Let $ABC$ be a triangle. Let $D$ be the midpoint of $\overline{BC}$, let $E$ be the midpoint of $\overline{AD}$, and let $F$ be the midpoint of $\overline{BE}$. Let $G$ be the point where the lines $AB$ and $CF$ intersect. What is the value of $\frac{AG}{AB}$?

A frog starts at $0$ on a number line and plays a game. On each turn the frog chooses at random to jump $1$ or $2$ integers to the right or left. It stops moving if it lands on a nonpositive number or a number on which it has already landed. If the expected number of times it will jump is $\tfrac{p}{q}$ for relatively prime positive integers $p$ and $q$, find $p+q$. [i]Proposed by Michael Kural[/i]

A plane $\rho$ and two points $M, N$ outside it are given. Determine the point $A$ on $\rho$ for which $\frac{AM}{AN}$ is minimal.

Let $ABC$ be a triangle such that $\angle BAC = 90^\circ$ and $AB < AC$. We divide the interior of the triangle into the following six regions: \begin{align*} S_1=\text{set of all points }\mathit{P}\text{ inside }\triangle ABC\text{ such that }PA<PB<PC \\ S_2=\text{set of all points }\mathit{P}\text{ inside }\triangle ABC\text{ such that }PA<PC<PB \\ S_3=\text{set of all points }\mathit{P}\text{ inside }\triangle ABC\text{ such that }PB<PA<PC \\ S_4=\text{set of all points }\mathit{P}\text{ inside }\triangle ABC\text{ such that }PB<PC<PA \\ S_5=\text{set of all points }\mathit{P}\text{ inside }\triangle ABC\text{ such that }PC<PA<PB \\ S_6=\text{set of all points }\mathit{P}\text{ inside }\triangle ABC\text{ such that }PC<PB<PA\end{align*} Suppose that the ratio of the area of the largest region to the area of the smallest non-empty region is $49 : 1$. Determine the ratio $AC : AB$.

At Typico High School, $60\%$ of the students like dancing, and the rest dislike it. Of those who like dancing, $80\%$ say that they like it, and the rest say that they dislike it. Of those who dislike dancing, $90\%$ say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it? $\textbf{(A) } 10\%\qquad \textbf{(B) } 12\%\qquad \textbf{(C) } 20\%\qquad \textbf{(D) } 25\%\qquad \textbf{(E) } 33\frac{1}{3}\%$

In this diagram the center of the circle is $O$, the radius is $a$ inches, chord $EF$ is parallel to chord $CD, O, G, H, J$ are collinear, and $G$ is the midpoint of $CD$. Let $K$ (sq. in.) represent the area of trapezoid $CDFE$ and let $R$ (sq. in.) represent the area of rectangle $ELMF$. Then, as $CD$ and $EF$ are translated upward so that $OG$ increases toward the value $a$, while $JH$ always equals $HG$, the ratio $K:R$ become arbitrarily close to: [asy] size((270)); draw((0,0)--(10,0)..(5,5)..(0,0)); draw((5,0)--(5,5)); draw((9,3)--(1,3)--(1,1)--(9,1)--cycle); draw((9.9,1)--(.1,1)); label("O", (5,0), S); label("a", (7.5,0), S); label("G", (5,1), SE); label("J", (5,5), N); label("H", (5,3), NE); label("E", (1,3), NW); label("L", (1,1), S); label("C", (.1,1), W); label("F", (9,3), NE); label("M", (9,1), S); label("D", (9.9,1), E); [/asy] $\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{1}{\sqrt{2}}+\frac{1}{2} \qquad\textbf{(E)}\ \frac{1}{\sqrt{2}}+1$

Prove that in triangle $ABC$, radical center of its excircles lies on line $GI$, which $G$ is Centroid of triangle $ABC$, and $I$ is the incenter.

There are $n\ge3$ integers around a circle. We know that for each of these numbers the ratio between the sum of its two neighbors and the number is a positive integer. Prove that the sum of the $n$ ratios is not greater than $3n$.

