Consider a circle $C$ with diameter $AB=1$. A point $P_0$ is chosen on $C$, $P_0 \ne A$, and starting in $P_0$ a sequence of points $P_1, P_2, \dots, P_n, \dots$ is constructed on $C$, in the following way: $Q_n$ is the symmetrical point of $A$ with respect of $P_n$ and the straight line that joins $B$ and $Q_n$ cuts $C$ at $B$ and $P_{n+1}$ (not necessary different). Prove that it is possible to choose $P_0$ such that: [b]i[/b] $\angle {P_0AB} < 1$. [b]ii[/b] In the sequence that starts with $P_0$ there are $2$ points, $P_k$ and $P_j$, such that $\triangle {AP_kP_j}$ is equilateral.

Let $I$ be the center of the incircle of non-isosceles triangle $ABC,A_{1}=AI\cap BC$ and $B_{1}=BI\cap AC.$ Let $l_{a}$ be the line through $A_{1}$ which is parallel to $AC$ and $l_{b}$ be the line through $B_{1}$ parallel to $BC.$ Let $l_{a}\cap CI=A_{2}$ and $l_{b}\cap CI=B_{2}.$ Also $N=AA_{2}\cap BB_{2}$ and $M$ is the midpoint of $AB.$ If $CN\parallel IM$ find $\frac{CN}{IM}$.

Moe uses a mower to cut his rectangular $ 90$-foot by $ 150$-foot lawn. The swath he cuts is $ 28$ inches wide, but he overlaps each cut by $ 4$ inches to make sure that no grass is missed. He walks at the rate of $ 5000$ feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow his lawn? $ \textbf{(A)}\ 0.75 \qquad \textbf{(B)}\ 0.8 \qquad \textbf{(C)}\ 1.35 \qquad \textbf{(D)}\ 1.5 \qquad \textbf{(E)}\ 3$

A point $P{}$ is considered on the incircle of the triangle $ABC$. We draw the tangent segments from $P{}$ to the three excircles of $ABC$. Prove that from the obtained three tangent segments it is possible to make a right triangle if and only if the point $P{}$ lies on one of the lines connecting two of the midpoints of the sides of $ABC$.

Let $ABCD$ be a cyclic quadrilateral in which internal angle bisectors $\angle ABC$ and $\angle ADC$ intersect on diagonal $AC$. Let $M$ be the midpoint of $AC$. Line parallel to $BC$ which passes through $D$ cuts $BM$ at $E$ and circle $ABCD$ in $F$ ($F \neq D$ ). Prove that $BCEF$ is parallelogram [i]Proposed by Steve Dinh[/i]

[b]Problem Section #4 a) There is a $6 * 6$ grid, each square filled with a grasshopper. After the bell rings, each grasshopper jumps to an adjacent square (A square that shares a side). What is the maximum number of empty squares possible?

Consider an arbitrary equilateral triangle $KLM$, whose vertices $K, L$ and $M$ lie on the sides $AB, BC$ and $CD$, respectively, of a given square $ABCD$. Find the locus of the midpoints of the sides $KL$ of all such triangles $KLM$.

In the quadrilateral $ABCD$, the diagonal $AC$ is the bisector $\angle BAD$ and $\angle ADC = \angle ACB$. The points $X, \, \, Y$ are the feet of the perpendiculars drawn from the point $A$ on the lines $BC, \, \, CD$, respectively. Prove that the orthocenter $\Delta AXY$ lies on the line $BD$.

Given a triangle $ABC$ with $BC=a$, $CA=b$, $AB=c$, $\angle BAC = \alpha$, $\angle CBA = \beta$, $\angle ACB = \gamma$. Prove that $$ a \sin(\beta-\gamma) + b \sin(\gamma-\alpha) +c\sin(\alpha-\beta) = 0.$$

Let $H$ be the orthocenter of an acute-angled triangle $ABC$; $E$, $F$ be points on $AB, AC$ respectively, such that $AEHF$ is a parallelogram; $X, Y$ be the common points of the line $EF$ and the circumcircle $\omega$ of triangle $ABC$; $Z$ be the point of $\omega$ opposite to $A$. Prove that $H$ is the orthocenter of triangle $XYZ$.

Let $P$ and $Q$ be the points on the diagonal $BD$ of a quadrilateral $ABCD$ such that $BP = PQ = QD$. Let $AP$ and $BC$ meet at $E$, and let $AQ$ meet $DC$ at $F$. (a) Prove that if $ABCD$ is a parallelogram, then $E$ and $F$ are the midpoints of the corresponding sides. (b) Prove the converse of (a).

