A coordinate system was drawn on the board and points $A (1,2)$ and $B (3,1)$ were marked. The coordinate system was erased. Restore it by the two marked points.

Let $ ABC$ be an acute triangle with the incircle $ C(I,r)$ and the circumcircle $ C(O,R)$ . Denote $ D\in BC$ for which $ AD\perp BC$ and $ AD \equal{} h_a$ . Prove that $ DI^2 \equal{} (2R \minus{} h_a)(h_a \minus{} 2r)$ .

Let $AD$ be an altitude of triangle $ABC$, and let $M$, $N$ and $P$ be midpoints of $AB$, $AD$ and $BC$, respectively. Furthermore let $K$ be a foot of perpendicular from point $D$ to line $AC$, and let $T$ be point on extension of line $KD$ (over point $D$) such that $\mid DT \mid = \mid MN \mid + \mid DK \mid$. If $\mid MP \mid = 2 \cdot \mid KN \mid$, prove that $\mid AT \mid = \mid MC \mid$.

2 curves $ y \equal{} x^3 \minus{} x$ and $ y \equal{} x^2 \minus{} a$ pass through the point $ P$ and have a common tangent line at $ P$. Find the area of the region bounded by these curves.

Let $ABCD$ be a convex quadrilateral. Prove that area $(ABCD) \le \frac{AB^2 + BC^2 + CD^2 + DA^2}{4}$

Let $ABC$ be a triangle with $AB=AC$ and let $I$ be its incenter. Let $\Gamma$ be the circumcircle of $ABC$. Lines $BI$ and $CI$ intersect $\Gamma$ in two new points, $M$ and $N$ respectively. Let $D$ be another point on $\Gamma$ lying on arc $BC$ not containing $A$, and let $E,F$ be the intersections of $AD$ with $BI$ and $CI$, respectively. Let $P,Q$ be the intersections of $DM$ with $CI$ and of $DN$ with $BI$ respectively. (i) Prove that $D,I,P,Q$ lie on the same circle $\Omega$ (ii) Prove that lines $CE$ and $BF$ intersect on $\Omega$

Let us consider an infinite grid plane as shown below. We start with 4 points $A$, $B$, $C$, $D$, that form a square. We perform the following operation: We pick two points $X$ and $Y$ from the currant points. $X$ is reflected about $Y$ to get $X'$. We remove $X$ and add $X'$ to get a new set of 4 points and treat it as our currant points. For example in the figure suppose we choose $A$ and $B$ (we can choose any other pair too). Then reflect $A$ about $B$ to get $A'$. We remove $A$ and add $A'$. Thus $A'$, $B$, $C$, $D$ is our new 4 points. We may again choose $D$ and $A'$ from the currant points. Reflect $D$ about $A'$ to obtain $D'$ and hence $A'$, $B$, $C$, $D'$ are now new set of points. Then similar operation is performed on this new 4 points and so on. Starting with $A$, $B$, $C$, $D$ can you get a bigger square by some sequence of such operations?

Point $L$ inside triangle $ABC$ is such that $CL = AB$ and $ \angle BAC + \angle BLC = 180^{\circ}$. Point $K$ on the side $AC$ is such that $KL \parallel BC$. Prove that $AB = BK$

Given $\triangle ABC$ with circumcircle $\ell$. Point $M$ in $\triangle ABC$ such that $AM$ is the angle bisector of $\angle BAC$. Circle with center $M$ and radius $MB$ intersects $\ell$ and $BC$ at $D$ and $E$ respectively, $(B \not= D, B \not= E)$. Let $P$ be the midpoint of arc $BC$ in $\ell$ that didn't have $A$. Prove that $AP$ angle bisector of $\angle DPE$ if and only if $\angle B = 90^{\circ}$.

Pentagon $ABCDE$ is such that all five diagonals $AC, BD, CE, DA,$ and $EB$ lie entirely within the pentagon. If the area of each of the triangles $ABC, BCD, CDE,$ and $DEA$ is equal to $1$ and the area of triangle $EAB$ is equal to $2$, the area of the pentagon $ABCDE$ is closest to $ \mathrm{(A) \ } 4.42 \qquad \mathrm{(B) \ } 4.44 \qquad \mathrm {(C) \ } 4.46 \qquad \mathrm{(D) \ } 4.48 \qquad \mathrm{(E) \ } 4.5$

Given triangle $ABC$ with $AB<AC$. The line passing through $B$ and parallel to $AC$ meets the external angle bisector of $\angle BAC$ at $D$. The line passing through $C$ and parallel to $AB$ meets this bisector at $E$. Point $F$ lies on the side $AC$ and satisfies the equality $FC=AB$. Prove that $DF=FE$.