On the sides $PQ, QR, RP$ of $\vartriangle PQR$ segments $AB, CD, EF$ are drawn. Given a point $S_0$ inside triangle $\vartriangle PQR$, find the locus of points $S$ for which the sum of the areas of triangles $\vartriangle SAB$, $\vartriangle SCD$ and $\vartriangle SEF$ is equal to the sum of the areas of triangles $\vartriangle S_0AB$, $\vartriangle S_0CD$, $\vartriangle S0EF$. Consider separately the case $$\frac{AB}{PQ }= \frac{CD}{QR} = \frac{EF}{RP}.$$

Consider the triangle $ABC$ and the points $D \in (BC)$ and $M \in (AD)$. Lines $BM$ and $AC$ meet at $E$, lines $CM$ and $AB$ meet at $F$, and lines $EF$ and $AD$ meet at $N$. Prove that $$\frac{AN}{DN}=\frac{1}{2}\cdot \frac{AM}{DM}$$

Show that there exists a convex polyhedron with all its vertices on the surface of a sphere and with all its faces congruent isosceles triangles whose ratio of sides are $\sqrt{3} :\sqrt{3} :2$.

Six-hundred fifty students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti? [asy] size(200); defaultpen(linewidth(0.7)); defaultpen(fontsize(8)); draw(origin--(0,250)); int i; for(i=0; i<6; i=i+1) { draw((0,50*i)--(5,50*i)); } filldraw((25,0)--(75,0)--(75,150)--(25,150)--cycle, gray, black); filldraw((75,0)--(125,0)--(125,100)--(75,100)--cycle, gray, black); filldraw((125,0)--(175,0)--(175,150)--(125,150)--cycle, gray, black); filldraw((225,0)--(175,0)--(175,250)--(225,250)--cycle, gray, black); label("$50$", (0,50), W); label("$100$", (0,100), W); label("$150$", (0,150), W); label("$200$", (0,200), W); label("$250$", (0,250), W); label(rotate(90)*"Lasagna", (50,0), S); label(rotate(90)*"Manicotti", (100,0), S); label(rotate(90)*"Ravioli", (150,0), S); label(rotate(90)*"Spaghetti", (200,0), S); label(rotate(90)*"$\mbox{Number of People}$", (-40,140), W);[/asy] $\textbf{(A)} \: \frac25\qquad \textbf{(B)} \: \frac12\qquad \textbf{(C)} \: \frac54\qquad \textbf{(D)} \: \frac53\qquad \textbf{(E)} \: \frac52$

Triangle $ABC$ is inscribed in circle $\omega$ with $AB = 5$, $BC = 7$, and $AC = 3$. The bisector of angle $A$ meets side $BC$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $DE$. Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$. Then $AF^2 = \frac mn$, where m and n are relatively prime positive integers. Find $m + n$.

The vertices of a triangle have integer coordinates and one of its sides is of length $\sqrt{n}$, where $n$ is a square-free natural number. Prove that the ratio of the circumradius and the inradius is an irrational number.

Suppose $F$ is a polygon with lattice vertices and sides parralell to x-axis and y-axis.Suppose $S(F),P(F)$ are area and perimeter of $F$. Find the smallest k that: $S(F) \leq k.P(F)^2$

Let $L$ and $N$ be the mid-points of the diagonals $[AC]$ and $[BD]$ of the cyclic quadrilateral $ABCD$, respectively. If $BD$ is the bisector of the angle $ANC$, then prove that $AC$ is the bisector of the angle $BLD$.

[u][b]The author of this posting is : Peter VDD[/b][/u] ____________________________________________________________________ most of us didn't really expect to get this, but here it goes (flanders mathematical olympiad 2004, today) triangle with sides 501m, 668m, 835m how many lines can be draws so that the line halves both area and circumference?

The sum of three numbers is $ 98$. The ratio of the first to the second is $ \frac {2}{3}$, and the ratio of the second to the third is $ \frac {5}{8}$. The second number is: $ \textbf{(A)}\ 15 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 32 \qquad\textbf{(E)}\ 33$

Let be given a triangle $ ABC$ and its internal angle bisector $ BD$ $ (D\in BC)$. The line $ BD$ intersects the circumcircle $ \Omega$ of triangle $ ABC$ at $ B$ and $ E$. Circle $ \omega$ with diameter $ DE$ cuts $ \Omega$ again at $ F$. Prove that $ BF$ is the symmedian line of triangle $ ABC$.

