Found problems: 844
$A,B,C$ are on circle $\mathcal C$. $I$ is incenter of $ABC$ , $D$ is midpoint of arc $BAC$. $W$ is a circle that is tangent to $AB$ and $AC$ and tangent to $\mathcal C$ at $P$. ($W$ is in $\mathcal C$)
Prove that $P$ and $I$ and $D$ are on a line.
Equilateral triangles $AMB,BNC,CKA$ are constructed on the exterior of a triangle $ABC$. The perpendiculars from the midpoints of $MN, NK, KM$ to the respective lines $CA, AB, BC$ are constructed. Prove that these three perpendiculars pass through a single point.
Let $ABC$ be a triangle, and let the tangent to the circumcircle of the triangle $ABC$ at $A$ meet the line $BC$ at $D$. The perpendicular to $BC$ at $B$ meets the perpendicular bisector of $AB$ at $E$. The perpendicular to $BC$ at $C$ meets the perpendicular bisector of $AC$ at $F$. Prove that the points $D$, $E$ and $F$ are collinear.
[i]Valentin Vornicu[/i]
An isosceles trapezoid has legs and shorter base of length $1$. Find the maximum possible value of its area
Let $ABCD$ be a trapezoid with $AD \parallel BC$, $AD < BC$. Let $E$ be a point on the side $AB$ and $F$ be point on the side $CD$. The circle $(AEF)$ intersects the segment $AD$ again at $A_1$ and the circle $(CEF)$ intersects these segment $BC$ again at $C_1$. Prove that the lines $A_1 C_1$, $BD$ and $EF$ are concurrent.
[i]Proposed by A. Kuznetsov[/i]
Let $ ABCD$ be a trapezoid with $ AB\parallel{}CD$. Let $ a \equal{} AB$ and $ b \equal{} CD$. For $ MN\parallel{}AB$ such that $ M$ lies on $ AD,$ $ N$ lies on $ BC$, and the trapezoids $ ABNM$ and $ MNCD$ have the same area, the length of $ MN$ equals
[img]http://i250.photobucket.com/albums/gg265/geometry101/NielsHenrikAbel1997Number6.jpg[/img]
A. $ \sqrt{ab}$
B. $ \frac{a\plus{}b}{2}$
C. $ \frac{a^2 \plus{} b^2}{a\plus{}b}$
D. $ \sqrt{\frac{a^2 \plus{} b^2}{2}}$
E. $ \frac{a^2 \plus{} (2 \sqrt{2} \minus{} 2)ab \plus{} b^2}{\sqrt{2} (a\plus{}b)}$
Let $ABCD$ be a rectangle and $E$ the reflection of $A$ with respect to the diagonal $BD$. If $EB = EC$, what is the ratio $\frac{AD}{AB}$ ?
Two parallel chords in a circle have lengths $10$ and $14$, and the distance between them is $6$. The chord parallel to these chords and midway between them is of length $\sqrt{a}$ where $a$ is
[asy]
// note: diagram deliberately not to scale -- azjps
void htick(pair A, pair B, real r){ D(A--B); D(A-(r,0)--A+(r,0)); D(B-(r,0)--B+(r,0)); }
size(120); pathpen = linewidth(0.7); pointpen = black+linewidth(3);
real min = -0.6, step = 0.5;
pair[] A, B; D(unitcircle);
for(int i = 0; i < 3; ++i) {
A.push(intersectionpoints((-9,min+i*step)--(9,min+i*step),unitcircle)[0]); B.push(intersectionpoints((-9,min+i*step)--(9,min+i*step),unitcircle)[1]);
D(D(A[i])--D(B[i]));
}
MP("10",(A[0]+B[0])/2,N);
MP("\sqrt{a}",(A[1]+B[1])/2,N);
MP("14",(A[2]+B[2])/2,N);
htick((B[1].x+0.1,B[0].y),(B[1].x+0.1,B[2].y),0.06); MP("6",(B[1].x+0.1,B[0].y/2+B[2].y/2),E);[/asy]
$\textbf{(A)}\ 144 \qquad
\textbf{(B)}\ 156 \qquad
\textbf{(C)}\ 168 \qquad
\textbf{(D)}\ 176 \qquad
\textbf{(E)}\ 184$
A trapezium in the plane is a quadrilateral in which a pair of opposite sides are parallel. A trapezium is said to be non-degenerate if it has positive area. Find the number of mutually non-congruent, non-degenerate trapeziums whose sides are four distinct integers from the set $\{5,6,7,8,9,10\}$
Rectangle $ABCD$ is divided into four parts of equal area by five segments as shown in the figure, where $XY = YB + BC + CZ = ZW = WD + DA + AX$, and $PQ$ is parallel to $AB$. Find the length of $AB$ (in cm) if $BC = 19$ cm and $PQ = 87$ cm.
