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29

Suppose we define the corner-power of a triangle as follows: A triangle gets $1$ corner-power point for each corner of the square that it contains excluding the vertices of the triangle. A triangle gets $\frac{1}2$ of a corner-power point for each corner of the square that is also a vertex of the triangle. It should be clear that, in order for a ...


6

$540^\circ$ is coterminal with $180^\circ$, so I would just say it's "coterminal with a straight angle". But angles with more than $360^\circ$ are not called straight.


5

Maybe the following will help. It's not an answer, but it's too long for a comment. Suppose the given distance is $d$. Then $L$ must be simultaneously tangent to three spheres of radius $d$, one centered at each of the given points. Consider the case where the distances between some pair of the points is $D>2d$. The radius $d$ spheres can be nestled ...


4

Let me briefly explain the case of $3$ dimension: Find the plane contains those $3$ points. In this plane, find the circumcenter of the triangle formed by these $3$ points, then the straight line perpendicular to this plane and passing through the circumcenter is the required line.


3

Here is something to get you started: Place a point $Q$ on the segment $AB$ such that $AB$ and $PQ$ meet at right angles: Let us introduce the variables $s=|AD|,x=|AQ|,y=|PQ|$. Then $s$ is the side length of the square. We then must have $$ \begin{align} x^2+y^2&=4\\ (s-y)^2+x^2&=1\\ (s-x)^2+y^2&=9 \end{align} $$ From the first equation we ...


3

Thomas Kalinowski has done the hard work in proving the result for finite $I$. To complete the argument, let $\mathscr{F}=\{X_i:i\in I\}$, where $I$ is arbitrary and $X_i$ is as in his answer. Each $X_i$ is a closed strip of constant width and slope $-x_i$. Fix distinct $i,j\in I$. Then $x_i\ne x_j$, so $X_i\cap X_j$ is a closed rhomboid $R$. Let ...


3

Manually seems fine to me. First draw the triangle. Consider the integral points on the line $x=1$. The point on the hypotenuse is $(1,40)$. The number of points inside the triangle are $39$, that is, $\{1,2\dots,39\}$. Similarly number of integral points on $x=2$ are $38$, that is $\{1,2,\dots,38\}$. So basically you have to sum up $39+38+\dots+1 = ...


3

I am not sure if you would agree that this is in the same "style" as Euclid's Elements -- style is something of an ill-defined term -- but I would nominate Hilbert's Foundations of Geometry (or, for purists, Grundlagen der Geometrie). It is definitely "synthetic" (as opposed to "analytic") and unequivocally rigorous. On the other hand, while it definitely ...


2

Though it is much more inclusive than Euclid, I have a fondness for Coxeter's "Introduction to Geometry." It is definitely not elementary.


2

Put $t=0.7$ in $$B + (A-B)t$$ to get the new point.


2

It should depend on your definition of straight angle. The standard definition of straight angle, insofar as I'm aware, is one of measure $180^\circ$. In that sense, no, $540^\circ$ is not a straight angle simply due to being a strictly different rotation than $180^\circ$. However, as others have pointed out, $180^\circ$ and $540^\circ$ are coterminal. That ...


2

Really, 540˚ is not an angle, but a measure of rotation, as an angle is the measurement of the final product, after the rotation has been applied. A rotation of 540˚ will give an angle of 180˚, but is not the same thing. If you were to rotate a dial by 540˚, It would give you a different result than if you only rotated it 180˚ - a straight angle.


2

Ultimately this is (as Thomas Andrews comments) a question of definitions, like asking "Is $1$ prime?" The "right answer" depends on what definition (of "numerical value of angle") best conforms to the desired usage, or is most convenient for stating and proving theorems. Here are sample arguments in favor of each answer: According to Wikipedia, an angle ...


2

In any dimension, this is a sphere. Suppose WLOG that $u=(1,0\ldots,0)$. The condition that a vector makes angle $\alpha$ with you is that the dot product is $\cos\alpha$ (this is essentially the definition of the angle). If $x=(x_1,\ldots,x_n)$ this condition simply means that $x_1=\cos\alpha$. Therefore the set we are talking about is defined by the ...


2

The angle is the same as 180, unless you are using some device that counts turns. Such are known in the more esoteric parts of geometry, but for ordinary geometry, you just done a 360 spin and facing the right direction.


2

You basically have it, but $\|u\|^2 \ge 0$ is not "given by definition". Rather, it follows from the fact that the square of a real number is nonnegative, and $\|u\|$ is a real number . It would be just fine if you said that $\|v\|^2 + \|u\|^2 \ge \|v\|^2$ since $\|u\|^2 \ge 0$.


2

For finite $I$ this is a consequence of Helly's Theorem. Without loss of generality your given lines are parallel to the $y$-axis, i.e., they have equations $x=x_i$ for $i\in I$. The segments have then the form $$s_i=\{(x_i,y)\ :\ y_i'\leqslant y\leqslant y_i''\}$$ The set of lines $y=ax+b$ intersecting segment $s_i$ is parameterized by the set ...


