Found problems: 970
In convex quadrilateral $ABCD$, the diagonals $AC$ and $BD$ meet at the point $P$. We know that $\angle DAC = 90^o$ and $2 \angle ADB = \angle ACB$. If we have $ \angle DBC + 2 \angle ADC = 180^o$ prove that $2AP = BP$.
Proposed by Iman Maghsoudi
Let $ABCD$ be a square with $AB=6.$ A point $P$ in the interior is $2$ units away from side $BC$ and $3$ units away from side $CD.$ What is the distance from $P$ to $A?$
If circular arcs $ AC$ and $ BC$ have centers at $ B$ and $ A$, respectively, then there exists a circle tangent to both $ \stackrel{\frown}{AC}$ and $ \stackrel{\frown}{BC}$, and to $ \overline{AB}$. If the length of $ \stackrel{\frown}{BC}$ is $ 12$, then the circumference of the circle is
[asy]unitsize(4cm);
defaultpen(fontsize(8pt)+linewidth(.8pt));
dotfactor=3;
pair O=(0,.375);
pair A=(-.5,0);
pair B=(.5,0);
pair C=shift(-.5,0)*dir(60);
draw(Arc(A,1,0,60));
draw(Arc(B,1,120,180));
draw(A--B);
draw(Circle(O,.375));
dot(A);
dot(B);
dot(C);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);[/asy]$ \textbf{(A)}\ 24 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 26 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 28$
The map below shows an east/west road connecting the towns of Acorn, Centerville, and Midland, and a north/south road from Centerville to Drake. The distances from Acorn to Centerville, from Centerville to Midland, and from Centerville to Drake are each 60 kilometers. At noon Aaron starts at Acorn and bicycles east at 17 kilometers per hour, Michael starts at Midland and bicycles west at 7 kilometers per hour, and David starts at Drake and bicycles at a constant rate in a straight line across an open field. All three bicyclists arrive at exactly the same time at a point along the road from Centerville to Midland. Find the number of kilometers that David bicycles.
[center][img]https://i.snag.gy/Ik094i.jpg[/img][/center]
Let $ABC$ be a triangle with $CA \neq CB$. Let $D$, $F$, and $G$ be the midpoints of the sides $AB$, $AC$, and $BC$ respectively. A circle $\Gamma$ passing through $C$ and tangent to $AB$ at $D$ meets the segments $AF$ and $BG$ at $H$ and $I$, respectively. The points $H'$ and $I'$ are symmetric to $H$ and $I$ about $F$ and $G$, respectively. The line $H'I'$ meets $CD$ and $FG$ at $Q$ and $M$, respectively. The line $CM$ meets $\Gamma$ again at $P$. Prove that $CQ = QP$.
[i]Proposed by El Salvador[/i]
$MD$ is a chord of length $2$ in a circle of radius $1,$ and $L$ is chosen on the circle so that the area of $\triangle MLD$ is the maximized. Find $\angle MLD.$
In $\triangle ABC$, $AB = AC$. Its circumcircle, $\Gamma$, has a radius of 2. Circle $\Omega$ has a radius of 1 and is tangent to $\Gamma$, $\overline{AB}$, and $\overline{AC}$. The area of $\triangle ABC$ can be expressed as $\frac{a\sqrt{b}}{c}$ for positive integers $a, b, c$, where $b$ is squarefree and $\gcd (a, c) = 1$. Compute $a + b + c$.
[i]Proposed by Aaron Lin[/i]
A round table has radius $ 4$. Six rectangular place mats are placed on the table. Each place mat has width $ 1$ and length $ x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $ x$. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $ x$?