[asy]
size(250);
pair A=origin, B=(96,0), C=(96,22), D=(0,22), W=(16,22), X=(20,0), Y=(80,0), Z=(76,22), P=(24,11), Q=(72,11);
draw(P--X--A--D--W--P--Q--Y--B--C--Z--Q^^W--Z^^X--Y);
dot(A^^B^^C^^D^^P^^Q^^W^^X^^Y^^Z);
pair point=(48,11);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, NE);
label("$D$", D, NW);
label("$P$", P, dir(point--P));
label("$Q$", Q, dir(point--Q));
label("$W$", W, N);
label("$X$", X, S);
label("$Y$", Y, S);
label("$Z$", Z, N);[/asy]
In a triangle $ABC$, it is drawn a circumference with center in the incenter $I$ and that meet twice each of the sides of the triangle: the segment $BC$ on $D$ and $P$ (where $D$ is nearer two $B$); the segment $CA$ on $E$ and $Q$ (where $E$ is nearer to $C$); and the segment $AB$ on $F$ and $R$ ( where $F$ is nearer to $A$).
Let $S$ be the point of intersection of the diagonals of the quadrilateral $EQFR$. Let $T$ be the point of intersection of the diagonals of the quadrilateral $FRDP$. Let $U$ be the point of intersection of the diagonals of the quadrilateral $DPEQ$.
Show that the circumcircle to the triangle $\triangle{FRT}$, $\triangle{DPU}$ and $\triangle{EQS}$ have a unique point in common.
Let $ABCD$ be a quadrilateral with $\measuredangle DAB=\measuredangle ABC=90^{\circ}$. Denote by $M$ the midpoint of the side $AB$, and assume that $\measuredangle CMD=90^{\circ}$. Let $K$ be the foot of the perpendicular from the point $M$ to the line $CD$. The line $AK$ meets $BD$ at $P$, and the line $BK$ meets $AC$ at $Q$. Show that $\angle{AKB}=90^{\circ}$ and $\frac{KP}{PA}+\frac{KQ}{QB}=1$.
[color=red][Moderator edit: The proposed solution can be found at http://erdos.fciencias.unam.mx/mexproblem3.pdf .][/color]
Let $ABCD$ be a trapezoid such that $AB \parallel CD$, $|AB|<|CD|$, and $\text{Area}(ABC)=30$. Let the line through $B$ parallel to $AD$ meet $[AC]$ at $E$. If $|AE|:|EC|=3:2$, then what is the area of trapezoid $ABCD$?
$
\textbf{(A)}\ 45
\qquad\textbf{(B)}\ 60
\qquad\textbf{(C)}\ 72
\qquad\textbf{(D)}\ 80
\qquad\textbf{(E)}\ 90
$
A trapezoid is given in which one base is twice as large as the other. Use one ruler (no divisions) to draw the midline of this trapezoid.
Prove that the sum of angles at the longer base of a trapezoid is less than the sum of angles at the shorter base.
Two planes, $P$ and $Q$, intersect along the line $p$. The point $A$ is given in the plane $P$, and the point $C$ in the plane $Q$; neither of these points lies on the straight line $p$. Construct an isosceles trapezoid $ABCD$ (with $AB \parallel CD$) in which a circle can be inscribed, and with vertices $B$ and $D$ lying in planes $P$ and $Q$ respectively.
We call an isosceles trapezoid $PQRS$ [i]interesting[/i], if it is inscribed in the unit square $ABCD$ in such a way, that on every side of the square lies exactly one vertex of the trapezoid and that the lines connecting the midpoints of two adjacent sides of the trapezoid are parallel to the sides of the square.
