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33

Here's a friendly equilateral triangle: The sides are all of the same length - let's say $a$. The angles are all the same too, and since the angles must add up to $180^\circ$, we conclude that the three angles in the equilateral triangle are equal to $180^\circ/3=60^\circ$. Now we do something sneaky. We draw a line all the way down from the top ...


9

Consider the areas of triangles. Let $[ABC]$ be the area of a triangle $ABC$. Since $[OAB]=[OAC]+[OBC]$, one has $$\frac{1}{2}ab\sin120^\circ=\frac{1}{2}ac\sin60^\circ+\frac 12bc\sin60^\circ.$$ This leads $$ab=ac+bc.$$ Now divide the both sides by $abc$.


6

Here's a geometric argument. Construct the equilateral triangles $ODC,OEC$, like in the figure. Then you have $$ \frac{c}{a}+\frac{c}{b} = \frac{OD}{OA}+\frac{OE}{OB} = \frac{BC}{AB}+\frac{AC}{AB} = 1$$ (note that $DC || OB$ and $CE || OA$)


5

This statement follows from the theorem: If $BC$ is the hypotenuse of a right-angled triangle $\triangle ABC$, it follows that the median $AM$ (which corresponds to the hypotenuse) is $AM = \dfrac{BC}{2}$. Try to apply some basic geometry to the triangles, which are created.


5

We have $$\dfrac{\sqrt3}2 \cos(x) + \dfrac12 \sin(x) = \dfrac14 \implies \sin(x+\pi/3) = \dfrac14$$ We hence obtain $x+\pi/3 = \arcsin(1/4)$ or $x+\pi/3 = \pi - \arcsin(1/4)$ Note that $\arcsin(1/4) < \arcsin(1/2) = \dfrac{\pi}6 < \dfrac{\pi}3$. Since $x \in (0,\pi)$, we have $$x+\pi/3 = \pi - \arcsin(1/4) \implies x = \pi - ...


4

By the law of cosines, one has $$\begin{align}a\cos A=b\cos B&\Rightarrow a\cdot\frac{b^2+c^2-a^2}{2bc}=b\cdot\frac{c^2+a^2-b^2}{2ca}\\&\Rightarrow a^2(b^2+c^2-a^2)=b^2(c^2+a^2-b^2)\\&\Rightarrow (a^2-b^2)(a^2+b^2-c^2)=0\end{align}$$


3

In general, if the vertex angle is $2\theta$ and $OC$ is the angle bisector, since $$\text{Area of triangle }OAB = \text{Area of triangle }OAC + \text{Area of triangle }OBC$$ we have $$\dfrac{OA \cdot OB \cdot \sin(2\theta)}2 = \dfrac{OA \cdot OC \cdot \sin(\theta)}2 + \dfrac{OC \cdot OB \cdot \sin(\theta)}2$$ $$ab\sin(2\theta) = ac\sin(\theta) + ...


3

Each path with finitely many steps has a finite number of small wiggles, with the size of the latter being inversely proportional to the number. So in a very informal sense, the "limit path" has infinitely many infinitesimally small wiggles, such that the total amount of "wiggle" is positive but finite.


3

Let BC = 1 unit. Then, $4 = y = [\triangle CBQ] = \frac {1.QC}{2}$ yields QC = 8 units Let DQ = s units. Then AB = s + 8 units $5 = z = [\triangle ABP] = \frac {(s+8).AP}{2}$ yields $AP = \frac {10}{s + 8}$ units $DP = 1 – AP = … = \frac {s – 2}{s + 8}$ Similarly, $3 = (\frac {1}{2}) {s}{\frac{ s – 2}{s + 8}}$. The above result is a solvable quadratic ...


3

Hint: Consider the line $\ell$ orthogonal to the segment $AB$ and passing through its midpoint. What can you tell about the triangle $\Delta ABC$ when $C$ is a point of $\ell$? Can you use this to find what you're looking for?


3

Yes. Choose an arbitrary value for a side of the triangle. Use the Law of Sines to find the other two sides. Then use Heron's formula $$A=\sqrt{s(s-a)(s-b)(s-c)}$$ where $$s=\frac{a+b+c}{2}$$ to find the area of your test triangle. Or you could use $$A=\frac 12ab\sin\theta$$ Either way, use that area to rescale your test triangle to the proper size. ...


3

I'd proceed like this: Prove that, in 2D, the set of all points defined by a fixed ratio $\rho\neq1$ of their distances to two distinct given points (which I will call poles for reasons not explained here) is a circle $K$. (I'd use equations with cartesian coordinates for that, but if you find a coordinate-free reasoning, that would be interesting.) ...


3

We must have: $$ a\leq b\leq c\leq N,\quad 2a^2+2b^2-c^2 = (2d)^2 $$ but $c$ must be even, so $c=2C$ and: $$ a^2+b^2=2(d^2+C^2).\tag{1}$$ Now it is useful to recall that the elements of the set $\square+\square$ are exactly the $n$s for which the multiplicity of every prime divisor of the form $4k+3$ is even. Moreover, $n=a^2+b^2$ with $\gcd(a,b)=1$ (also ...


3

Let $r$ be the radius of the circle. The area of the triangle is $\frac12 (base \times height)$, if you split each of the triangles down the middle (to give a right angled triangle) then the height will be given by $$height = r \cos(\tfrac\pi n),$$ notice how we have halfed the angle as we split the triangle in 2. Similarly the base will be given by $$ base ...


3

Pascal's triangle is based on an identity for choosing unordered samples taken without replacement (no item sampled more than once). Binomial coefficients. The number of unordered samples of size $k$ from a larger set of size $n$ is $$C(n, k) = {}_n C_k = {n \choose k} = \frac{n!}{k!(n-k)!},$$ for integers $0 \le k \le n.$ I have shown three common ...


