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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

set $t_a=\tan A$ etc. for any pair of angles $A,B$ we have $$ \tan(A+B) = \frac{t_a+t_b}{1-t_at_b} $$ so for any three angles $A,B,C$: $$ \tan(A+B+C)= \frac{ \frac{t_a+t_b}{1-t_at_b} +t_c}{1-\frac{t_a+t_b}{1-t_at_b}t_c} $$ $$ =\frac{t_a+t_b+t_c -t_at_bt_c}{1-(t_at_b+t_bt_c+t_ct_a)} \tag{1} $$ if, in addition, we know that: $$ A+B+C = n\pi \tag{2} $$ then ...


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

Your problem is in stating that $\cos \beta = \frac {\sqrt {7}} {14}$. This is incorrect. Although it is true that $\cos^2 \beta = \frac {7} {196}$, there are two possible values for $\cos \beta$. If you were to make a scale drawing, you would find that AB is over twice as long as Ac, making $\beta$ an obtuse angle. Thus $\cos \beta = -\frac {\sqrt {7}} ...


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

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

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

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 ...


2

According to this, it was open when Unsolved Problems in Number Theory was published, which was probably in the eighties, although I can't find the publication date for the life of me. Two rational medians, rational sides, and rational area is possible according to this source. Your problem though in fact was still open in 2004 for the third edition of ...


2

Reflect at the green line to obtain a triangle, which must be equilateral.


2

Use Law of sines, $$\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}$$ Let, $\angle{A}=60^o$ $\angle{B}=90^o$ $\angle{C}=30^o$ $a=?$ $b=?$ $c=10cm$ From Law of sines we have, $$\frac{a}{\sin{A}}=\frac{c}{\sin{C}}$$ Put the values and find $a$. Now, from Law of sines we have, $$\frac{a}{\sin{A}}=\frac{b}{\sin{B}}$$ Put the known values and ...


2

In the first case, we may assume that the angle between the $a$ and $b$ sides is $\theta$. Then: $$ h_a = b\sin\theta,\qquad h_b = a\sin\theta \tag{1}$$ and by the law of cosines: $$ c^2 = a^2+b^2-2ab\cos\theta = \frac{h_a^2+h_b^2-2h_a h_b\cos\theta}{\sin^2\theta}\tag{2}$$ so we just have to find the stationary points of the function: $$ p(\theta) = ...


2

$$ 0 = \tan 180^\circ = \tan (A+B+C) $$ Apply to the line above the formula for the tangent of a sum of two numbers. You can take the two numbers to be $A$ and $B+C$, so $$ \tan(A+(B+C)) = \frac{\tan A + \tan(B+C)}{1-\tan A\tan(B+C)}. $$ Then apply it again to $\tan(B+C)$. Do the routine simplifications. You get a fraction. A fraction is zero only if the ...


2

Without loss of generality, we only need to show that $|z_2-z_1|\le |z_2|+|z_1|$. We will use the notation that $z_1=|z_1|e^{i\theta_1}$ and $z_2=|z_2|e^{i\theta_2}$ and $z^{*}$ will denote the complex conjugate of $z$. So, we write $$\begin{align} |z_2-z_1|^2&=(z_2-z_1)(z_2-z_1)^*\\\\ &=|z_2|^2+|z_1|^2-2\text{Re}\{z_1z_2^*\}\\\\ ...


2

Your question actually has a lot of answers. Because we can! Having unrestricted $\sin$ and $\cos$ functions does no harm to people who only need the restricted versions, so why not? As noted by other answers and comments, they do work fine for obtuse, or reversed, triangles, even if it is possible to manage without them. For a thought experiment, you ...


2

If you think of the graph of $\sin(x)$ it's a nice periodic function, the graph is a wave. It's very useful in physics (for example) to have functions that model wave behavior. For this you need to allow the angle to go for multiple cycles. Without such functions things like Fourier analysis would be impossible.


2

What is triangle $\triangle ABC$ called in relation to triangle $\triangle DEF$? According to Wikipedia, the inner triangle is called the Gergonne triangle, contact triangle or intouch triangle of the outer. What is triangle $\triangle DEF$ called in relation to triangle $\triangle ABC$? I don't know an answer to this yet, but searching the web ...


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

I think we get $$a+b<c+h$$ Squaring we obtain $$a^2+b^2+2ab<c^2+h^2+2hc$$ thus we have $$2ab<h^2+2hc$$ $$2hc<h^2+2hc$$ because $ab=ch$ (area formulas) and we get $$h^2>0,$$ which is true.


2

As you wrote the radius $R$ of the circumcircle of $\triangle{ABC}$ is $\sqrt{21}$. Since the radius of the circumcircle of $\triangle{ADC}$ is also $R=\sqrt{21}$, by the law of sines, $$2R=\frac{AD}{\sin{\angle{ACD}}}\Rightarrow AD=\sqrt{21}.$$ Then, applying the law of cosines to $\triangle{ACD}$ gives you $$21=3^2+CD^2-2\cdot 3\cdot CD\cdot ...


2

Not an answer, but information that may help lead to one. With the help of Mathematica to push some symbols, I've determined that the coordinates of the "Draksis Triangle" $\triangle A^\prime B^\prime C^\prime$ is obtained from those of $\triangle ABC$ thusly: $$\begin{align} A^\prime &= \frac{1}{2}\;A + \frac{\cos(\beta/2) ...


2

Candidate quantities can be found in inequalities for triangles in which equality occurs just for equilateral triangles. For example, the isoperimetric inequality for triangles asserts that the ratio $p^2/A$ (where $p$ is the perimeter and $A$ the area) is minimal for the equilateral triangle (among triangles). This quantity, then, measures deviation from ...


2

I would imagine that the quantity "circumradius/inradus" would be a good candidate. This quantity will be larger for slender triangles. Intuitively one would expect this to be minimized by an equilateral triangle at a measurement of 2.


2

Set up the circles so that their centers are on a horizontal line, one to the left and one to the right, and the centers $A$ and $B$ are distance $R$ apart, with $A$ being the left one. There are two intersection points between the circles, an upper, $U$, and a lower, $V$. Then the distance from $A$ to $U$ is $R$, as is the distance from $B$ to $U$. And ...


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 $\angle$ ...


1

Since $CD \perp AB$ and $AE \perp BC$, we have that $ADEC$ to be a cyclic quadrilateral. This means $\angle{DCE} = \angle{DAE}$ and $\angle{DCA} = \angle{DEA}$. This means $$\angle{C} = \angle{DCE} + \angle{DCA} = \underbrace{\angle{DAE} + \angle{DEA} = \angle{EDB}}_{\text{Exterior angle of }\Delta ADE}$$ By a similar argument, $\angle{A} = \angle{BED}$. ...


1

Possible way: let vectors $a,\ b,\ a-b$ be the sides of the triangle. Then the two altitudes, corresponding to $a,b$ are $-a+\frac{(a,b)}{b^2}b,\ -b+\frac{(a,b)}{a^2}a$. Let's find the third as $ta+(1-t)b$: $$(ta+(1-t)b)(a-b)=0$$ $$ta^2-t(a,b)+(1-t)(a,b)-(1-t)b^2=0$$ $$t(a^2-(a,b)-(a,b)+b^2)+(a,b)-b^2=0$$ $$t=\frac{-b(a-b)}{(a-b)^2}$$ And the altitude: ...


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

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*} ...



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