Let $ ABC$ be a triangle. Circle $ \omega$­ passes through points $ B$ and $ C.$ Circle $ \omega_{1}$ is tangent internally to $ \omega$­ and also to sides $ AB$ and $ AC$ at $ T,\, P,$ and $ Q,$ respectively. Let $ M$ be midpoint of arc $ BC\, ($containing $ T)$ of ­$ \omega.$ Prove that lines $ PQ,\,BC,$ and $ MT$ are concurrent.

$ ABC$ is an acute-angled triangle. $ M$ is the midpoint of $ BC$ and $ P$ is the point on $ AM$ such that $ MB \equal{} MP$. $ H$ is the foot of the perpendicular from $ P$ to $ BC$. The lines through $ H$ perpendicular to $ PB$, $ PC$ meet $ AB, AC$ respectively at $ Q, R$. Show that $ BC$ is tangent to the circle through $ Q, H, R$ at $ H$. [i]Original Formulation: [/i] For an acute triangle $ ABC, M$ is the midpoint of the segment $ BC, P$ is a point on the segment $ AM$ such that $ PM \equal{} BM, H$ is the foot of the perpendicular line from $ P$ to $ BC, Q$ is the point of intersection of segment $ AB$ and the line passing through $ H$ that is perpendicular to $ PB,$ and finally, $ R$ is the point of intersection of the segment $ AC$ and the line passing through $ H$ that is perpendicular to $ PC.$ Show that the circumcircle of $ QHR$ is tangent to the side $ BC$ at point $ H.$

We have twenty-seven $1$ by $1$ cubes. Each face of every cube is marked with a natural number so that two opposite faces (top and bottom, front and back, left and right) are always marked with an even number and an odd number where the even number is twice that of the odd number. The twenty-seven cubes are put together to form one $3$ by $3$ cube as shown. When two cubes are placed face-to-face, adjoining faces are always marked with an odd number and an even number where the even number is one greater than the odd number. Find the sum of all of the numbers on all of the faces of all the $1$ by $1$ cubes. [asy] import graph; size(7cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); draw((-1,7)--(-1,4)); draw((-1,9.15)--(-3.42,8.21)); draw((-1,9.15)--(1.42,8.21)); draw((-1,7)--(1.42,8.21)); draw((1.42,7.21)--(-1,6)); draw((1.42,6.21)--(-1,5)); draw((1.42,5.21)--(-1,4)); draw((1.42,8.21)--(1.42,5.21)); draw((-3.42,8.21)--(-3.42,5.21)); draw((-3.42,7.21)--(-1,6)); draw((-3.42,8.21)--(-1,7)); draw((-1,4)--(-3.42,5.21)); draw((-3.42,6.21)--(-1,5)); draw((-2.61,7.8)--(-2.61,4.8)); draw((-1.8,4.4)--(-1.8,7.4)); draw((-0.2,7.4)--(-0.2,4.4)); draw((0.61,4.8)--(0.61,7.8)); label("2",(-1.07,9.01),SE*labelscalefactor); label("9",(-1.88,8.65),SE*labelscalefactor); label("1",(-2.68,8.33),SE*labelscalefactor); label("3",(-0.38,8.72),SE*labelscalefactor); draw((-1.8,7.4)--(0.63,8.52)); draw((-0.27,8.87)--(-2.61,7.8)); draw((-2.65,8.51)--(-0.2,7.4)); draw((-1.77,8.85)--(0.61,7.8)); label("7",(-1.12,8.33),SE*labelscalefactor); label("5",(-1.9,7.91),SE*labelscalefactor); label("1",(0.58,8.33),SE*labelscalefactor); label("18",(-0.36,7.89),SE*labelscalefactor); label("1",(-1.07,7.55),SE*labelscalefactor); label("1",(-0.66,6.89),SE*labelscalefactor); label("5",(-0.68,5.8),SE*labelscalefactor); label("1",(-0.68,4.83),SE*labelscalefactor); label("2",(0.09,7.27),SE*labelscalefactor); label("1",(0.15,6.24),SE*labelscalefactor); label("2",(0.11,5.26),SE*labelscalefactor); label("1",(0.89,7.61),SE*labelscalefactor); label("3",(0.89,6.63),SE*labelscalefactor); label("9",(0.92,5.62),SE*labelscalefactor); label("18",(-3.18,7.63),SE*labelscalefactor); label("2",(-3.07,6.61),SE*labelscalefactor); label("2",(-3.09,5.62),SE*labelscalefactor); label("1",(-2.29,7.25),SE*labelscalefactor); label("3",(-2.27,6.22),SE*labelscalefactor); label("5",(-2.29,5.2),SE*labelscalefactor); label("7",(-1.49,6.89),SE*labelscalefactor); label("34",(-1.52,5.81),SE*labelscalefactor); label("1",(-1.41,4.86),SE*labelscalefactor); [/asy]

The diagonals of a rectangle $ABCD$ meet at point $E$. A circle centered at $E$ lies inside the rectangle. Let $CF$, $DG$, $AH$ be the tangents to this circle from $C$, $D$, $A$; let $CF$ meet $DG$ at point $I$, $EI$ meet $AD$ at point $J$, and $AH$ meet $CF$ at point $L$. Prove that $LJ$ is perpendicular to $AD$.

