Found problems: 259
In a triangle, the edges are extended past both vertices by the length of the edge opposite to the respective vertex.
Show that the area of the resulting hexagon is at least $13$ times the area of the original triangle.
For $i=1,2,$ let $T_i$ be a triangle with side length $a_i,b_i,c_i,$ and area $A_i.$ Suppose that $a_1\le a_2, b_1\le b_2, c_1\le c_2,$ and that $T_2$ is an acute triangle. Does it follow that $A_1\le A_2$?
Given a rectangle $ABCD$ such that $AB = b > 2a = BC$, let $E$ be the midpoint of $AD$. On a line parallel to $AB$ through point $E$, a point $G$ is chosen such that the area of $GCE$ is
$$(GCE)= \frac12 \left(\frac{a^3}{b}+ab\right)$$
Point $H$ is the foot of the perpendicular from $E$ to $GD$ and a point $I$ is taken on the diagonal $AC$ such that the triangles $ACE$ and $AEI$ are similar. The lines $BH$ and $IE$ intersect at $K$ and the lines $CA$ and $EH$ intersect at $J$. Prove that $KJ \perp AB$.
In the rectangle $ABCD, M, N, P$ and $Q$ are the midpoints of the sides. If the area of the shaded triangle is $1$, calculate the area of the rectangle $ABCD$.
[img]https://2.bp.blogspot.com/-9iyKT7WP5fc/XNYuXirLXSI/AAAAAAAAKK4/10nQuSAYypoFBWGS0cZ5j4vn_hkYr8rcwCK4BGAYYCw/s400/may3.gif[/img]
Give an acute-angled triangle $ABC$ with area $S$, let points $A',B',C'$ be located as follows: $A'$ is the point where altitude from $A$ on $BC$ meets the outwards facing semicirle drawn on $BX$ as diameter.Points $B',C'$ are located similarly. Evaluate the sum $T=($area $\vartriangle BCA')^2+($area $\vartriangle CAB')^2+($area $\vartriangle ABC')^2$.
Median $AM$ and the angle bisector $CD$ of a right triangle $ABC$ ($\angle B=90^o$) intersect at the point $O$. Find the area of the triangle $ABC$ if $CO=9, OD=5$.
Prove that inside any convex hexagon with pairs of parallel sides of area $1$, you can draw a triangle of area $1/2$.
Let $ABCD$ be a trapezium (trapezoid) with $AD$ parallel to $BC$ and $J$ be the intersection of the diagonals $AC$ and $BD$. Point $P$ a chosen on the side $BC$ such that the distance from $C$ to the line $AP$ is equal to the distance from $B$ to the line $DP$.
[i]The following three questions 1, 2 and 3 are independent, so that a condition in one question does not apply in another question.[/i]
1.Suppose that $Area( \vartriangle AJB) =6$ and that $Area(\vartriangle BJC) = 9$. Determine $Area(\vartriangle APD)$.
2. Find all points $Q$ on the plane of the trapezium such that $Area(\vartriangle AQB) = Area(\vartriangle DQC)$.
3. Prove that $PJ$ is the angle bisector of $\angle APD$.
In a triangle $ABC$ the area is $18$, the length $AB$ is $5$, and the medians from $A$ and $B$ are orthogonal. Find the lengths of the sides $BC,AC$.
Triangle $ABC$ has $AB=27$, $AC=26$, and $BC=25$. Let $I$ denote the intersection of the internal angle bisectors of $\triangle ABC$. What is $BI$?
$ \textbf{(A)}\ 15\qquad\textbf{(B)}\ 5+\sqrt{26}+3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{26}\qquad\textbf{(D)}\ \frac{2}{3}\sqrt{546}\qquad\textbf{(E)}\ 9\sqrt{3} $
Let $ABC$ be an equilateral triangle with side length $10$ and $P$ be a point inside the triangle such that $|PA|^2+ |PB|^2 + |PC|^2 = 128$. What is the area of a triangle with side lengths $|PA|,|PB|,|PC|$?
$
\textbf{(A)}\ 6\sqrt 3
\qquad\textbf{(B)}\ 7 \sqrt 3
\qquad\textbf{(C)}\ 8 \sqrt 3
\qquad\textbf{(D)}\ 9 \sqrt 3
\qquad\textbf{(E)}\ 10 \sqrt 3
$
Prove that any rigid flat triangle $T$ of area less than $4$ can be inserted through a triangular hole $Q$ with area $3$.
Let $p, q$, and $r$ be the angles of a triangle, and let $a = \sin2p, b = \sin2q$, and $c = \sin2r$. If $s = \frac{(a + b + c)}2$, show that
\[s(s - a)(s - b)(s -c) \geq 0.\]
When does equality hold?
Point $P$ is inside $\triangle ABC$. Line segments $APD$, $BPE$, and $CPF$ are drawn with $D$ on $BC$, $E$ on $AC$, and $F$ on $AB$ (see the figure at right). Given that $AP=6$, $BP=9$, $PD=6$, $PE=3$, and $CF=20$, find the area of $\triangle ABC$.