Quadrilateral $ABCD$ is inscribed in circle $\Gamma$. Segments $AC$ and $BD$ intersect at $E$. Circle $\gamma$ passes through $E$ and is tangent to $\Gamma$ at $A$. Suppose the circumcircle of triangle $BCE$ is tangent to $\gamma$ at $E$ and is tangent to line $CD$ at $C$. Suppose that $\Gamma$ has radius $3$ and $\gamma$ has radius $2$. Compute $BD$.

Let $O$ be the center of a circle inscribed in a convex quadrilateral $ABCD$ and $|AB|= a$, $|CD|=$c. Prove that $$\frac{a}{c}=\frac{AO\cdot BO}{CO\cdot DO}.$$

$(BEL 1)$ A parabola $P_1$ with equation $x^2 - 2py = 0$ and parabola $P_2$ with equation $x^2 + 2py = 0, p > 0$, are given. A line $t$ is tangent to $P_2.$ Find the locus of pole $M$ of the line $t$ with respect to $P_1.$

Prove that there are no four points $A, B, C, D$ on a plane such that all triangles $\vartriangle ABC,\vartriangle BCD, \vartriangle CDA, \vartriangle DAB$ are acute ones.

A square with sidelength $1$ is covered by $3$ congruent disks. Find the smallest possible value of the radius of the disks.

Tweaking a convex $n$-gon means the following: choose two adjacent sides $AB$ and $BC$ and replaces them with the line segment $AM$, $MN$, $NC$, where $M \in AB$ and $N \in BC$ are arbitrary points inside these segments. In other words, you cut off a corner and get an $(n+1)$-corner. Starting from a regular hexagon $P_6$ with area $1$, by continuous Tweaks a sequence $P_6,P_7,P_8, ...$ convex polygons. Show that Area of $​​P_n$ for all $n\ge 6$ greater than $\frac1 2$ is, regardless of how tweaks takes place.

The figure $ABCD$ is bounded by a semicircle $CDA$ and a quarter circle $ABC$. Given that the distance from $A$ to $C$ is $18$, find the area of the figure. [asy] size(200); defaultpen(linewidth(0.8)); pair A=(-9,0),B=(0,9*sqrt(2)-9),C=(9,0),D=(0,9); dot(A^^B^^C^^D); draw(arc(origin,9,0,180)^^arc((0,-9),9*sqrt(2),45,135)); label("$A$",A,S); label("$B$",B,N); label("$C$",C,S); label("$D$",D,N); [/asy]

The bisectors of angles $A$ and $C$ of triangle $ABC$ intersect the circumcircle of this triangle at points $A_0$ and $C_0$, respectively. A straight line passing through the center of the inscribed circle of a triangle $ABC$ is parallel to side $AC$ and intersects line $A_0C_0$ at point $P$. Prove that line $PB$ is tangent to the circumcircle of the triangle $ABC$.

Let $BD$ be an altitude of $\triangle ABC$ with $AB < BC$ and $\angle B > 90^\circ$. Let $M$ be the midpoint of $AC$, and point $K$ be symmetric to point $D$ with respect to point $M$. A perpendicular drawn from point $M$ to the line $BC$ intersects line $AB$ at point $L$. Prove that $\angle MBL = \angle MKL$. [i]Proposed by Oleksandra Yakovenko[/i]

Squares of an $n \times n$ table ($n \ge 3$) are painted black and white as in a chessboard. A move allows one to choose any $2 \times 2$ square and change all of its squares to the opposite color. Find all such n that there is a finite number of the moves described after which all squares are the same color.

