Let $N$ be a point on the longest side $AC$ of a triangle $ABC$. The perpendicular bisectors of $AN$ and $NC$ intersect $AB$ and $BC$ respectively in $K$ and $M$. Prove that the circumcenter $O$ of $\triangle ABC$ lies on the circumcircle of triangle $KBM$.

The centers of two circles are $ 41$ inches apart. The smaller circle has a radius of $ 4$ inches and the larger one has a radius of $ 5$ inches. The length of the common internal tangent is: $ \textbf{(A)}\ 41\text{ inches} \qquad\textbf{(B)}\ 39\text{ inches} \qquad\textbf{(C)}\ 39.8\text{ inches} \qquad\textbf{(D)}\ 40.1\text{ inches}\\ \textbf{(E)}\ 40\text{ inches}$

$\vartriangle ABC \sim \vartriangle A'B'C'$. $\vartriangle ABC$ has sides $a,b,c$ and $\vartriangle A'B'C'$ has sides $a',b',c'$. Two sides of $\vartriangle ABC$ are equal to sides of $\vartriangle A'B'C'$. Furthermore, $a < a'$, $a < b < c$, $a = 8$. Prove that there is exactly one pair of such triangles with all sides integers.

The point $M$ is inside the convex quadrilateral $ABCD$, such that \[ MA = MC, \hspace{0,2cm} \widehat{AMB} = \widehat{MAD} + \widehat{MCD} \quad \textnormal{and} \quad \widehat{CMD} = \widehat{MCB} + \widehat{MAB}. \] Prove that $AB \cdot CM = BC \cdot MD$ and $BM \cdot AD = MA \cdot CD.$

Let $U$ be the center of the circumcircle of the acute-angled triangle $ABC$. Let $M_A, M_B$ and $M_C$ be the circumcenters of triangles $UBC, UAC$ and $UAB$ respecrively. For which triangles $ABC$ is the triangle $M_AM_BM_C$ similar to the starting triangle (with a suitable order of the vertices)?

For an isosceles triangle $ABC$ where $AB=AC$, it is possible to construct, using only compass and straightedge, an isosceles triangle $PQR$ where $PQ=PR$ such that triangle $PQR$ is similar to triangle $ABC$, point $P$ is in the interior of line segment $AC$, point $Q$ is in the interior of line segment $AB$, and point $R$ is in the interior of line segment $BC$. Describe one method of performing such a construction. Your method should work on every isosceles triangle $ABC$, except that you may choose an upper limit or lower limit on the size of angle $BAC$. [asy] defaultpen(linewidth(0.7)); pair a= (79,164),b=(19,22),c=(138,22),p=(109,91),q=(38,67),r=(78,22); pair point = ((p.x+q.x+r.x)/3,(p.y+q.y+r.y)/3); draw(a--b--c--cycle); draw(p--q--r--cycle); label("$A$",a,dir(point--a)); label("$B$",b,dir(point--b)); label("$C$",c,dir(point--c)); label("$P$",p,dir(point--p)); label("$Q$",q,dir(point--q)); label("$R$",r,dir(point--r));[/asy]

In a rhombus, $ ABCD$, line segments are drawn within the rhombus, parallel to diagonal $ BD$, and terminated in the sides of the rhombus. A graph is drawn showing the length of a segment as a function of its distance from vertex $ A$. The graph is: $ \textbf{(A)}\ \text{A straight line passing through the origin.} \\ \textbf{(B)}\ \text{A straight line cutting across the upper right quadrant.} \\ \textbf{(C)}\ \text{Two line segments forming an upright V.} \\ \textbf{(D)}\ \text{Two line segments forming an inverted V.} \\ \textbf{(E)}\ \text{None of these.}$

On the sides of triangle $ABC$, three similar triangles are constructed with triangle $YBA$ and triangle $ZAC$ in the exterior and triangle $XBC$ in the interior. (Above, the vertices of the triangles are ordered so that the similarities take vertices to corresponding vertices, for example, the similarity between triangle $YBA$ and triangle $ZAC$ takes $Y$ to $Z, B$ to $A$ and $A$ to $C$). Prove that $AYXZ$ is a parallelogram

Given triangle $ABC$, let $D$, $E$, $F$ be the midpoints of $BC$, $AC$, $AB$ respectively and let $G$ be the centroid of the triangle. For each value of $\angle BAC$, how many non-similar triangles are there in which $AEGF$ is a cyclic quadrilateral?

Let $ABCD$ be a trapezoid, with $AB \parallel CD$ (the vertices are listed in cyclic order). The diagonals of this trapezoid are perpendicular to one another and intersect at $O$. The base angles $\angle DAB$ and $\angle CBA$ are both acute. A point $M$ on the line sgement $OA$ is such that $\angle BMD = 90^o$, and a point $N$ on the line segment $OB$ is such that $\angle ANC = 90^o$. Prove that triangles $OMN$ and $OBA$ are similar.