2

Let angle BCD be $\theta$ and let $x$ be the distance between the two poles. Then, $tan(\theta) = x/20$ $(1)$ Let $\alpha$ be the angle ACB. Then, $tan(\theta+ \alpha) = x/18$ We can evaluate $\alpha$, since $tan(\alpha) = \dfrac{2}{77}$, thus $\alpha = 1.488$ Therefore, $tan(\theta + 1.488) = x/18$ $(2)$ Equating (1) and (2) gives, $20 ...


2

My suggestion is to use the following: Your triangle is isosceles, so the hieght issued from the vertex C is the perpendicular bisector of the base AB. The midpoint of AB has the following coordinates I = ($-\frac{1}{2}$, 0). So the point C pass through the st line (D) passing through I and perpendicular to $(AB)$, and (D) has equation ...


2

This is a tricky question because it did not specify which side should be considered as base. Separating into 2 different cases, two sets of answers are therefore expected. Case 1 (figure 1) AB = 5, and M, the midpoint of AB is at $(\frac {-1}{2}, 0)$ $10 = \frac {5.CM}{2}$ yields $CM = 4$ $CM$ is the perpendicular bisector of $AB$ and its equation is ...


2

Here's one way to do it: Step 1: Determine the planes determined by the two triangles. Step 2a: If they are the same plane, this has become a two dimensional problem. Step 2b: If the planes are different there are many options. Step 3: The vertices of triangle 1 cannot all be on the same side of the plane determined by triangle 2. Similarly, the ...


2

If an equilateral triangle has side length $d$, the length of every heigth is given by $\frac{\sqrt{3}}{2}d$ by the pythagorean theorem, hence the area is $\frac{\sqrt{3}}{4}d^2$ and the perimeter is $3d$. Hence, if the area equals $\sqrt{3}$, the perimeter equals $6$.


2

Draw DE where E lies on AB at a length of $x$ units from A and $y$ units from B. Then, triangles ABD and AED will be congruent. (2 equal sides and the angle between them is also equal). Now, consider the triangle DEC. The sides EC and ED are equal (both equal to $y$). Thus, angles EDC and ECD will be equal. $\angle$ EDC = 2$\alpha - 120^{\circ}$ and $\angle$ ...


2

You have rediscovered one of the paradoxes of limits, namely, that a sequence of paths in the plane can converge to a limit path in the sense that the maximum distance to the limit path from any point on your path converges to zero, yet the lengths of the paths do not converge to the length of the limit path. To put it simply, the detour through $C$ gives a ...


2

First locate the centroid and then draw the cevian through it from the given vertex, which should meet the opposite side at its midpoint, which can be found because the centroid divides the cevian in a ratio of 2 to 1. The opposite side is perpendicular to the altitude from the given vertex through the orthocentre, and the two vertices on that side can be ...


1

Since $B ,C , B+C < \frac{\pi}{2} $ we have $\frac{\cos B }{\cos (B +C )} >1$ and $\frac{\cos C }{\cos (B+C )} >1$ hence $$|\tan A|=\tan (B+C) =\frac{\cos B}{\cos (B+C)} \cdot \sin C + \frac{\cos C}{\cos (B+C)} \cdot \sin B >1\cdot \sin C +1\cdot \sin B =\sin C +\sin B .$$


1

Seems to be an embedding of the Fano plane (7 points) into a larger projective plane (with 13 points). A projective plane consists of a set of lines, a set of points, and a relation between points and lines called incidence, having the following properties: Given any two distinct points, there is exactly one line incident with both of them. Given any two ...


1

The reason why you keep on getting a + b as a result is that the formula that explains your procedure is: Let n be the number of steps $$ c = {\frac {a*n}{n}}+{\frac {b*n}{n}} $$ With n=2 we have your first example. The bigger n, the smaller the step, but the higher the amount of steps. If it approaches infinite, we have your "infinite number of steps, ...


1

Another approach: Use the law of sines and a double angle identity: \begin{align*} a\cos{A} &= b\cos{B} \\ \Rightarrow \frac{a\cos{A}}{\sin{A}} &= \frac{b\cos{B}}{\sin{A}} \\ \Rightarrow \frac{b\cos{A}}{\sin{B}} &= \frac{b\cos{B}}{\sin{A}} \\ \Rightarrow \cos{A}\sin{A} &= \cos{B}\sin{B} \\ \Rightarrow \sin{2A} &= \sin{2B} \end{align*} ...


1

Let $G$ be the centroid of $\triangle ABC$ and let $M$ be the midpoint of the side $BC$. Now since one has $AG:GM=2:1$ (do you need the proof?), one has $$\vec{OG}=\frac{1\times \vec{OA}+2\times \vec{OM}}{2+1}=\frac{\vec{OA}+2\times \frac{\vec{OB}+\vec{OC}}{2}}{3}=\frac{\vec{OA}+\vec{OB}+\vec{OC}}{3}.$$


1

First of all, note that $EA=AB$. Take a point $G$ on $DB$ such that $\angle{GAB}=20^\circ$. Since $\angle{ABG}=\angle{AGB}=80^\circ$, one has $AB=AG$. So, since $\triangle EAG$ is an equilateral triangle, one has $AG=GE$. Since $\angle{GAD}=\angle{GDA}=40^\circ$, one has $AG=GD$. Hence, one has $GE=GD$. So, since ...


1

Fantastic problem! Thanks for sharing. I solved it with Euler Math Toolbox (see below). However, I'd like to see a more geometric solution. Probably, I am just blind right now. The method of the solution is straightforward. $xh$ is $x$ depending on $a$ and $b$ (computing the area as side times side divided by 2). $fh$ is $f$ depending on $a$, $b$, $z$ and ...



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