[asy]
size(200);
pair A=origin, B=(7,0), C=(3.2,15), D=midpoint(B--C), F=(3,0), P=intersectionpoint(C--F, A--D), ex=B+40*dir(B--P), E=intersectionpoint(B--ex, A--C);
draw(A--B--C--A--D^^C--F^^B--E);
pair point=P;
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("$P$", P, dir(0));[/asy]
Given two acute triangles $\vartriangle ABC, \vartriangle DEF$. If $AB \ge DE, BC \ge EF$ and $CA \ge FD$, show that the area of $\vartriangle ABC$ is not less than the area of $\vartriangle DEF$
Points $A$, $B$, and $C$ lie on a circle of radius $5$ such that $AB=6$ and $AC=8$. Find the smaller of the two possible values of $BC$.
A rectangle has sides of length $a$ and $36$. A hinge is installed at each vertex of the rectangle and at the midpoint of each side of length $36$. The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length $a$ parallel and separated by a distance of $24,$ the hexagon has the same area as the original rectangle. Find $a^2$.
[asy]
pair A,B,C,D,E,F,R,S,T,X,Y,Z;
dotfactor = 2;
unitsize(.1cm);
A = (0,0);
B = (0,18);
C = (0,36);
// don't look here
D = (12*2.236, 36);
E = (12*2.236, 18);
F = (12*2.236, 0);
draw(A--B--C--D--E--F--cycle);
dot(" ",A,NW);
dot(" ",B,NW);
dot(" ",C,NW);
dot(" ",D,NW);
dot(" ",E,NW);
dot(" ",F,NW);
//don't look here
R = (12*2.236 +22,0);
S = (12*2.236 + 22 - 13.4164,12);
T = (12*2.236 + 22,24);
X = (12*4.472+ 22,24);
Y = (12*4.472+ 22 + 13.4164,12);
Z = (12*4.472+ 22,0);
draw(R--S--T--X--Y--Z--cycle);
dot(" ",R,NW);
dot(" ",S,NW);
dot(" ",T,NW);
dot(" ",X,NW);
dot(" ",Y,NW);
dot(" ",Z,NW);
// sqrt180 = 13.4164
// sqrt5 = 2.236
[/asy]
Let $ABC$ be a triangle, and $M$ be the midpoint of $BC$, Let $N$ be the point on the segment $AM$ with $AN = 3NM$, and $P$ be the intersection point of the lines $BN$ and $AC$. What is the area in cm$^2$ of the triangle $ANP$ if the area of the triangle $ABC$ is $40$ cm$^2$?
In a convex pentagon $ABCDE$, the area of the triangles $ABC, ABD, ACD$ and $ADE$ are equal and have the value $F$. What is the area of the triangle $BCE$ ?
Points $A, B,$ and $C$ are on a circle such that $\triangle ABC$ is an acute triangle. $X, Y ,$ and $Z$ are on the circle such that $AX$ is perpendicular to $BC$ at $D$, $BY$ is perpendicular to $AC$ at $E$, and $CZ$ is perpendicular to $AB$ at $F$. Find the value of \[ \frac{AX}{AD}+\frac{BY}{BE}+\frac{CZ}{CF}, \] and prove that this value is the same for all possible $A, B, C$ on the circle such that $\triangle ABC$ is acute.
[asy]
pathpen = linewidth(0.7);
pair B = (0,0), C = (10,0), A = (2.5,8); path cir = circumcircle(A,B,C);
pair D = foot(A,B,C), E = foot(B,A,C), F = foot(C,A,B), X = IP(D--2*D-A,cir), Y = IP(E--2*E-B,cir), Z = IP(F--2*F-C,cir); D(MP("A",A,N)--MP("B",B,SW)--MP("C",C,SE)--cycle); D(cir); D(A--MP("X",X)); D(B--MP("Y",Y,NE)); D(C--MP("Z",Z,NW)); D(rightanglemark(B,F,C,12)); D(rightanglemark(A,D,B,12)); D(rightanglemark(B,E,C,12));[/asy]
Determine all integers $ n > 3$ for which there exist $ n$ points $ A_{1},\cdots ,A_{n}$ in the plane, no three collinear, and real numbers $ r_{1},\cdots ,r_{n}$ such that for $ 1\leq i < j < k\leq n$, the area of $ \triangle A_{i}A_{j}A_{k}$ is $ r_{i} \plus{} r_{j} \plus{} r_{k}$.
Suppose $a,b,c$ are three rela numbers such that the quadratic equation \[ x^2 - (a +b +c )x + (ab +bc +ca) = 0 \] has roots of the form $\alpha + i \beta$ where $\alpha > 0$ and $\beta \not= 0$ are real numbers. Show that
(i) The numbers $a,b,c$ are all positive.
(ii) The numbers $\sqrt{a}, \sqrt{b} , \sqrt{c}$ form the sides of a triangle.
Point $M$ is the midpoint of the side $AB$ of an acute triangle $ABC$. Circle with center $M$ passing through point $ C$, intersects lines $AC ,BC$ for the second time at points $P,Q$ respectively. Point $R$ lies on segment $AB$ such that the triangles $APR$ and $BQR$ have equal areas. Prove that lines $PQ$ and $CR$ are perpendicular.
The sides of a triangle are $25,39,$ and $40$. The diameter of the circumscribed circle is:
$ \textbf{(A)}\ \frac{133}{3}\qquad\textbf{(B)}\ \frac{125}{3}\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 41\qquad\textbf{(E)}\ 40 $
Let a, b, c be the sidelengths and S the area of a triangle ABC. Denote $x=a+\frac{b}{2}$, $y=b+\frac{c}{2}$ and $z=c+\frac{a}{2}$. Prove that there exists a triangle with sidelengths x, y, z, and the area of this triangle is $\geq\frac94 S$.