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5

This is NOT an answer but is an as-accurate-as-possible re-sketch of the original figure after guessing. Please let me know if there is any mis-interpretation. [Note: The previous diagrams have been incorrectly drawn and were therefore deleted. The one below is the most updated version. Sorry for giving some misleading info.] This time Geogebra shows ...

4

Notations: Write $a:=GH$, $b:=HF$, $c:=FG$, and $s:=\frac{a+b+c}{2}$. Let $\Omega$ and $\omega$ be the circumcircle and the incircle of $FGH$, respectively. The circle internally tangent to $FG$, $FH$, and $\Omega$ is denoted by $\Gamma$. Suppose that $\Gamma$ intersects $HF$ and $FG$ at $P$ and $Q$, respectively. Denote by $\omega_a$ the excircle ...

3

Using the Evan Chen's mixtilinear incircle article in here, the results become trivial. I will change some notations. In $\triangle ABC$, let the incircle hit $BC$ at $D$ and the $A$-mixtilinear incircle hit the circumcircle of $\triangle ABC$ at $E$. Prove that $\angle DEB = \angle ABC$. Since (9) holds, we have $\angle DTM_A = \angle AFB=180-\angle B - ... 3 I do not think that the length of BE is determined completely by the data. For one thing, the fact that CE=DE already follows from the other data. On the other hand, one can freely vary the magnitude of the angle CAE=EAD while keeping AC=AD constant at 8 cm, and still obtain a variety of lengths for BE. Try a hand drawing of the two extreme cases where ... 3 Let the tangent of circle$O$at$AB$be$M$, also let$O$'s tangent at$BD$be$X$. Let the tangent of circle$O_1$at$AC$be$N$, also let$O_1$'s tangent at$DC$be$Y$. Then obviously$IX=IM$and$IY=IN$as triangles$IXB$and$IMB$are congruent and$IY$part similarly. Connect$AI$, since$I$is the incenter of$ABC$,$AI$is also an angle ... 3 First, observe that just like @Nicholas said, the equation$\,x^2-y^2+2y=1\,defines two lines: \begin{align} x^2-y^2+2y=1 \iff x^2 = (y-1)^2 \implies \begin{cases} l_1: & y = x + 1 \\ l_2: & y = -x + 1 \end{cases} \end{align} The slope of the first one is\,\dfrac{\pi}{4} = 45º,\,$and for the second one the slope is$\,\dfrac{3\pi}{4} = ...

3

Note that the $x$ coordinate is supported on $[0:n]$. I'll assume that the points are selected uniformly and independently, and you should edit the question in case this is not true. Suppose the first point is $(X_1, Y_1)$, and the second $(X_2, Y_2)$. It's easy to see that the area is \begin{align} A &= \frac{1}{2}|X_1 Y_2 - X_2 Y_1| &\text{(half ...

