On the sides $AD$ and $BC$ of a rectangle $ABCD$ select points $M, N$ and $P, Q$ respectively such that $AM = MN = ND = BP = PQ = QC$. On segment $QC$ selected point $X$, different from the ends of the segment. Prove that the perimeter of $\vartriangle ANX$ is more than the perimeter of $\vartriangle MDX$.

A rectangular sheet of paper is folded so that one corner lies on top of the corner diagonally opposite. The resulting shape is a pentagon whose area is $20\%$ one-sheet thick, and $80\%$ two-sheets-thick. Determine the ratio of the two sides of the original sheet of paper.

Circles $A, B,$ and $C$ each have radius 1. Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}$. What is the area inside Circle $C$ but outside circle $A$ and circle $B$ ? [asy] pathpen = linewidth(.7); pointpen = black; pair A=(-1,0), B=-A, C=(0,1); fill(arc(C,1,0,180)--arc(A,1,90,0)--arc(B,1,180,90)--cycle, gray(0.5)); D(CR(D("A",A,SW),1)); D(CR(D("B",B,SE),1)); D(CR(D("C",C,N),1));[/asy] ${ \textbf{(A)}\ 3 - \frac{\pi}{2} \qquad \textbf{(B)}\ \frac{\pi}{2} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \frac{3\pi}{4} \qquad \textbf{(E)}\ 1+\frac{\pi}{2}} $

Mustafa bought a big rug. The seller measured the rug with a ruler that was supposed to measure one meter. As it turned out to be $30$ meters long by $20$ meters wide, he charged Rs $120.000$ Rs. When Mustafa arrived home, he measured the rug again and realized that the seller had overcharged him by $9.408$ Rs. How many centimeters long is the ruler used by the seller?

Let $S$ be the set of ordered pairs $(x, y)$ such that $0<x\le 1$, $0<y\le 1$, and $\left[\log_2{\left(\frac 1x\right)}\right]$ and $\left[\log_5{\left(\frac 1y\right)}\right]$ are both even. Given that the area of the graph of $S$ is $m/n$, where $m$ and $n$ are relatively prime positive integers, find $m+n$. The notation $[z]$ denotes the greatest integer that is less than or equal to $z$.

If $ABCD$ is a cyclic quadrilateral, then prove that the incenters of the triangles $ABC$, $BCD$, $CDA$, $DAB$ are the vertices of a rectangle.

Consider the $2\times3$ rectangle below. We fill in the small squares with the numbers $1,2,3,4,5,6$ (one per square). Define a [i]tasty[/i] filling to be one such that each row is [b]not[/b] in numerical order from left to right and each column is [b]not[/b] in numerical order from top to bottom. If the probability that a randomly selected filling is tasty is $\frac{m}{n}$ for relatively prime positive integers $m$ and $n$, what is $m+n$? \begin{tabular}{|c|c|c|c|} \hline & & \\ \hline & & \\ \hline \end{tabular} [i]2016 CCA Math Bonanza Lightning #4.2[/i]

A rectangle $ABCD$ is given. On the side $AB$, n different points are chosen strictly between $A$ and $B$. Similarly, $m$ different points are chosen on the side $AD$ between $A$ and $D$. Lines are drawn from the points parallel to the sides. How many rectangles are formed in this way? An example of a particular rectangle $ABCD$ is shown with a shaded one rectangle that may be formed in this way. [img]https://cdn.artofproblemsolving.com/attachments/e/4/f7a04300f0c846fb6418d12dc23f5c74b54242.png[/img]

In rectangle $ABCD$, the points $M,N,P, Q$ lie on $AB$, $BC$, $CD$, $DA$ respectively such that the area of triangles $AQM$, $BMN$, $CNP$, $DPQ$ are equal. Prove that the quadrilateral $MNPQ$ is parallelogram. by Mahdi Etesami Fard

Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC=CB=BA$. The area of the "bat wings" is [asy] size(180); defaultpen(fontsize(11pt)); draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("$D$",(0,4),NW); label("$C$",(1,4),N); label("$B$",(2,4),N); label("$A$",(3,4),NE); label("$E$",(0,0),SW); label("$F$",(3,0),SE);[/asy] $\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5$

For a regular $n$-gon, let $M$ be the set of the lengths of the segments joining its vertices. Show that the sum of the squares of the elements of $M$ is greater than twice the area of the polygon.

Rectangle $ABCD$ measures $70$ by $40$. Eighteen points (including $A$ and $C$) are marked on the diagonal $AC$ dividing the diagonal into $17$ congruent pieces. Twenty-two points (including A and B) are marked on the side $AB$ dividing the side into $21$ congruent pieces. Seventeen non-overlapping triangles are constructed as shown. Each triangle has two vertices that are two of these adjacent marked points on the side of the rectangle, and one vertex that is one of the marked points along the diagonal of the rectangle. Only the left $17$ of the $21$ congruent pieces along the side of the rectangle are used as bases of these triangles. Find the sum of the areas of these $17$ triangles. [asy] size(200); defaultpen(linewidth(0.8)); pair A=origin,B=(21,0),C=(21,12),D=(0,12); path P=origin; draw(A--B--C--D--cycle--C); for (int r = 1; r <= 17;++r) { P=P--(21*r/17,12*r/17)--(r,0); } P=P--cycle; filldraw(P,gray(0.7)); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); [/asy]

Given two concentric circles and a pair of parallel lines. Find the locus of the fourth vertices of all rectangles with three vertices on the concentric circles, two vertices on one circle and the third on the other and with sides parallel to the given lines.