Given is an acute triangle $ABC$ $\left(AB<AC<BC\right)$,inscribed in circle $c(O,R)$.The perpendicular bisector of the angle bisector $AD$ $\left(D\in BC\right)$ intersects $c$ at $K,L$ ($K$ lies on the small arc $\overarc{AB}$).The circle $c_1(K,KA)$ intersects $c$ at $T$ and the circle $c_2(L,LA)$ intersects $c$ at $S$.Prove that $\angle{BAT}=\angle{CAS}$. [hide=Diagram][asy]import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.94236331697463, xmax = 15.849400903703716, ymin = -5.002235438802758, ymax = 7.893104843949444; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); pen qqqqtt = rgb(0.,0.,0.2); draw((1.8318261909633622,3.572783369254345)--(0.,0.)--(6.,0.)--cycle, aqaqaq); draw(arc((1.8318261909633622,3.572783369254345),0.6426249310341638,-117.14497824050169,-101.88970202103212)--(1.8318261909633622,3.572783369254345)--cycle, qqqqtt); draw(arc((1.8318261909633622,3.572783369254345),0.6426249310341638,-55.85706977865775,-40.60179355918817)--(1.8318261909633622,3.572783369254345)--cycle, qqqqtt); /* draw figures */ draw((1.8318261909633622,3.572783369254345)--(0.,0.), uququq); draw((0.,0.)--(6.,0.), uququq); draw((6.,0.)--(1.8318261909633622,3.572783369254345), uququq); draw(circle((3.,0.7178452373968209), 3.0846882800136055)); draw((2.5345020274407277,0.)--(1.8318261909633622,3.572783369254345)); draw(circle((-0.01850947366601585,1.3533783539547308), 2.889550258039566)); draw(circle((5.553011501106743,2.4491551634556963), 3.887127532933951)); draw((-0.01850947366601585,1.3533783539547308)--(5.553011501106743,2.4491551634556963), linetype("2 2")); draw((1.8318261909633622,3.572783369254345)--(0.7798408954511686,-1.423695174396108)); draw((1.8318261909633622,3.572783369254345)--(5.22015910454883,-1.4236951743961088)); /* dots and labels */ dot((1.8318261909633622,3.572783369254345),linewidth(3.pt) + dotstyle); label("$A$", (1.5831274347452782,3.951671933606579), NE * labelscalefactor); dot((0.,0.),linewidth(3.pt) + dotstyle); label("$B$", (-0.6,0.05), NE * labelscalefactor); dot((6.,0.),linewidth(3.pt) + dotstyle); label("$C$", (6.188606107156787,0.07450151636712989), NE * labelscalefactor); dot((2.5345020274407277,0.),linewidth(3.pt) + dotstyle); label("$D$", (2.3,-0.7), NE * labelscalefactor); dot((-0.01850947366601585,1.3533783539547308),linewidth(3.pt) + dotstyle); label("$K$", (-0.3447473583572136,1.6382221818835927), NE * labelscalefactor); dot((5.553011501106743,2.4491551634556963),linewidth(3.pt) + dotstyle); label("$L$", (5.631664500260511,2.580738747400365), NE * labelscalefactor); dot((0.7798408954511686,-1.423695174396108),linewidth(3.pt) + dotstyle); label("$T$", (0.5977692071595602,-1.960477431907719), NE * labelscalefactor); dot((5.22015910454883,-1.4236951743961088),linewidth(3.pt) + dotstyle); label("$S$", (5.160406217502124,-1.8747941077698307), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/hide]

Let $T$ be the intersection of the common internal tangents of circles $C_1$, $C_2$ with centers $O_1$, $O_2$ respectively. Let $P$ be one of the points of tangency on $C_1$ and let line $\ell$ bisect angle $O_1TP$ . Label the intersection of $\ell$ with $C_1$ that is farthest from $T$, $R$, and label the intersection of $\ell$ with $C_2$ that is closest to $T$, $S$. If $C_1$ has radius $4$, $C_2$ has radius $6$, and $O_1O_2= 20$ , calculate $(TR)(TS) $. [img]https://cdn.artofproblemsolving.com/attachments/3/c/284f17bb0dd73eab93132e41f27ecc121f496d.png[/img]

Find all positive integers $ n$ having the following properties:in two-dimensional Cartesian coordinates, there exists a convex $ n$ lattice polygon whose lengths of all sides are odd numbers, and unequal to each other. (where lattice polygon is defined as polygon whose coordinates of all vertices are integers in Cartesian coordinates.)

Let $\vartriangle A_0B_0C_0$ be an equilateral triangle with area $1$, and let $A_1$, $B_1$, $C_1$ be the midpoints of $\overline{A_0B_0}$, $\overline{B_0C_0}$, and $\overline{C_0A_0}$, respectively. Furthermore, set $A_2$, $B_2$, $C_2$ as the midpoints of segments $\overline{A_0A_1}$, $\overline{B_0B_1}$, and $\overline{C_0C_1}$ respectively. For $n \ge 1$, $A_{2n+1}$ is recursively defined as the midpoint of $A_{2n}A_{2n-1}$, and $A_{2n+2}$ is recursively defined as the midpoint of $\overline{A_{2n+1}A_{2n-1}}$. Recursively define $B_n$ and $C_n$ the same way. Compute the value of $\lim_{n \to \infty }[A_nB_nC_n]$, where $[A_nB_nC_n]$ denotes the area of triangle $\vartriangle A_nB_nC_n$.