3

Obviously, $\angle BAC = \angle BCA = 20^{\circ}$, and $\angle DBA = 40^{\circ}$. Choose point $E$ such that $\triangle EAB$ forms an equilateral triangle. Since $\angle DBA = 40^{\circ}$, and $\angle ABE = 60^{\circ}$, then $\angle DBE = 20^{\circ}$. Now notice: $\overline {AC} = \overline {DB}$; $\angle ACB = \angle DBE = 20^{\circ}$; and $\overline ... 3 Making a chain of$\geq2$hexagons you can realize all$k$of the form$k=10+4m$,$m\geq0$. Starting with your figure and attaching a chain of hexagons you can realize all$k=13+4m$vertices, and starting with a blob of$4$, resp.$5$, hexagons one shows that all$k$of the form$k=16+4m$or$k=19+4m$,$m\geq0$, can be realized in this way. Note that$10$, ... 3 Let$I=\{i:x_i>0\}, J=\{j:x_j<0\}$,$S=\sum_{i \in I}x_i, T=\sum_{j \in J}|x_j|$,$A=\sum_{i \in I}\frac{x_i}{i}, B=\sum_{j \in J}\frac{|x_j|}{j}$. Then we have$S=T=\frac{1}{2}$, hence$A \leq \frac{1}{2}$and$B \geq \frac{1}{2n}$and the value of the expression is$A-B$(if$A>B$as we may assume w.l.o.g). 3 Using Abel's rearrangement and$\displaystyle \sum\limits_{k=1}^{n}x_k = 0$: $$\sum\limits_{k=1}^{n}\frac{x_k}{k} = \sum\limits_{k=1}^{n-1}\left(\frac{1}{k} - \frac{1}{k+1}\right)\sum\limits_{j=1}^{k}x_j + \left(\frac{1}{n}\sum\limits_{k=1}^{n}x_k\right) = \sum\limits_{k=1}^{n-1}\frac{x_1+\cdots +x_k}{k(k+1)}$$ Thus, using Triangle inequality: ... 3 A right triangle with a 60 degree angle would be half an equilateral triangle. Hence$BC$would be half as long as$AC$, so$AC=8$. But$7^2+4^2\ne 8^2$so indeed Pythagoras tells us that this is not the case. 2 Remind that: $$\forall\vec{u},\|\vec{u}\|^2=\vec{u}\cdot\vec{u},$$ $$\forall\vec{u},\vec{v},\vec{u}\cdot\vec{v}=\|\vec{u}\|\|\vec{v}\|\cos\left(\widehat{\vec{u},\vec{v}}\right).$$ From there, notice that: ... 2 Let$\alpha = \angle AOB, \beta = \angle BOC, \gamma = \angle COA, \vec{u} = \dfrac{\vec{OA}}{||\vec{OA}||}, \vec{v} = \dfrac{\vec{OB}}{||\vec{OB}||}, \vec{w} = \dfrac{\vec{OC}}{||\vec{OC}||}\Rightarrow ||\vec{u}||=||\vec{v}||=||\vec{w}||=1, \alpha+\beta+\gamma = 2\pi \Rightarrow ||\vec{u}+\vec{v}+\vec{w}|| \leq 1\iff ||\vec{u}+\vec{v}+\vec{w}||^2 \leq ...

2

[I may have some things off by $90^\circ$ here or have $\sin$ and $\cos$ swapped, but I think you’ll get the idea.] If you rotate the stationary pipe through a full $360^\circ$, you would get this sort of funnel-like shape, where the angle between the horizontal pipe and the funnel is $180^\circ$ minus the angle by which your pipe has been bent. That’s the ...

2

Just a hint : when calculating analytically you should exact value of the distance, Which is $\sqrt{10}$, and not use the approximation. To solve your problem: A vector space is equipped with an inner product in order to being able to calculate angles between elements of that space. For $\mathbb{R}^2$ (and higher dimensions) one can use the relation ...