Find all interesting isosceles trapezoids and their areas.
In convex hexagon $AXBYCZ$, sides $AX$, $BY$ and $CZ$ are parallel to diagonals $BC$, $XC$ and $XY$, respectively. Prove that $\triangle ABC$ and $\triangle XYZ$ have the same area.
[i]Proposed by Evan Chen[/i]
Let $H$ be the orthocenter of $\triangle ABC$. The points $A_{1} \not= A$, $B_{1} \not= B$ and $C_{1} \not= C$ lie, respectively, on the circumcircles of $\triangle BCH$, $\triangle CAH$ and $\triangle ABH$ and satisfy $A_{1}H=B_{1}H=C_{1}H$. Denote by $H_{1}$, $H_{2}$ and $H_{3}$ the orthocenters of $\triangle A_{1}BC$, $\triangle B_{1}CA$ and $\triangle C_{1}AB$, respectively. Prove that $\triangle A_{1}B_{1}C_{1}$ and $\triangle H_{1}H_{2}H_{3}$ have the same orthocenter.
The diagonals of a trapezoid $ ABCD$ intersect at point $ P$. Point $ Q$ lies between the parallel lines $ BC$ and $ AD$ such that $ \angle AQD \equal{} \angle CQB$, and line $ CD$ separates points $ P$ and $ Q$. Prove that $ \angle BQP \equal{} \angle DAQ$.
[i]Author: Vyacheslav Yasinskiy, Ukraine[/i]
Let $M, N$ be points inside the angle $\angle BAC$ usch that $\angle MAB\equiv \angle NAC$. If $M_{1}, M_{2}$ and $N_{1}, N_{2}$ are the projections of $M$ and $N$ on $AB, AC$ respectively then prove that $M, N$ and $P$ the intersection of $M_{1}N_{2}$ with $N_{1}M_{2}$ are collinear.
Two circles $ C_1,C_2$ with different radii are given in the plane, they touch each other externally at $ T$. Consider any points $ A\in C_1$ and $ B\in C_2$, both different from $ T$, such that $ \angle ATB \equal{} 90^{\circ}$.
(a) Show that all such lines $ AB$ are concurrent.
(b) Find the locus of midpoints of all such segments $ AB$.
A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. THe remainder of the yard has a trapezoidal shape, as shown. The parallel sides of the trapezoid have lengths $ 15$ and $ 25$ meters. What fraction of the yard is occupied by the flower beds?
[asy]unitsize(2mm);
defaultpen(linewidth(.8pt));
fill((0,0)--(0,5)--(5,5)--cycle,gray);
fill((25,0)--(25,5)--(20,5)--cycle,gray);
draw((0,0)--(0,5)--(25,5)--(25,0)--cycle);
draw((0,0)--(5,5));
draw((20,5)--(25,0));[/asy]$ \textbf{(A)}\ \frac18\qquad
\textbf{(B)}\ \frac16\qquad
\textbf{(C)}\ \frac15\qquad
\textbf{(D)}\ \frac14\qquad
\textbf{(E)}\ \frac13$
Let $H$ be the orthocenter of $\triangle ABC$. The points $A_{1} \not= A$, $B_{1} \not= B$ and $C_{1} \not= C$ lie, respectively, on the circumcircles of $\triangle BCH$, $\triangle CAH$ and $\triangle ABH$ and satisfy $A_{1}H=B_{1}H=C_{1}H$. Denote by $H_{1}$, $H_{2}$ and $H_{3}$ the orthocenters of $\triangle A_{1}BC$, $\triangle B_{1}CA$ and $\triangle C_{1}AB$, respectively. Prove that $\triangle A_{1}B_{1}C_{1}$ and $\triangle H_{1}H_{2}H_{3}$ have the same orthocenter.
In triangle $ABC$, $\omega$ is its circumcircle and $O$ is the center of this circle. Points $M$ and $N$ lie on sides $AB$ and $AC$ respectively. $\omega$ and the circumcircle of triangle $AMN$ intersect each other for the second time in $Q$. Let $P$ be the intersection point of $MN$ and $BC$. Prove that $PQ$ is tangent to $\omega$ iff $OM=ON$.
[i]proposed by Mr.Etesami[/i]