2

let $AB = DC = 1, AD = a.$ the by the rule sine applied to the triangle $ABD,$ we have $$\frac a{\sin x} = \frac 1{\sin 3x}\to a = \frac{\sin x}{\sin 3x} = \frac{\sin x}{\sin x \cos 2x + \sin 2x \cos x} = \frac{1}{\cos 2x + 2\cos^2x} $$ apply the rule of sine to the triangle $ABC,$ then $$\frac{1+a}{\sin 7x} =\frac1{\sin 9x}$$ so we need to solve $$\sin ...


2

Express α, β, γ in terms of a, b, c According to the Encyclopedia of Triangle Centers, the orthocenter $X(4)$ has barycentric coordinates $[\tan A:\tan B:\tan C]$. From the cosine law you have $\cos C=\frac{a^2+b^2-c^2}{2ab}$ and likewise for the other angles. So you get \begin{align*} \tan C&=\frac{\sin C}{\cos C}=\frac{\sqrt{1-\cos C^2}}{\cos C} ...


2

This is a different way to obtain the key equality $2a\alpha = c^2-b^2$. We will use notion of power of a point with respect to a circle. On the one hand, power of $G$ with respect to a circle centered at $I$ and radius $\frac b2$ equals $GI^2-\left(\frac b2\right)^2=\frac{c^2-b^2}{4}$. On the other hand, this circle passes through $C$ and $D$ so the ...


2

So turns out it is possible to do this question with Pythagoras theorem alone! I found the proof online here However I do not imagine my self ever coming up with this in an exam. Regardless thank you String and Archaick for helping me out.


2

Construction: CH is produced to cut the circumference at X. BX, when joined, cuts QP produced at M. The pink angles are all $30^0$. OM is the perpendicular bisector of BX. This makes the angles marked red are all equal to $30^0$. The above is sufficient to say $\angle AQH = \angle APQ = 60^0$


2

Yes. In fact, $\operatorname{SO}(2)$ is isomorphic to $S^1$ (the circle) as a subset of $\mathbb{C}$. An amusing way of seeing this is to consider all the $2\times 2$ matrices of the form $$Z:=\begin{bmatrix}a&-b\\b&a\end{bmatrix}$$ for any choice of $a,b\in\mathbb{R}$. Exercise: check that the set of all such matrices is closed under addition and ...


2

Note that you have an oriented angle: $(u,v)$ and $(v,u)$ will be distinct angles unless $v = \pm u.$ That problem disappears if you start considering pairs of unit vectors in $\mathbb R^3$ and $SO(3).$ I believe the notion originates with Xavi Hernandez of FC Barcelona.


1

In the depicted configuration, we have that $OJ$ bisects $\widehat{COA}$ and $OK$ bisects $\widehat{COB}$, hence: $$\widehat{JOK}=\frac{1}{2}\widehat{AOB}=50^\circ.$$


1

(This was intended to be a comment but was too long.) Here is another approach, but a bit messy. We can calculate how much of the perimeter of the square an equilateral triangle can hold. There are two cases, one where a side of the triangle is a side of the square which gives that one equilateral triangle can hold only length 1. The other case is where a ...


1

This is NOT a proof. As suggested by @Blue, I think the following sketch meets what the post describes. In which, I is the in-center of ⊿ABC with solid colored lines as the angle bisectors. The dotted color lines are the corresponding parallels. The object is to prove that these dotted lines are concurrent at I’. The strategy is to “shrink” ⊿ACB to ...


1

The main idea is to express everything in terms of the triangle sides and then obtain something trivial (or at least easy-to-prove). For this, just use the property that the angle bisector divides the opposite side in the ratio of the neighbour sides i.e. $\frac{AE}{EC}=\frac{AB}{BC}$. Then (with the usual notations for triangle sides) you get ...


1

Pick $G$ on $AB$ such that $AG=AE$. It follows that the angles $\angle AFG$ and $\angle AFE$ are equal. They are labelled $\alpha$ in the figure. Other angles $\beta$ and $\mu$ are also labelled. Now, $\angle ACB=60$ implies that $\alpha+\beta=120$: $$ \alpha+\beta=180-(\angle FAB+\angle FBA)=180-\frac{1}{2}(\angle CAB+\angle CBA)=180-\frac{1}{2}(180-60). ...


1

Given two points $a=(a_1,\dots,a_n),b=(b_1,\dots,b_n)\in L$, there is always another $c=(c_1,\dots,c_n)\in L$ with all $c_k\in\mathbb R^d$ different from all $a_i,b_j\in\mathbb R^d$. One can draw a path in $L$ from $a$ to $(c_1,a_2,\dots,a_n)$, then another to $(c_1,c_2,a_3,\dots)$, and so on till one reaches $c=(c_1,\dots,c_n)$. Note that all intermediate ...



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