19/2,0)--(10,0)--cycle,linewidth(1)); label("Change 1",(3,3*sqrt(3)/4),N); label("$\Longrightarrow $",(3,5*sqrt(3)/8),S); label("Change 2",(7,3*sqrt(3)/4),N); label("$\Longrightarrow $",(7,5*sqrt(3)/8),S); [/asy] $\text{(A)}\ \frac{1}{1024} \qquad \text{(B)}\ \frac{15}{64} \qquad \text{(C)}\ \frac{243}{1024} \qquad \text{(D)}\ \frac{1}{4} \qquad \text{(E)}\ \frac{81}{256}$

Let $x \neq y$ be positive reals satisfying $x^3+2013y=y^3+2013x$, and let $M = \left( \sqrt{3}+1 \right)x + 2y$. Determine the maximum possible value of $M^2$. [i]Proposed by Varun Mohan[/i]

Doug constructs a square window using $8$ equal-size panes of glass, as shown. The ratio of the height to width for each pane is $5:2$, and the borders around and between the panes are $2$ inches wide. In inches, what is the side length o the square window? [asy] fill((0,0)--(25,0)--(25,25)--(0,25)--cycle,grey); for(int i = 0; i < 4; ++i){ for(int j = 0; j < 2; ++j){ fill((6*i+2,11*j+3)--(6*i+5,11*j+3)--(6*i+5,11*j+11)--(6*i+2,11*j+11)--cycle,white); } }[/asy] $\textbf{(A) }26\qquad\textbf{(B) }28\qquad\textbf{(C) }30\qquad\textbf{(D) }32\qquad\textbf{(E) }34$

Let $ABCD$ be a rectangle of centre $O$, such that $\angle DAC=60^{\circ}$. The angle bisector of $\angle DAC$ meets $DC$ at $S$. Lines $OS$ and $AD$ meet at $L$, and lines $BL$ and $AC$ meet at $M$. Prove that lines $SM$ and $CL$ are parallel.

[b]i.)[/b] Let $ABC$ be a triangle with $AB = 12$ and $AC = 16.$ Suppose $M$ is the midpoint of side $BC$ and points $E$ and $F$ are chosen on sides $AC$ and $AB$, respectively, and suppose that lines $EF$ and $AM$ intersect at $G.$ If $AE = 2 \cdot AF$ then find the ratio \[ \frac{EG}{GF} \] [b]ii.)[/b] Let $E$ be a point external to a circle and suppose that two chords $EAB$ and $EDC$ meet at angle of $40^{\circ}.$ If $AB = BC = CD$ find the size of angle $ACD.$

The first four terms of an arithmetic sequence are $a, x, b, 2x$. The ratio of $a$ to $b$ is $ \textbf{(A)}\ \frac{1}{4} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{1}{2} \qquad\textbf{(D)}\ \frac{2}{3} \qquad\textbf{(E)}\ 2 $

If an arc of $ 45^\circ$ on circle $ A$ has the same length as an arc of $ 30^\circ$ on circle $ B$, then the ratio of the area of circle $ A$ to the area of circle $ B$ is $ \textbf{(A)}\ \frac {4}{9} \qquad \textbf{(B)}\ \frac {2}{3} \qquad \textbf{(C)}\ \frac {5}{6} \qquad \textbf{(D)}\ \frac {3}{2} \qquad \textbf{(E)}\ \frac {9}{4}$

The ratio of $w$ to $x$ is $4 : 3$, the ratio of $y$ to $z$ is $3 : 2$, and the ratio of $z$ to $x$ is $1 : 6$. What is the ratio of $w$ to $y$? $\textbf{(A) }4:3 \qquad \textbf{(B) }3:2 \qquad \textbf{(C) } 8:3 \qquad \textbf{(D) } 4:1 \qquad \textbf{(E) } 16:3 $

Let $ABC$ be a triangle with circumcenter $O$ and let $X$, $Y$, $Z$ be the midpoints of arcs $BAC$, $ABC$, $ACB$ on its circumcircle. Let $G$ and $I$ denote the centroid of $\triangle XYZ$ and the incenter of $\triangle ABC$. Given that $AB = 13$, $BC = 14$, $CA = 15$, and $\frac {GO}{GI} = \frac mn$ for relatively prime positive integers $m$ and $n$, compute $100m+n$. [i]Proposed by Evan Chen[/i]

An urn contains $4$ green balls and $6$ blue balls. A second urn contains $16$ green balls and $N$ blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is $0.58$. Find $N$.

Find the ratio between the sum of the areas of the circles and the area of the fourth circle that are shown in the figure Each circle passes through the center of the previous one and they are internally tangent. [img]https://cdn.artofproblemsolving.com/attachments/d/2/29d2be270f7bcf9aee793b0b01c2ef10131e06.jpg[/img]

In isosceles triangle $ABC$ ($AC=BC$), angle bisector $\angle BAC$ and altitude $CD$ from point $C$ intersect at point $O$, such that $CO=3 \cdot OD$. In which ratio does altitude from point $A$ on side $BC$ divide altitude $CD$ of triangle $ABC$

The ratio of the perimeter of an equilateral triangle having an altitude equal to the radius of a circle, to the perimeter of an equilateral triangle inscribed in the circle is: $ \textbf{(A)}\ 1: 2 \qquad\textbf{(B)}\ 1: 3 \qquad\textbf{(C)}\ 1: \sqrt {3} \qquad\textbf{(D)}\ \sqrt {3}: 2 \qquad\textbf{(E)}\ 2: 3$

The admission fee for an exhibition is $ \$25$ per adult and $ \$12$ per child. Last Tuesday, the exhibition collected $ \$1950$ in admission fees from at least one adult and at least one child. Of all the possible ratios of adults to children at the exhibition last Tuesday, which one is closest to $ 1$?

Consider five points in the plane, no three collinear. The convex hull of this points has area $S$. Prove that there exist three points of them that form a triangle with area at most $\frac{5-\sqrt 5}{10}S$