[asy]unitsize(4mm);
defaultpen(linewidth(.8)+fontsize(8));
draw(Circle((0,0),4));
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;
draw(mat);
draw(rotate(60)*mat);
draw(rotate(120)*mat);
draw(rotate(180)*mat);
draw(rotate(240)*mat);
draw(rotate(300)*mat);
label("$x$",(-2.687,0),E);
label("$1$",(-3.187,1.5513),S);[/asy]$ \textbf{(A)}\ 2\sqrt {5} \minus{} \sqrt {3} \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ \frac {3\sqrt {7} \minus{} \sqrt {3}}{2} \qquad \textbf{(D)}\ 2\sqrt {3} \qquad \textbf{(E)}\ \frac {5 \plus{} 2\sqrt {3}}{2}$
In a rectangular plot of land, a man walks in a very peculiar fashion. Labeling the corners $ABCD$, he starts at $A$ and walks to $C$. Then, he walks to the midpoint of side $AD$, say $A_1$. Then, he walks to the midpoint of side $CD$ say $C_1$, and then the midpoint of $A_1D$ which is $A_2$. He continues in this fashion, indefinitely. The total length of his path if $AB=5$ and $BC=12$ is of the form $a + b\sqrt{c}$. Find $\displaystyle\frac{abc}{4}$.
A square with area 4 is inscribed in a square with area 5, with one vertex of the smaller square on each side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length $a$, and the other of length $b$. What is the value of $ab$ ?
[asy]
draw((0,2)--(2,2)--(2,0)--(0,0)--cycle);
draw((0,0.3)--(0.3,2)--(2,1.7)--(1.7,0)--cycle);
label("$a$",(-0.1,0.15));
label("$b$",(-0.1,1.15));
[/asy]
$\textbf{(A)}\hspace{.05in}\dfrac15 \qquad \textbf{(B)}\hspace{.05in}\dfrac25 \qquad \textbf{(C)}\hspace{.05in}\dfrac12 \qquad \textbf{(D)}\hspace{.05in}1 \qquad \textbf{(E)}\hspace{.05in}4 $
There exists a quadrilateral $\mathcal{Q} _{1}$ such that the midpoints of its sides lie on a circle. Prove that there exists a cyclic quadrilateral $\mathcal{Q} _{2}$ with the same sides as $\mathcal{Q} _{1}$ with two of the same angles.
In trapezoid $ABCD,$ leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD},$ and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001},$ find $BC^2.$
The lengths in inches of the three sides of each of four triangles I, II, III, and IV are as follows:
\[ \begin{array}{rlrl} \hbox {I}& 3,\ 4,\ \hbox{and}\ 5 \qquad & \hbox{III}& 7,\ 24,\ \hbox{and}\ 25 \\ \hbox{II}& 4,\ 7\frac{1}{2},\ \hbox{and}\ 8\frac{1}{2} \qquad & \hbox{IV}& 3\frac{1}{2},\ 4\frac{1}{2},\ \hbox{and}\ 5\frac{1}{2}. \end{array} \]
Of these four given triangles, the only right triangles are
\[ \begin{tabular}{rlrlrl} (A) & I and II \qquad & (B) & I and III \qquad & (C) & I and IV \\ (D) & I, II, and III \qquad & (E) & I, II, and IV & \end{tabular} \]
A cube with side length $2$ is inscribed in a sphere. A second cube, with faces parallel to the first, is inscribed between the sphere and one face of the first cube. What is the length of a side of the smaller cube?
Find all integers $n \geq 3$ such that among any $n$ positive real numbers $a_1, a_2, \hdots, a_n$ with $\text{max}(a_1,a_2,\hdots,a_n) \leq n \cdot \text{min}(a_1,a_2,\hdots,a_n)$, there exist three that are the side lengths of an acute triangle.
Hexagon $ABCDEF$ is divided into four rhombuses, $\mathcal{P, Q, R, S,}$ and $\mathcal{T,}$ as shown. Rhombuses $\mathcal{P, Q, R,}$ and $\mathcal{S}$ are congruent, and each has area $\sqrt{2006}$. Let $K$ be the area of rhombus $\mathcal{T}$. Given that $K$ is a positive integer, find the number of possible values for $K$.
