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 $AL$ and $BK$ be angle bisectors in the non-isosceles triangle $ABC$ ($L$ lies on the side $BC$, $K$ lies on the side $AC$). The perpendicular bisector of $BK$ intersects the line $AL$ at point $M$. Point $N$ lies on the line $BK$ such that $LN$ is parallel to $MK$. Prove that $LN = NA$.

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]

$\triangle ABC$, $\angle A=23^{\circ},\angle B=46^{\circ}$. Let $\Gamma$ be a circle with center $C$, radius $AC$. Let the external angle bisector of $\angle B$ intersects $\Gamma$ at $M,N$. Find $\angle MAN$.

Let $ABC$ be an acute-angled triangle inscribed in circle $(O)$. Let $G$ be a point on the small arc $AC$ of $(O)$ and $(K)$ be a circle passing through $A$ and $G$. Bisector of $\angle BAC$ cuts $(K)$ again at $P$. The point $E$ is chosen on $(K)$ such that $AE$ is parallel to $BC$. The line $PK$ meets the perpendicular bisector of $BC$ at $F$. Prove that $\angle EGF = 90^o$.

Let \( n \geq 3 \) be a positive integer. In a convex polygon with \( n \) sides, all the internal bisectors of its \( n \) internal angles are drawn. Determine, as a function of \( n \), the smallest possible number of distinct lines determined by these bisectors.

Let $\Gamma$ be the circumcircle of a triangle $ABC$ and let $E$ and $F$ be the intersections of the bisectors of $\angle ABC$ and $\angle ACB$ with $\Gamma$. If $EF$ is tangent to the incircle $\gamma$ of $\triangle ABC$, then find the value of $\angle BAC$.

Let $l,m$ be two parallel lines in the plane. Let $P$ be a fixed point between them. Let $E,F$ be variable points on $l,m$ such that the angle $EPF$ is fixed to a number like $\alpha$ where $0<\alpha<\frac{\pi}2$. (By angle $EPF$ we mean the directed angle) Show that there is another point (not $P$) such that it sees the segment $EF$ with a fixed angle too.

In triangle $ABC$ point $M$ is the midpoint of side $AB$, and point $D$ is the foot of altitude $CD$. Prove that $\angle A = 2\angle B$ if and only if $AC = 2 MD$.

Suppose $ABCD$ is a trapezoid with $AB\parallel CD$ and $AB\perp BC$. Let $X$ be a point on segment $\overline{AD}$ such that $AD$ bisects $\angle BXC$ externally, and denote $Y$ as the intersection of $AC$ and $BD$. If $AB=10$ and $CD=15$, compute the maximum possible value of $XY$.

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 $A, B,C$ be colinear points in this order, $\omega$ an arbitrary circle passing through $B$ and $C$, and $l$ an arbitrary line different from $BC$, passing through A and intersecting $\omega$ at $M$ and $N$. The bisectors of the angles $\angle CMB$ and $\angle CNB$ intersect $BC$ at $P$ and $Q$. Prove that $AP\cdot AQ = AB \cdot AC$.

Let $ABC$ be a triangle and let $\Omega$ be its circumcircle. The internal bisectors of angles $A, B$ and $C$ intersect $\Omega$ at $A_1, B_1$ and $C_1$, respectively, and the internal bisectors of angles $A_1, B_1$ and $C_1$ of the triangles $A_1 A_2 A_ 3$ intersect $\Omega$ at $A_2, B_2$ and $C_2$, respectively. If the smallest angle of the triangle $ABC$ is $40^{\circ}$, what is the magnitude of the smallest angle of the triangle $A_2 B_2 C_2$ in degrees?

Let $ M,N $ be points on the segments $ AB,AC, $ respectively, of the triangle $ ABC. $ Also, let $ P,Q, $ be the midpoints of the segments $ MN,BC, $ respectively. Knowing that $ PQ $ is parallel to the bisector of $ \angle BAC , $ show that $ BM=CN. $ [i]Gheorghe Duță[/i]

The angles of an acute triangle $ABC$ at $\alpha , \beta, \gamma$. Let $AD$ be a height, $CF$ a median, and $BE$ the bisector of $\angle B$. Show that $AD,CF$ and $BE$ are concurrent if and only if $\cos \gamma \tan\beta = \sin \alpha$ .