2

As in the other answer, draw the triangle so that its vertices have coordinates $A(0,0)$, $B(3,0)$, and $C(3,4)$. Let $P$ be the point on the hypotenuse $2$ units away from $C$. This means that $\overline{AP}$ has length $3$. Draw the vertical line down from $P$ and let $Q$ be the point at which this line intersects $\overline{AB}$. Notice that $\triangle ... 2 “$YK$is the perpendicular bisector of$AB$” implies$\alpha = \alpha_1$. Together with$\beta = \beta_1$, we have$AB // YE$. Similarly,$BC // DX$. as hinted by @HoseynHeydari . Together with the given parallels, we have$AQEY$and$XDEC$as parallelograms. In particular,$\beta = \gamma$means$B, D,Y, E$are con-cyclic.$B, D, X, E$are con-cyclic ... 2 The black lines are your dates. at first we draw green lines which are parallels to the rays in the point$M$, so we get the points$J,K$. with center in$J$and radio$AJ$we draw the red circle. similar with the point$K$so we get the purple points$B,C$and the segment$BC$is your goal. 2 The symmetric of$B$with respect to$M$is$C$, hence in order to find our triangle it is enough to intersect the$AC$line with the symmetric of the$AB$line with respect to$M$. 2 Hint : $$sin(140-A)=sin(140)cos(A)-cos(140)sin(A)=ksin(A)$$ with$k=\frac{1}{2sin(70)}$can be transformed in $$\frac{sin(140)}{k+cos(140)}=tan(A)$$ 2$\vec{AC}=c-a\vec{AR}={1\over2}(c-a)\vec{OR}={1\over2}(c-a)+a={1\over2}(c+a)\vec{AB}=b-a\vec{RB}=(b-a)-{1\over2}(c-a)=b-{1\over2}(c+a)\vec{RG}={1\over3}(b-{1\over2}(c+a))$This comes from$RG={1\over3}RB$as$\large{{RG\over GB}= {RF\over AB} = {1\over2}}\vec{OG}={1\over2}(c+a)+{1\over3}(b-{1\over2}(c+a))={1\over3}(a+b+c)$2 For fun we give a proof that uses no number-theoretic machinery. Let the sides of our Pythagorean triangle be$x,y,z$, with$z$the hypotenuse. If the area is equal to the hypotenuse, then$\frac{1}{2}xy=z$, or equivalently$2xy=4z$. Then from$x^2+y^2=z^2$we obtain $$(x+y)^2=x^2+y^2+2xy=z^2+4z.$$ It follows that$z^2+4z$is a perfect square. We show this ... 2 Have you heard of Heron's formula? Especially on the form $$A = \sqrt{\frac S2\left(\frac S2-a\right)\left(\frac S2-b\right)\left(\frac S2-c\right)},\quad a, b, c\text{ are the sides of the triangle}$$we come very close to a full solution. What we need to get all the way is the AM-GM inequality, which states that for any three positive numbers$k,l,m$, we ... 2 WLOG, we may consider$T$to be the origin. Then consider$g(z) = f(z) f(\omega z) f(\omega^2 z)$, where$\omega$is a complex cube root of$1$. Note than that$g$when evaluated on any side of the triangle, it involves the product of values of$f$from all three sides. So$|g(z)| \le 8$on the sides of the triangle, and analytic inside. Then use the ... 2 First let's assume$AB=1$and call$PB=a;PC=b;PA=c$, we have to prove that$a^2+b^2=c^2$. Applying the law of cosines to$PBC$with respect to angle$\angle BPC=\pi/6$we get $$a^2+b^2-\sqrt3ab=1$$ Now, call$\angle PCB=y$. Applying the law of cosines to$PBC$with respect to angle$y$we get $$1+b^2-2b\cos y=a^2$$ hence$\cos y=\frac{1+b^2-a^2}{2b}$. ... 2 The pictures I made with Cinderella Geometry show that the problem is stated correctly. The answer is that indeed$x=y$. (Of course, a proof is needed.) A different triangle, angles again are equal. 2 Lets say you are given one of the angles and lets call this angle$\alpha$. Lets call the side opposite$\alpha$,$a$and the side adjacent to$\alpha$,$b$. Finally lets call the area of this triangle$A$. We can then say that:$$\tan(\alpha)=\frac{a}{b}\tag{1}$$ and:$$A=\frac{1}{2}ab\tag{2}$$ From (2) we get:$$b=\frac{2A}{a}\tag{3}$$and if we substitute ... 2 You can think of a spherical triangle being on the surface of a sphere and having angles that sum greater than$\pi$, or$180^\circ$, but less than$2\pi$, or$360^{\circ}\$. That means you can have two right angles in the triangle. So one possible solution would be to have two sides perpendicular to another (visually this might be like two lines of ...

2

In general, the excess of a spherical triangle (i.e., the difference between the sum of the angles and the "ordinary" pi for planar triangles) is equal to its surface area on the unit sphere. If the sum is 5pi/3 then the area is 2pi/3. The complete sphere is 4pi and a hemisphere is 2pi, so you could use one-third of a hemisphere. This is an elaborate way of ...

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