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7

The triangle with vertexes $(-3,0),(3,0),(0,4)$ Note 1: Consider any Pythagorician triple $(a,b,c)$, then $a,b,c \in \mathbb{N}$ and $a^2+b^2=c^2$. Now consider the triangle with vertexes of coordinates $(0,0),(b,0),(0,a)$. Finally to avoid the right angle consider the triangle with vertexes $(-b,0),(b,0),(0,a)$. Clearly all coordinates are integers and ...


5

Let $O$ be the incenter of $\triangle ABC$, and let $X$, $Y$, $Z$ be its excenters. Let $O^\prime$ be the incenter of the target $\triangle A^\prime B^\prime C^\prime$. The keys to this solution these observations: $\overleftrightarrow{XA^\prime}$, $\overleftrightarrow{YB^\prime}$, $\overleftrightarrow{ZC^\prime}$ concur at $O^\prime$ $O^\prime$, as well ...


4

The key is that two reflections with respect to two non parallel lines is a rotation, where the rotation angle is twice the angle between the two non parallel lines. Consider image 1. A radial line is associated with an angle $\xi$ and we can denote a radial line as $r(\xi)$. A reflection with respect to a radial line with associated angle $\xi$ can be ...


4

You can get arbitrarily small area, since you can get arbitrarily close to the degenerate triangle $a=b=1$, $c=2$, which has zero area. The maximal possible area is obtained when forming equilateral triangle with $a=b=c=2$. (I.e., the area is $\sqrt3$.) One of possible arguments is the following: If two sides are shorter than $2$ units, you can increase ...


3

Polynomial proof Here is a rather ugly computational proof. I'm using homogeneous coordinates in the projective plane $\mathbb{RP}^2$. W.l.o.g. the incircle is the unit circle, represented by the matrix $$M=\begin{pmatrix}1&0&0\\0&1&0\\0&0&-1\end{pmatrix}$$ For some parameters $a,b,c\in\mathbb R$, the vectors $$ ...


3

The angle in question does not seem to be a rational number of degrees. It is approximately $$x\approx17.877987144333100702669230434486544610886743725521319531393624°$$ It is characterized by $$1216\,\cos^6x - 3264\,\cos^4x + 2916\,\cos^2x - 867 = 0$$ Since this is a cubic equation in $\cos^2x$ you could write down an explicit formula for the solution of ...


2

Yes, its completely logical. Alternative approach would be to find the point of intersection of two perpendiculars drawn from the vertices onto opposite sides, by finding solution of two linear equations but it'll be long of course.


1

Use AM-GM inequality,we have $$\cos^3{x}+\dfrac{\cos{x}}{4}\ge 2\sqrt{\cos^3{x}\cdot\dfrac{\cos{x}}{4}}=\cos^2{x}$$ then we have $$x_{1}+x_{3}\ge\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=2x_{2}$$ so $$x_{1}+x_{2}+x_{3}\ge 3x_{2}=\dfrac{3}{2}$$ because we have use this follow well know $$\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=1$$


1

Using complex numbers for the rational coordinates, take an initial triangle with $$A=a+0\ i,\ B=b+0\ i,\ C=0+c\ i.$$ This triangle works provided each of $a^2+c^2,\ b^2+c^2$ is a rational square. This triangle could be rescaled by multiplying all by a common denominator to make the coordinates all integers. However for now just keep the coordinates ...


1

You want to solve for $\sin(c)$ from $$ \sin(a) \sin(b) \sin(c) + \cos(a) \cos(b) = 1 $$ There are multiple approaches, but let's algebra first. Just rearanging we get $$ \sin(c) = \frac{1-\cos(a) \cos(b)}{\sin(a) \sin(b)} $$ From here I would use product-to-sum identities, noting that $$ \cos(a) \cos(b) = \frac{\cos(a-b) + \cos(a+b)}{2} $$ and $$ \sin(a) ...


1

The curves are definitely not ellipses. I'll just discuss the bottom corner, $a$ (which I'll call $F$). Coordinatize, with $A = (-p,0)$, $B = (p, 0 )$, $C = (c, h )$. Let the "lower" trisectors from $A$ and $B$ meet at the "bottom" Morley vertex, $F = (x,y)$. Then $$\tan \angle CAB = \frac{h}{p+c} \qquad \tan\angle CBA = \frac{h}{p-c}$$ $$\tan \angle FAB ...


1

Refer to the figure. 2x = 14y = 7z (given) After dividing throughout by 14, we have x/7 = y/1 = z/2 = k for some non-zero k. Then, x = 7k; y = k; and z = 2k [note:- all measures are in degrees] Angle DBC = [1] = z = 2k [2] = 180 – 2z = 180 – 4k [3] = 180 – z – [1] – y – x = … = 180 – 12k [4] = (180 – [3]) / 2 = … = 6k [5] = [4] = 6k [6] = x – ...


