A geometric series is one where the ratio between each two consecutive terms is constant (ex. 3,6,12,24,...). The fifth term of a geometric series is 5!, and the sixth term is 6!. What is the fourth term?

A student performed an experiment to determine the ratio of $\text{H}_2\text{O}$ to $\text{CuSO}_4$ in a sample of hydrated copper(II) sulfate by heating it to drive off the water and weighing the solid before and after heating. The formula obtained experimentally was $\text{CuSO}_4 \bullet 5.5\text{H}_2\text{O}$ but the accepted formula is $\text{CuSO}_4 \bullet 5 \text{H}_2\text{O}$. Which error best accounts for the difference in results? $ \textbf{(A)}\ \text{During heating some of the hydrated copper(II) sulfate was lost} \qquad$ $\textbf{(B)}\ \text{The hydrated sample was not heated long enough to drive off all the water}\qquad$ $\textbf{(C)}\ \text{The student weighed out too much sample initially.} \qquad$ $\textbf{(D)}\ \text{The student used a balance that gave weights that were consistently too high by 0.10 g }\qquad$

In the figure shown below, $ABCDE$ is a regular pentagon and $AG=1$. What is $FG+JH+CD$? [asy] import cse5;pathpen=black;pointpen=black; size(2inch); pair A=dir(90), B=dir(18), C=dir(306), D=dir(234), E=dir(162); D(MP("A",A,A)--MP("B",B,B)--MP("C",C,C)--MP("D",D,D)--MP("E",E,E)--cycle,linewidth(1.5)); D(A--C--E--B--D--cycle); pair F=IP(A--D,B--E), G=IP(B--E,C--A), H=IP(C--A,B--D), I=IP(D--B,E--C), J=IP(C--E,D--A); D(MP("F",F,dir(126))--MP("I",I,dir(270))--MP("G",G,dir(54))--MP("J",J,dir(198))--MP("H",H,dir(342))--cycle); [/asy] $\textbf{(A) } 3 \qquad\textbf{(B) } 12-4\sqrt5 \qquad\textbf{(C) } \dfrac{5+2\sqrt5}{3} \qquad\textbf{(D) } 1+\sqrt5 \qquad\textbf{(E) } \dfrac{11+11\sqrt5}{10} $

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 $3/2$.

Given an acute triangle $ABC$. Let $A'$ be a point symmetric to $A$ with respect to $BC, O_A$ is the center of the circle passing through $A$ and the midpoints of the segments $A'B$ and $A'C. O_B$ and $O_C$ points are defined similarly. Find the ratio of the radii of the circles circumscribed around the triangles $ABC$ and $O_AO_BO_C$.

Two triangles are said to be [i]twins [/i] if one of them is an image of the other one under a parallel projection. Prove that two triangles are twins if and only if either at least a side of one of them equals a side of another or both the triangles have equal segments that connect the corresponding vertices with some points on the opposite sides which divide these sides in the same ratio. (E. Barabanov)

In rectangle $ ABCD$, $ AB\equal{}5$ and $ BC\equal{}3$. Points $ F$ and $ G$ are on $ \overline{CD}$ so that $ DF\equal{}1$ and $ GC\equal{}2$. Lines $ AF$ and $ BG$ intersect at $ E$. Find the area of $ \triangle{AEB}$. [asy]unitsize(6mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair A=(0,0), B=(5,0), C=(5,3), D=(0,3), F=(1,3), G=(3,3); pair E=extension(A,F,B,G); draw(A--B--C--D--A--E--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",E,N); label("$F$",F,SE); label("$G$",G,SW); label("$B$",B,SE); label("1",midpoint(D--F),N); label("2",midpoint(G--C),N); label("3",midpoint(B--C),E); label("3",midpoint(A--D),W); label("5",midpoint(A--B),S);[/asy]$ \textbf{(A)}\ 10 \qquad \textbf{(B)}\ \frac{21}{2} \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ \frac{25}{2} \qquad \textbf{(E)}\ 15$

In rectangle $ABCD$, $DC = 2CB$ and points $E$ and $F$ lie on $\overline{AB}$ so that $\overline{ED}$ and $\overline{FD}$ trisect $\angle ADC$ as shown. What is the ratio of the area of $\triangle DEF$ to the area of rectangle $ABCD$? [asy] draw((0, 0)--(0, 1)--(2, 1)--(2, 0)--cycle); draw((0, 0)--(sqrt(3)/3, 1)); draw((0, 0)--(sqrt(3), 1)); label("A", (0, 1), N); label("B", (2, 1), N); label("C", (2, 0), S); label("D", (0, 0), S); label("E", (sqrt(3)/3, 1), N); label("F", (sqrt(3), 1), N); [/asy] ${ \textbf{(A)}\ \ \frac{\sqrt{3}}{6}\qquad\textbf{(B)}\ \frac{\sqrt{6}}{8}\qquad\textbf{(C)}\ \frac{3\sqrt{3}}{16}\qquad\textbf{(D)}}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$

Triangle $ABC$ is given. The points $D, E$, and $F$ are located on the sides $BC, CA$ and $AB$ respectively so that the lines $AD, BE$ and $CF$ intersect at point $O$. Prove that $\frac{AO}{AD} + \frac{BO}{BE} + \frac{CO}{ CF}=2$

Let $a,b,c$ be the sides of triangle $ABC$ and let $\alpha,\beta,\gamma$ be the corresponding angles. (a) If $\alpha=3\beta$, prove that $\left(a^2-b^2\right)(a-b)=bc^2$. (b) Is the converse true?