Where is AMC 10a No.19? Thanks

Points $D$ and $E$ are taken on sides $BC$ and $CA$ of a triangle $ BD\equal{}AE$. Segments $AD$ and $BE$ meet at $P$. The bisector of $\angle ACB$ intersects $AD$ and $BE$ at $Q$ and $R$ respectively. Prove that $ \frac{PQ}{PR}\equal{}\frac{AD}{BE}$.

Let $O$ be the circumcenter of an acute triangle $ABC$. Line $OA$ intersects the altitudes of $ABC$ through $B$ and $C$ at $P$ and $Q$, respectively. The altitudes meet at $H$. Prove that the circumcenter of triangle $PQH$ lies on a median of triangle $ABC$.

Two different points $A$ and $B$ and a circle $\omega$ that passes through $A$ and $B$ are given. $P$ is a variable point on $\omega$ (different from $A$ and $B$). $M$ is a point such that $MP$ is the bisector of the angle $\angle{APB}$ ($M$ lies outside of $\omega$) and $MP=AP+BP$. Find the geometrical locus of $M$.

In triangle $ABC$, $BC = 23$, $CA = 27$, and $AB = 30$. Points $V$ and $W$ are on $\overline{AC}$ with $V$ on $\overline{AW}$, points $X$ and $Y$ are on $\overline{BC}$ with $X$ on $\overline{CY}$, and points $Z$ and $U$ are on $\overline{AB}$ with $Z$ on $\overline{BU}$. In addition, the points are positioned so that $\overline{UV} \parallel \overline{BC}$, $\overline{WX} \parallel \overline{AB}$, and $\overline{YZ} \parallel \overline{CA}$. Right angle folds are then made along $\overline{UV}$, $\overline{WX}$, and $\overline{YZ}$. The resulting figure is placed on a level floor to make a table with triangular legs. Let $h$ be the maximum possible height of a table constructed from triangle $ABC$ whose top is parallel to the floor. Then $h$ can be written in the form $\tfrac{k \sqrt{m}}{n}$, where $k$ and $n$ are relatively prime positive integers and $m$ is a positive integer that is not divisible by the square of any prime. Find $k + m + n$. [asy] unitsize(1 cm); pair translate; pair[] A, B, C, U, V, W, X, Y, Z; A[0] = (1.5,2.8); B[0] = (3.2,0); C[0] = (0,0); U[0] = (0.69*A[0] + 0.31*B[0]); V[0] = (0.69*A[0] + 0.31*C[0]); W[0] = (0.69*C[0] + 0.31*A[0]); X[0] = (0.69*C[0] + 0.31*B[0]); Y[0] = (0.69*B[0] + 0.31*C[0]); Z[0] = (0.69*B[0] + 0.31*A[0]); translate = (7,0); A[1] = (1.3,1.1) + translate; B[1] = (2.4,-0.7) + translate; C[1] = (0.6,-0.7) + translate; U[1] = U[0] + translate; V[1] = V[0] + translate; W[1] = W[0] + translate; X[1] = X[0] + translate; Y[1] = Y[0] + translate; Z[1] = Z[0] + translate; draw (A[0]--B[0]--C[0]--cycle); draw (U[0]--V[0],dashed); draw (W[0]--X[0],dashed); draw (Y[0]--Z[0],dashed); draw (U[1]--V[1]--W[1]--X[1]--Y[1]--Z[1]--cycle); draw (U[1]--A[1]--V[1],dashed); draw (W[1]--C[1]--X[1]); draw (Y[1]--B[1]--Z[1]); dot("$A$",A[0],N); dot("$B$",B[0],SE); dot("$C$",C[0],SW); dot("$U$",U[0],NE); dot("$V$",V[0],NW); dot("$W$",W[0],NW); dot("$X$",X[0],S); dot("$Y$",Y[0],S); dot("$Z$",Z[0],NE); dot(A[1]); dot(B[1]); dot(C[1]); dot("$U$",U[1],NE); dot("$V$",V[1],NW); dot("$W$",W[1],NW); dot("$X$",X[1],dir(-70)); dot("$Y$",Y[1],dir(250)); dot("$Z$",Z[1],NE); [/asy]

Let $ABC$ be a triangle where $M$ is the midpoint of $\overline{AC}$, and $\overline{CN}$ is the angle bisector of $\angle ACB$ with $N$ on $\overline{AB}$. Let $X$ be the intersection of the median $\overline{BM}$ and the bisector $\overline{CN}$. In addition $\bigtriangleup BXN$ is equilateral and $AC=2$. What is $BN^2$? $\textbf{(A)}\ \frac{10-6\sqrt{2}}{7} \qquad\textbf{(B)}\ \frac{2}{9} \qquad\textbf{(C)}\ \frac{5\sqrt{2} - 3\sqrt{3}}{8} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{6} \qquad\textbf{(E)}\ \frac{3\sqrt{3} - 4}{5}$.