1

(The following solves a simpler problem; see Blue's comment.) Since both triangles have the same circle as incircle it is enough to verify that their angles agree. To this end connect the center $O$ of $ABC\,$'s incircle with the chosen tangency point $P$ and look at the angles formed between $OP$ and the angle bisectors of $ABC$. These angles reappear ...


1

As mentioned in my comment, triangles have a special connection with ellipses, namely, for every triangle there is a unique ellipse such that the ellipse circumscribes the triangle, and the triangle circumscribes another ellipse matching the first except smaller by a factor of $\frac12$ in each dimension. Each ellipse can circumscribe many triangles which ...


1

Using the Law of sines: $\dfrac {\sin A}{|BC|} = \dfrac{\sin B}{|AC|} = \dfrac{\sin C}{|AB|}=D$ ($D$ is the diameter of the circumcircle) we see that \begin{align}\dfrac{\sin A}{\sin B} =\dfrac{\vert BC\vert}{\vert AC\vert}= \dfrac56\\\dfrac{\sin B}{\sin C} =\dfrac{\vert AC\vert}{\vert AB\vert}= \dfrac45\end{align} So $\displaystyle \dfrac{\vert ...


1

Your conclusion that $\triangle PVY$ has area $40$ is correct, and your reasoning — that $\triangle PVY$ and $\triangle PYW$ have the same height and bases in ratio $4:3$ — is correct. (A detail: $PY$ is not necessarily the height of these triangles, since it is not necessarily perpendicular to $VW$. But the perpendicular distance from $P$ to $VW$ is the ...


1

Let $(d_a,d_b,d_c)$ be the distances of your point from the sides. You have to minimize the quantity: $$ d_a^2+d_b^2+d_c^2 $$ under the constraint: $$ a d_a + b d_b + c d_c = 2\Delta,$$ hence Lagrange multipliers gives that $(d_a,d_b,d_c)=\lambda(a,b,c)$, so your stationary point is the isogonal conjugate of the point having trilinear coordinates ...


1

HINT: Any point on the curve $y=x^2$ can be expressed as $(t,t^2)$ So, $\triangle ABC$ will be the absolute value of $$\frac12\begin{vmatrix}0 & 2 & t\\-10 &0 &t^2\\1 &1&1\end{vmatrix}$$ On simplification this will be a Quadratic equation in $t$ Now, for $\displaystyle A>0, ...


1

You have proved that the proposed condition is equivalent to $$ (s-a)^2+4(s-b)^2+9(s-c)^2=\frac{36}{49}s^2\tag 1 $$ Note that, by the Cauchy-Schwarz inequality, we have $$\eqalign{ s&=(s-a)+\frac{1}{2}(2(s-b))+\frac{1}{3}(3(s-c))\cr &\leq\sqrt{(s-a)^2+4(s-b)^2+9(s-c)^2}\sqrt{1+\frac{1}{4}+\frac{1}{9}}\cr ...


1

$a+b=x$, $a.b=y$ and it is found $m(\widehat{ACB})=\frac{2\pi }{3}$ since $% x^{2}-c^{2}=y$. Then area of the triangle is $S(ABC)=\frac{1}{2}ab\sin (% \frac{2\pi }{3})=\frac{abc}{4R}=\frac{a+b+c}{2}r$, where $R$ and r are circumradius and inradius respectively. Then you can easily calculate $R=% \frac{c}{\sqrt{3}}$ and $r=\frac{\sqrt{3}y}{2(x+c)}$. Finally, ...


1

Let L be a line through $P$ that crosses both $AC$ and $XY$. Let $S=L\cap AC$ and $T=L\cap XY$. Triangles $SCP$ and $PTY$ are similar, so are $ASP$ and $PXT$ and the ration of lengths is given by $SP/PT$. In particular $$\frac{AS}{SC}=\frac{XT}{TY}$$ Since the ratio $AS/SC$ runs from $0$ to $\infty$ as $L$ change among the lines crossing $AC$ and $XY$, ...


1

If by ''sides'' you mean the length of the sides: it's not possible. Imagine a triangle made out of cardboard (all sides/angles fixed) that's fixated with a nail on one end point. You can still rotate the triangle and thereby move the position of the other two endpoints. If you mean the vectors that make up the sides, you can calculate the missing points by ...


1

$2x + 17 < 5x + 5 \iff 12 < 3x \iff x > 4$. The question asks when this could be a possible answer, which it could (if $x > 4$). Note how the other answers are never possible ($x+2$ will never be greater than $x+3$, $5x+6$ will never be smaller than $5x+5$ for positive $x$)



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