Equilateral triangle $DEF$ is inscribed in equilateral triangle $ABC$ such that $\overline{DE} \perp \overline{BC}$. The ratio of the area of $\triangle DEF$ to the area of $\triangle ABC$ is [asy] size(180); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); pair B = (0,0), C = (1,0), A = dir(60), D = C*2/3, E = (2*A+C)/3, F = (2*B+A)/3; D(D("A",A,N)--D("B",B,SW)--D("C",C,SE)--cycle); D(D("D",D)--D("E",E,NE)--D("F",F,NW)--cycle); D(rightanglemark(C,D,E,1.5));[/asy] $\textbf{(A)}\ \dfrac{1}{6}\qquad \textbf{(B)}\ \dfrac{1}{4} \qquad \textbf{(C)}\ \dfrac{1}{3} \qquad \textbf{(D)}\ \dfrac{2}{5} \qquad \textbf{(E)}\ \dfrac{1}{2}$

Three unit circles have a common point $O.$ The other points of (pairwise) intersection are $A, B$ and $C$. Show that the points $A, B$ and $C$ are located on some unit circle.

Let $ABCDEF$ be a convex hexagon of area $1$, whose opposite sides are parallel. The lines $AB$, $CD$ and $EF$ meet in pairs to determine the vertices of a triangle. Similarly, the lines $BC$, $DE$ and $FA$ meet in pairs to determine the vertices of another triangle. Show that the area of at least one of these two triangles is at least $3/2$.

In the adjoining diagram, $BO$ bisects $\angle CBA$, $CO$ bisects $\angle ACB$, and $MN$ is parallel to $BC$. If $AB=12$, $BC=24$, and $AC=18$, then the perimeter of $\triangle AMN$ is [asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, C=(24,0), A=intersectionpoints(Circle(B,12), Circle(C,18))[0], O=incenter(A,B,C), M=intersectionpoint(A--B, O--O+40*dir(180)), N=intersectionpoint(A--C, O--O+40*dir(0)); draw(B--M--O--B--C--O--N--C^^N--A--M); label("$A$", A, dir(90)); label("$B$", B, dir(O--B)); label("$C$", C, dir(O--C)); label("$M$", M, dir(90)*dir(B--A)); label("$N$", N, dir(90)*dir(A--C)); label("$O$", O, dir(90));[/asy] $\textbf {(A) } 30 \qquad \textbf {(B) } 33 \qquad \textbf {(C) } 36 \qquad \textbf {(D) } 39 \qquad \textbf {(E) } 42$

In $ ABC$ we have $ AB \equal{} 25$, $ BC \equal{} 39$, and $ AC \equal{} 42$. Points $ D$ and $ E$ are on $ AB$ and $ AC$ respectively, with $ AD \equal{} 19$ and $ AE \equal{} 14$. What is the ratio of the area of triangle $ ADE$ to the area of quadrilateral $ BCED$? $ \textbf{(A)}\ \frac{266}{1521}\qquad \textbf{(B)}\ \frac{19}{75}\qquad \textbf{(C)}\ \frac{1}{3}\qquad \textbf{(D)}\ \frac{19}{56}\qquad \textbf{(E)}\ 1$

On a circle $k$, we marked four points $(A, B, C, D)$ and drew pairwise their connecting segments. We denoted angles as seen on the diagram. We know that $\alpha_1 : \alpha_2 = 2 : 5$, $\beta_1 : \beta_2 = 7 : 11$, and $\gamma_1 : \gamma_2 = 10 : 3$. If $\delta_1 : \delta_2 = p : q$, where $p$ and $q$ are coprime positive integers, then what is $p$? [img]https://cdn.artofproblemsolving.com/attachments/c/e/b532dd164a7cf99cea7b3b7d98f81796622da5.png[/img]