[color=blue][size=150]PERU TST IMO - 2006[/size] Saturday, may 20.[/color] [b]Question 04[/b] In an actue-angled triangle $ABC$ draws up: its circumcircle $w$ with center $O$, the circumcircle $w_1$ of the triangle $AOC$ and the diameter $OQ$ of $w_1$. The points are chosen $M$ and $N$ on the straight lines $AQ$ and $AC$, respectively, in such a way that the quadrilateral $AMBN$ is a parallelogram. Prove that the intersection point of the straight lines $MN$ and $BQ$ belongs the circumference $w_1.$ --- [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88513]Spanish version[/url] $\text{\LaTeX}{}$ed by carlosbr

Let ABCD be a convex quadrilateral and O a point inside it. Let the parallels to the lines BC, AB, DA, CD through the point O meet the sides AB, BC, CD, DA of the quadrilateral ABCD at the points E, F, G, H, respectively. Then, prove that $ \sqrt {\left|AHOE\right|} \plus{} \sqrt {\left|CFOG\right|}\leq\sqrt {\left|ABCD\right|}$, where $ \left|P_1P_2...P_n\right|$ is an abbreviation for the non-directed area of an arbitrary polygon $ P_1P_2...P_n$.

$ABC$ is a triangle. Take $a = BC$ etc as usual. Take points $T_1, T_2$ on the side $AB$ so that $AT_1 = T_1T_2 = T_2B$. Similarly, take points $T_3, T_4$ on the side BC so that $BT_3 = T_3T_4 = T_4C$, and points $T_5, T_6$ on the side $CA$ so that $CT_5 = T_5T_6 = T_6A$. Show that if $a' = BT_5, b' = CT_1, c'=AT_3$, then there is a triangle $A'B'C'$ with sides $a', b', c'$ ($a' = B'C$' etc). In the same way we take points $T_i'$ on the sides of $A'B'C' $ and put $a'' = B'T_6', b'' = C'T_2', c'' = A'T_4'$. Show that there is a triangle $A'' B'' C'' $ with sides $a'' b'' , c''$ and that it is similar to $ABC$. Find $a'' /a$.

In the adjoining figure, points $B$ and $C$ lie on line segment $AD$, and $AB$, $BC$, and $CD$ are diameters of circle $O$, $N$, and $P$, respectively. Circles $O$, $N$, and $P$ all have radius $15$ and the line $AG$ is tangent to circle $P$ at $G$. If $AG$ intersects circle $N$ at points $E$ and $F$, then chord $EF$ has length [asy] size(250); defaultpen(fontsize(10)); pair A=origin, O=(1,0), B=(2,0), N=(3,0), C=(4,0), P=(5,0), D=(6,0), G=tangent(A,P,1,2), E=intersectionpoints(A--G, Circle(N,1))[0], F=intersectionpoints(A--G, Circle(N,1))[1]; draw(Circle(O,1)^^Circle(N,1)^^Circle(P,1)^^G--A--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F^^G^^O^^N^^P); label("$A$", A, W); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, dir(0)); label("$P$", P, S); label("$N$", N, S); label("$O$", O, S); label("$E$", E, dir(120)); label("$F$", F, NE); label("$G$", G, dir(100));[/asy] $\textbf {(A) } 20 \qquad \textbf {(B) } 15\sqrt{2} \qquad \textbf {(C) } 24 \qquad \textbf{(D) } 25 \qquad \textbf {(E) } \text{none of these}$

Quadrilateral $ABCD$ is inscribed into circle $\omega$, $AC$ intersect $BD$ in point $K$. Points $M_1$, $M_2$, $M_3$, $M_4$-midpoints of arcs $AB$, $BC$, $CD$, and $DA$ respectively. Points $I_1$, $I_2$, $I_3$, $I_4$-incenters of triangles $ABK$, $BCK$, $CDK$, and $DAK$ respectively. Prove that lines $M_1I_1$, $M_2I_2$, $M_3I_3$, and $M_4I_4$ all intersect in one point.

We consider an acute triangle $ABC$. The altitude from $A$ is $AD$, the altitude from $D$ in triangle $ABD$ is $DE,$ and the altitude from $D$ in triangle $ACD$ is $DF$. a) Prove that the triangles $ABC$ and $AF E$ are similar. b) Prove that the segment $EF$ and the corresponding segments constructed from the vertices $B$ and $C$ all have the same length.

Three circles of equal radius have a common point $O$ and lie inside a given triangle. Each circle touches a pair of sides of the triangle. Prove that the incenter and the circumcenter of the triangle are collinear with the point $O$.

In triangle $ABC$, $AB=13$, $BC=14$, and $CA=15$. Distinct points $D$, $E$, and $F$ lie on segments $\overline{BC}$, $\overline{CA}$, and $\overline{DE}$, respectively, such that $\overline{AD}\perp\overline{BC}$, $\overline{DE}\perp\overline{AC}$, and $\overline{AF}\perp\overline{BF}$. The length of segment $\overline{DF}$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? ${ \textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D}}\ 27\qquad\textbf{(E)}\ 30 $

Let $ABC$ be a triangle and $D$ a point in $AC$. If $\angle{CBD} - \angle{ABD} = 60^{\circ}, \hat{BDC} = 30^{\circ}$ and also $AB \cdot BC = BD^{2}$, determine the measure of all the angles of triangle $ABC$.