[asy]
size(150);defaultpen(linewidth(0.7)+fontsize(10));
draw(rotate(45)*polygon(4));
pair F=(1+sqrt(2))*dir(180), C=(1+sqrt(2))*dir(0), A=F+sqrt(2)*dir(45), E=F+sqrt(2)*dir(-45), B=C+sqrt(2)*dir(180-45), D=C+sqrt(2)*dir(45-180);
draw(F--(-1,0)^^C--(1,0)^^A--B--C--D--E--F--cycle);
pair point=origin;
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$E$", E, dir(point--E));
label("$F$", F, dir(point--F));
label("$\mathcal{P}$", intersectionpoint( A--(-1,0), F--(0,1) ));
label("$\mathcal{S}$", intersectionpoint( E--(-1,0), F--(0,-1) ));
label("$\mathcal{R}$", intersectionpoint( D--(1,0), C--(0,-1) ));
label("$\mathcal{Q}$", intersectionpoint( B--(1,0), C--(0,1) ));
label("$\mathcal{T}$", point);
dot(A^^B^^C^^D^^E^^F);[/asy]
In triangle $ ABC$, angle $ C$ is a right angle and $ CB > CA$. Point $ D$ is located on $ \overline{BC}$ so that angle $ CAD$ is twice angle $ DAB$. If $ AC/AD \equal{} 2/3$, then $ CD/BD \equal{} m/n$, where $ m$ and $ n$ are relatively prime positive integers. Find $ m \plus{} n$.
$ \textbf{(A)}\ 10\qquad
\textbf{(B)}\ 14\qquad
\textbf{(C)}\ 18\qquad
\textbf{(D)}\ 22\qquad
\textbf{(E)}\ 26$
Given that $x$, $y$, and $z$ are real numbers that satisfy:
\[ x=\sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}} \]
\[ y=\sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}} \]
\[ z=\sqrt{x^2-\frac{1}{36}}+\sqrt{y^2-\frac{1}{36}} \]
and that $x+y+z=\frac{m}{\sqrt{n}}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime, find $m+n$.
Let $\Gamma$ be the circumcircle of triangle $ABC$. The line parallel to $AC$ passing through $B$ meets $\Gamma$ at $D$ ($D\neq B$), and the line parallel to $AB$ passing through $C$ intersects $\Gamma$ to $E$ ($E\neq C$). Lines $AB$ and $CD$ meet at $P$, and lines $AC$ and $BE$ meet at $Q$. Let $M$ be the midpoint of $DE$. Line $AM$ meets $\Gamma$ at $Y$ ($Y\neq A$) and line $PQ$ at $J$. Line $PQ$ intersects the circumcircle of triangle $BCJ$ at $Z$ ($Z\neq J$). If lines $BQ$ and $CP$ meet each other at $X$, show that $X$ lies on the line $YZ$.
A paper equilateral triangle $ABC$ has side length $12$. The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$. The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$, where $m$, $n$, and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$.
[asy]
import cse5;
size(12cm);
pen tpen = defaultpen + 1.337;
real a = 39/5.0;
real b = 39/7.0;
pair B = MP("B", (0,0), dir(200));
pair A = MP("A", (9,0), dir(-80));
pair C = MP("C", (12,0), dir(-20));
pair K = (6,10.392);
pair M = (a*B+(12-a)*K) / 12;
pair N = (b*C+(12-b)*K) / 12;
draw(B--M--N--C--cycle, tpen);
draw(M--A--N--cycle);
fill(M--A--N--cycle, mediumgrey);
pair shift = (-20.13, 0);
pair B1 = MP("B", B+shift, dir(200));
pair A1 = MP("A", K+shift, dir(90));
pair C1 = MP("C", C+shift, dir(-20));
draw(A1--B1--C1--cycle, tpen);[/asy]