Let $ABCD$ be a cyclic quadrilateral such that $AC =56, BD = 65, BC>DA$ and $AB: BC =CD: DA$. Find the ratio of areas $S (ABC): S (ADC)$.

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 $\sqrt{3} :\sqrt{3} :2$.

As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$. [asy] size(200); pair A=origin, B=(14,0), C=(9,12), D=foot(A, B,C), E=foot(B, A, C), F=foot(C, A, B), H=orthocenter(A, B, C); draw(F--C--A--B--C^^A--D^^B--E); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("84", centroid(H, C, E), fontsize(9.5)); label("35", centroid(H, B, D), fontsize(9.5)); label("30", centroid(H, F, B), fontsize(9.5)); label("40", centroid(H, A, F), fontsize(9.5));[/asy]

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]

Inscribed in a triangle $ABC$ with the center of the circle $I$ touch the sides $AB$ and $AC$ at points $C_{1}$ and $B_{1}$, respectively. The point $M$ divides the segment $C_{1}B_{1}$ in a 3:1 ratio, measured from $C_{1}$. $N$ - the midpoint of $AC$. Prove that the points $I, M, B_{1}, N$ lie on a circle, if you know that $AC = 3 (BC-AB)$.

Each side of the large square in the figure is trisected (divided into three equal parts). The corners of an inscribed square are at these trisection points, as shown. The ratio of the area of the inscribed square to the area of the large square is [asy] draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((1,0)--(1,0.2)); draw((2,0)--(2,0.2)); draw((3,1)--(2.8,1)); draw((3,2)--(2.8,2)); draw((1,3)--(1,2.8)); draw((2,3)--(2,2.8)); draw((0,1)--(0.2,1)); draw((0,2)--(0.2,2)); draw((2,0)--(3,2)--(1,3)--(0,1)--cycle); [/asy] $\textbf{(A)}\ \dfrac{\sqrt{3}}{3} \qquad \textbf{(B)}\ \dfrac{5}{9} \qquad \textbf{(C)}\ \dfrac{2}{3} \qquad \textbf{(D)}\ \dfrac{\sqrt{5}}{3} \qquad \textbf{(E)}\ \dfrac{7}{9}$

Point $D$ lies on side $AB$ of triangle $ABC$, and $$\frac{AD}{DC} = \frac{AB}{BC}.$$ Prove that angle $C$ is obtuse. (Sergey Berlov)

Let $D$ be a point different from the vertices on the side $BC$ of a triangle $ABC.$ Let $I, \: I_1$ and $I_2$ be the incenters of the triangles $ABC, \: ABD$ and $ADC,$ respectively. Let $E$ be the second intersection point of the circumcircles of the triangles $AI_1I$ and $ADI_2,$ and $F$ be the second intersection point of the circumcircles of the triangles $AII_2$ and $AI_1D.$ Prove that if $AI_1=AI_2,$ then \[ \frac{EI}{FI} \cdot \frac{ED}{FD}=\frac{{EI_1}^2}{{FI_1}^2}.\]

An infinite geometric series has sum $2005$. A new series, obtained by squaring each term of the original series, has $10$ times the sum of the original series. The common ratio of the original series is $\frac{m}{n}$ where $m$ and $n$ are relatively prime integers. Find $m+n$.

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$.

Triangle $ ABC$ has $ AC \equal{} 450$ and $ BC \equal{} 300$. Points $ K$ and $ L$ are located on $ \overline{AC}$ and $ \overline{AB}$ respectively so that $ AK \equal{} CK$, and $ \overline{CL}$ is the angle bisector of angle $ C$. Let $ P$ be the point of intersection of $ \overline{BK}$ and $ \overline{CL}$, and let $ M$ be the point on line $ BK$ for which $ K$ is the midpoint of $ \overline{PM}$. If $ AM \equal{} 180$, find $ LP$.

Let points $ A = (0,0) , \ B = (1,2), \ C = (3,3), $ and $ D = (4,0) $. Quadrilateral $ ABCD $ is cut into equal area pieces by a line passing through $ A $. This line intersects $ \overline{CD} $ at point $ \left (\frac{p}{q}, \frac{r}{s} \right ) $, where these fractions are in lowest terms. What is $ p + q + r + s $? $ \textbf{(A)} \ 54 \qquad \textbf{(B)} \ 58 \qquad \textbf{(C)} \ 62 \qquad \textbf{(D)} \ 70 \qquad \textbf{(E)} \ 75 $

A regular octahedron is formed by joining the centers of adjoining faces of a cube. The ratio of the volume of the octahedron to the volume of the cube is $ \textbf{(A)}\ \frac{\sqrt{3}}{12} \qquad\textbf{(B)}\ \frac{\sqrt{6}}{16} \qquad\textbf{(C)}\ \frac{1}{6} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{8} \qquad\textbf{(E)}\ \frac{1}{4} $

The vertices of a regular $2012$-gon are labeled $A_1,A_2,\ldots, A_{2012}$ in some order. It is known that if $k+\ell$ and $m+n$ leave the same remainder when divided by $2012$, then the chords $A_kA_{\ell}$ and $A_mA_n$ have no common points. Vasya walks around the polygon and sees that the first two vertices are labeled $A_1$ and $A_4$. How is the tenth vertex labeled? [i]Proposed by A. Golovanov[/i]

In a triangle $ABC$, $AB = AC$, and $D$ is on $BC$. A point $E$ is chosen on $AC$, and a point $F$ is chosen on $AB$, such that $DE = DC$ and $DF = DB$. It is given that $\frac{DC}{BD} = 2$ and $\frac{AF}{AE} = 5$. Determine that value of $\frac{AB}{BC}$.