Each of the following four large congruent squares is subdivided into combinations of congruent triangles or rectangles and is partially [b]bolded[/b]. What percent of the total area is partially bolded? [asy] import graph; size(7.01cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.42,xmax=14.59,ymin=-10.08,ymax=5.26; pair A=(0,0), B=(4,0), C=(0,4), D=(4,4), F=(2,0), G=(3,0), H=(1,4), I=(2,4), J=(3,4), K=(0,-2), L=(4,-2), M=(0,-6), O=(0,-4), P=(4,-4), Q=(2,-2), R=(2,-6), T=(6,4), U=(10,0), V=(10,4), Z=(10,2), A_1=(8,4), B_1=(8,0), C_1=(6,-2), D_1=(10,-2), E_1=(6,-6), F_1=(10,-6), G_1=(6,-4), H_1=(10,-4), I_1=(8,-2), J_1=(8,-6), K_1=(8,-4); draw(C--H--(1,0)--A--cycle,linewidth(1.6)); draw(M--O--Q--R--cycle,linewidth(1.6)); draw(A_1--V--Z--cycle,linewidth(1.6)); draw(G_1--K_1--J_1--E_1--cycle,linewidth(1.6)); draw(C--D); draw(D--B); draw(B--A); draw(A--C); draw(H--(1,0)); draw(I--F); draw(J--G); draw(C--H,linewidth(1.6)); draw(H--(1,0),linewidth(1.6)); draw((1,0)--A,linewidth(1.6)); draw(A--C,linewidth(1.6)); draw(K--L); draw((4,-6)--L); draw((4,-6)--M); draw(M--K); draw(O--P); draw(Q--R); draw(O--Q); draw(M--O,linewidth(1.6)); draw(O--Q,linewidth(1.6)); draw(Q--R,linewidth(1.6)); draw(R--M,linewidth(1.6)); draw(T--V); draw(V--U); draw(U--(6,0)); draw((6,0)--T); draw((6,2)--Z); draw(A_1--B_1); draw(A_1--Z); draw(A_1--V,linewidth(1.6)); draw(V--Z,linewidth(1.6)); draw(Z--A_1,linewidth(1.6)); draw(C_1--D_1); draw(D_1--F_1); draw(F_1--E_1); draw(E_1--C_1); draw(G_1--H_1); draw(I_1--J_1); draw(G_1--K_1,linewidth(1.6)); draw(K_1--J_1,linewidth(1.6)); draw(J_1--E_1,linewidth(1.6)); draw(E_1--G_1,linewidth(1.6)); dot(A,linewidth(1pt)+ds); dot(B,linewidth(1pt)+ds); dot(C,linewidth(1pt)+ds); dot(D,linewidth(1pt)+ds); dot((1,0),linewidth(1pt)+ds); dot(F,linewidth(1pt)+ds); dot(G,linewidth(1pt)+ds); dot(H,linewidth(1pt)+ds); dot(I,linewidth(1pt)+ds); dot(J,linewidth(1pt)+ds); dot(K,linewidth(1pt)+ds); dot(L,linewidth(1pt)+ds); dot(M,linewidth(1pt)+ds); dot((4,-6),linewidth(1pt)+ds); dot(O,linewidth(1pt)+ds); dot(P,linewidth(1pt)+ds); dot(Q,linewidth(1pt)+ds); dot(R,linewidth(1pt)+ds); dot((6,0),linewidth(1pt)+ds); dot(T,linewidth(1pt)+ds); dot(U,linewidth(1pt)+ds); dot(V,linewidth(1pt)+ds); dot((6,2),linewidth(1pt)+ds); dot(Z,linewidth(1pt)+ds); dot(A_1,linewidth(1pt)+ds); dot(B_1,linewidth(1pt)+ds); dot(C_1,linewidth(1pt)+ds); dot(D_1,linewidth(1pt)+ds); dot(E_1,linewidth(1pt)+ds); dot(F_1,linewidth(1pt)+ds); dot(G_1,linewidth(1pt)+ds); dot(H_1,linewidth(1pt)+ds); dot(I_1,linewidth(1pt)+ds); dot(J_1,linewidth(1pt)+ds); dot(K_1,linewidth(1pt)+ds); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy] $ \textbf{(A)}12\frac 12\qquad\textbf{(B)}20\qquad\textbf{(C)}25\qquad\textbf{(D)}33 \frac 13\qquad\textbf{(E)}37\frac 12 $

Suppose a square piece of paper is folded in half vertically. The folded paper is then cut in half along the dashed line. Three rectangles are formed-a large one and two small ones. What is the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle? $\text{(A)}\ \frac{1}{2} \qquad \text{(B)}\ \frac{2}{3} \qquad \text{(C)}\ \frac{3}{4} \qquad \text{(D)}\ \frac{4}{5} \qquad \text{(E)}\ \frac{5}{6}$ [asy] draw((0,0)--(0,8)--(6,8)--(6,0)--cycle); draw((0,8)--(5,9)--(5,8)); draw((3,-1.5)--(3,10.3),dashed); draw((0,5.5)..(-.75,4.75)..(0,4)); draw((0,4)--(1.5,4),EndArrow); [/asy]

Chim Tu has a large rectangular table. On it, there are finitely many pieces of paper with nonoverlapping interiors, each one in the shape of a convex polygon. At each step, Chim Tu is allowed to slide one piece of paper in a straight line such that its interior does not touch any other piece of paper during the slide. Can Chim Tu always slide all the pieces of paper off the table in finitely many steps?