In $\triangle ABC$, shown on the right, let $r$ denote the radius of the inscribed circle, and let $r_A$, $r_B$, and $r_C$ denote the radii of the smaller circles tangent to the inscribed circle and to the sides emanating from $A$, $B$, and $C$, respectively. Prove that $r \leq r_A + r_B + r_C$

In a triangle $ABC$ $A_0$,$B_0$ and $C_0$ are the midpoints of the sides $BC$,$CA$ and $AB$.$A_1$,$B_1$,$C_1$ are the midpoints of the broken lines $BAC,CAB,ABC$.Show that $A_0A_1,B_0B_1,C_0C_1$ are concurrent.

Let $ABCD$ be a cyclic quadrilatetral inscribed in a circle $k$. Let $M$ and $N$ be the midpoints of the arcs $AB$ and $CD$ which do not contain $C$ and $A$ respectively. If $MN$ meets side $AB$ at $P$, then show that $\frac{AP}{BP}=\frac{AC+AD}{BC+BD}$

Let $l,m$ be two parallel lines in the plane. Let $P$ be a fixed point between them. Let $E,F$ be variable points on $l,m$ such that the angle $EPF$ is fixed to a number like $\alpha$ where $0<\alpha<\frac{\pi}2$. (By angle $EPF$ we mean the directed angle) Show that there is another point (not $P$) such that it sees the segment $EF$ with a fixed angle too.

Let $ABCD$ be a cyclic quadrilateral.$E$ is the midpoint of $(AC)$ and $F$ is the midpoint of $(BD)$ {$G$}=$AB\cap CD$ and {$H$}=$AD\cap BC$. a)Prove that the intersections of the angle bisector of $\angle{AHB}$ and the sides $AB$ and $CD$ and the intersections of the angle bisector of$\angle{AGD}$ with $BC$ and $AD$ are the verticles of a rhombus b)Prove that the center of this rhombus lies on $EF$

Let $ABC$ be a triangle and $AA_1$ be the angle bisector of $A$ ($A_1 \in BC$). The point $P$ is on the segment $AA_1$ and $M$ is the midpoint of the side $BC$. The point $Q$ is on the line connecting $P$ and $M$ such that $M$ is the midpoint of $PQ$. Define $D$ and $E$ as the intersections of $BQ$, $AC$, and $CQ$, $AB$. Prove that $CD=BE$.

Let $\vartriangle ABC$ be an acute triangle, and let $P$ be the foot of altitude from $C$ to $AB$. Let $\omega$ be the circle with diameter $BC$. The tangents from $A$ to $\omega$ are drawn touching $\omega$ at $D$ and $E$. Lines $AD$ and $AE$ intersect line $BC$ at $M$ and $N$ respectively, so that $B$ lies between $M$ and $C$. Let $CP$ intersect $DE$ at $Q, ME$ intersect $ND$ at $R$, and let $QR$ intersect $BC$ at $S$. Show that $QS$ bisects $\angle DSE$

Let $\Delta ABC$ be a triangle with an inscribed circle centered at $I$. The line perpendicular to $AI$ at $I$ intersects $\odot (ABC)$ at $P,Q$ such that, $P$ lies closer to $B$ than $C$. Let $\odot (BIP) \cap \odot (CIQ) =S$. Prove that, $SI$ is the angle bisector of $\angle PSQ$

Two straight lines are given on a plane, intersecting at point $O$ at an angle $a$. Let $A$, $B$ and $C $ be three points on one of the lines, located on one side of$ O$ and following in the indicated order, $M$ be an arbitrary point on another line, different from $O$, Let $\angle AMB=\gamma$, $\angle BMC = \phi$. Consider the function $F(M) = ctg \gamma + ctg \phi$ . Prove that$ F(M)$ takes the smallest value on each of the rays into which $O$ divides the second straight line. (Each has its own.) Let us denote one of these smallest values by $q$, and the other by $p$. Prove that the exprseeion $\frac{p}{q}$ is independent of choice of points $A$, $B$ and $C$. Express this relationship in terms of $a$.

All angles of $ABC$ are in $(30,90)$. Circumcenter of $ABC$ is $O$ and circumradius is $R$. Point $K$ is projection of $O$ to angle bisector of $\angle B$, point $M$ is midpoint $AC$. It is known, that $2KM=R$. Find $\angle B$