Call a $ 3$-digit number [i]geometric[/i] if it has $ 3$ distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers.

Let $ \overline{AB}$ be a diameter of a circle and $ C$ be a point on $ \overline{AB}$ with $ 2 \cdot AC \equal{} BC$. Let $ D$ and $ E$ be points on the circle such that $ \overline{DC} \perp \overline{AB}$ and $ \overline{DE}$ is a second diameter. What is the ratio of the area of $ \triangle DCE$ to the area of $ \triangle ABD$? [asy]unitsize(2.5cm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0); pair D=dir(aCos(C.x)), E=(-D.x,-D.y); draw(A--B--D--cycle); draw(D--E--C); draw(unitcircle,white); drawline(D,C); dot(O); clip(unitcircle); draw(unitcircle); label("$E$",E,SSE); label("$B$",B,E); label("$A$",A,W); label("$D$",D,NNW); label("$C$",C,SW); draw(rightanglemark(D,C,B,2));[/asy]$ \textbf{(A)} \ \frac {1}{6} \qquad \textbf{(B)} \ \frac {1}{4} \qquad \textbf{(C)}\ \frac {1}{3} \qquad \textbf{(D)}\ \frac {1}{2} \qquad \textbf{(E)}\ \frac {2}{3}$

The area of $\triangle EBD$ is one third of the area of $3-4-5$ $ \triangle ABC$. Segment $DE$ is perpendicular to segment $AB$. What is $BD$? [asy] unitsize(10mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(5,0), C=(1.8,2.4), D=(5-4sqrt(3)/3,0), E=(5-4sqrt(3)/3,sqrt(3)); pair[] ps={A,B,C,D,E}; draw(A--B--C--cycle); draw(E--D); draw(rightanglemark(E,D,B)); dot(ps); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$E$",E,NE); label("$3$",midpoint(A--C),NW); label("$4$",midpoint(C--B),NE); label("$5$",midpoint(A--B),SW);[/asy] $ \textbf{(A)}\ \frac{4}{3} \qquad \textbf{(B)}\ \sqrt{5} \qquad \textbf{(C)}\ \frac{9}{4} \qquad \textbf{(D)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(E)}\ \frac{5}{2} $

Let $ABCD$ be a convex quadrilateral, and let $M$ and$ N$ be midpoints of sides $AB$ and $CD$ respectively. Point $P$ is chosen on $CD$ so that $MP \perp CD$, and point $Q$ is chosen on $AB$ so that $NQ \perp AB$. Show that $AD \parallel BC$ if and only if $\frac{AB}{CD} =\frac{MP}{NQ}$ .

Let $ABC$ be a triangle with area $1$ and $P$ the middle of the side $[BC]$. $M$ and $N$ are two points of $[AB]-\left \{ A,B \right \} $ and $[AC]-\left \{ A,C \right \}$ respectively such that $AM=2MB$ and$CN=2AN$. The two lines $(AP)$ and $(MN)$ intersect in a point $D$. Find the area of the triangle $ADN$.

In a $\triangle XYZ$, points $A,B$ lie on segment $ZX, C,D$ lie on segment $XY , E, F$ lie on segment $YZ$. $A, B, C, D$ lie on a circle, and $\frac{AZ \cdot EY \cdot ZB \cdot Y F}{EZ \cdot CY \cdot ZF \cdot Y D}= 1$ . Let $L = ZX \cap DE$, $M = XY \cap AF$, $N = Y Z \cap BC$. Prove that $L,M,N$ are collinear.

Given are two points $B,C$. Consider point $A$ not lying on the line $BC$ and draw the circles $C_1(K_1,R_1)$ (with center $K_1$ and radius $R_1$) and $C_2(K_2,R_2)$ with chord $AB, AC$ respectively such that their centers lie on the interior of the triangle $ABC$ and also $R_1 \cdot AC= R_2 \cdot AB$. Let $T$ be the intersection point of the two circles, different from $A$, and M be a random pointof line $AT$, prove that $TC \cdot S_{(MBT)}=TB \cdot S_{(MCT)}$

Let $ ABC$ be a triangle, let $ AB > AC$. Its incircle touches side $ BC$ at point $ E$. Point $ D$ is the second intersection of the incircle with segment $ AE$ (different from $ E$). Point $ F$ (different from $ E$) is taken on segment $ AE$ such that $ CE \equal{} CF$. The ray $ CF$ meets $ BD$ at point $ G$. Show that $ CF \equal{} FG$.

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}$.

In a triangle $ABC$, points $M$ and $N$ are on sides $AB$ and $AC$, respectively, such that $MB = BC = CN$. Let $R$ and $r$ denote the circumradius and the inradius of the triangle $ABC$, respectively. Express the ratio $MN/BC$ in terms of $R$ and $r$.

A point $(x,y)$ in the first quadrant lies on a line with intercepts $(a,0)$ and $(0,b)$, with $a,b > 0$. Rectangle $M$ has vertices $(0,0)$, $(x,0)$, $(x,y)$, and $(0,y)$, while rectangle $N$ has vertices $(x,y)$, $(x,b)$, $(a,b)$, and $(a,y)$. What is the ratio of the area of $M$ to that of $N$? [i]Proposed by Eugene Chen[/i]

Given an acute angles triangle $ABC$, and $H$ is its orthocentre. The external bisector of the angle $\angle BHC$ meets the sides $AB$ and $AC$ at the points $D$ and $E$ respectively. The internal bisector of the angle $\angle BAC$ meets the circumcircle of the triangle $ADE$ again at the point $K$. Prove that $HK$ is through the midpoint of the side $BC$.

Prove that for any positive integer $n$ and any real numbers $a_1,a_2,\cdots,a_n,b_1,b_2,\cdots,b_n$ we have that the equation \[a_1 \sin(x) + a_2 \sin(2x) +\cdots+a_n\sin(nx)=b_1 \cos(x)+b_2\cos(2x)+\cdots +b_n \cos(nx)\] has at least one real root.

In a convex quadrilateral $ABCD$ we have $\angle ADB=\angle ACD$ and $AC=CD=DB$. If the diagonals $AC$ and $BD$ intersect at $X$, prove that $\frac{CX}{BX}-\frac{AX}{DX}=1$.

Let $ABCDE$ be a convex pentagon with $AB\parallel CE$, $BC\parallel AD$, $AC\parallel DE$, $\angle ABC=120^\circ$, $AB=3$, $BC=5$, and $DE=15$. Given that the ratio between the area of triangle $ABC$ and the area of triangle $EBD$ is $m/n$, where $m$ and $n$ are relatively prime positive integers, find $m+n$.