Peter is a pentacrat and spends his time drawing pentagrams. With the abbreviation $\left|XYZ\right|$ for the area of an arbitrary triangle $XYZ$, he notes that any convex pentagon $ABCDE$ satisfies the equality $\left|EAC\right|\cdot\left|EBD\right|=\left|EAB\right|\cdot\left|ECD\right|+\left|EBC\right|\cdot\left|EDA\right|$. Guess what you are supposed to do and do it.

A function $f$ is defined for all real numbers and satisfies \[f(2 + x) = f(2 - x)\qquad\text{and}\qquad f(7 + x) = f(7 - x)\] for all real $x$. If $x = 0$ is a root of $f(x) = 0$, what is the least number of roots $f(x) = 0$ must have in the interval $-1000 \le x \le 1000$?

Parallelograms $ABGF$, $CDGB$ and $EFGD$ are drawn so that $ABCDEF$ is a convex hexagon, as shown. If $\angle ABG = 53^o$ and $\angle CDG = 56^o$, what is the measure of $\angle EFG$, in degrees? [img]https://cdn.artofproblemsolving.com/attachments/9/f/79d163662e02bc40d2636a76b73f632e59d584.png[/img]

In a non-isosceles $\Delta ABC$ with angle bisectors $AL_a$, $BL_b$, and $CL_c$ we have that $L_aL_c=L_bL_c$. Prove that $\angle C$ is smaller than $120^\circ$.

In rectangular coodinate system $Oxy$, consider the line $y=3x$ and point $A(4,3)$. Find on the line $y=3x$, point $B\ne O$ such that the area of triangle $OBC$ is the minimum possible, where $C= AB\cap Ox$.

One thousand unit cubes are fastened together to form a large cube with edge length 10 units; this is painted and then separated into the original cubes. The number of these unit cubes which have at least one face painted is $ \textbf{(A)}\ 600 \qquad \textbf{(B)}\ 520 \qquad \textbf{(C)}\ 488 \qquad \textbf{(D)}\ 480 \qquad \textbf{(E)}\ 400$

In $\triangle AEF$, let $B$ and $D$ be on segments $AE$ and $AF$ respectively, and let $ED$ and $FB$ intersect at $C$. Define $K,L,M,N$ on segments $AB,BC,CD,DA$ such that $\frac{AK}{KB}=\frac{AD}{BC}$ and its cyclic equivalents. Let the incircle of $\triangle AEF$ touch $AE,AF$ at $S,T$ respectively; let the incircle of $\triangle CEF$ touch $CE,CF$ at $U,V$ respectively. Prove that $K,L,M,N$ concyclic implies $S,T,U,V$ concyclic.

In a triangle $ABC$, $\angle C = 60^o$, $D, E, F$ are points on the sides $BC, AB, AC$ respectively, and $M$ is the intersection point of $AD$ and $BF$. Suppose that $CDEF$ is a rhombus. Prove that $DF^2 = DM \cdot DA$

Two circles $ \gamma_{1}$ and $ \gamma_{2}$ meet at points $ X$ and $ Y$. Consider the parallel through $ Y$ to the nearest common tangent of the circles. This parallel meets again $ \gamma_{1}$ and $ \gamma_{2}$ at $ A$, and $ B$ respectively. Let $ O$ be the center of the circle tangent to $ \gamma_{1},\gamma_{2}$ and the circle $ AXB$, situated outside $ \gamma_{1}$ and $ \gamma_{2}$ and inside the circle $ AXB.$ Prove that $ XO$ is the bisector line of the angle $ \angle{AXB}.$ [b]Radu Gologan[/b]

Let $ABC$ be an acute-angled triangle such that $AB+BC=4AC$. Let $D$ in $AC$ such that $BD$ is angle bisector of $\angle ABC$. In the segment $BD$, points $P$ and $Q$ are marked such that $BP=2DQ$. The perpendicular line to $BD$, passing by $Q$, cuts the segments $AB$ and $BC$ in $X$ and $Y$, respectively. Let $L$ be the parallel line to $AC$ passing by $P$. The point $B$ is in a different half-plane(with respect to the line $L$) of the points $X$ and $Y$. An ant starts a run in the point $X$, goes to a point in the line $AC$, after that goes to a point in the line $L$, returns to a point in the line $AC$ and finishes in the point $Y$. Prove that the least length of the ant's run is equal to $4XY$.