# Tag Info

6

For acute triangles the relation $\tan(A)+\tan(B)+\tan(C)=\tan(A)\tan(B)\tan(C)$ can be combined with the AM-GM inequality. If $s$ is the sum of tangents and $p$ the product one gets $(s/3)^3 \geq p = s$ so that $s^2 \geq 3^3$ with equality for equilateral triangles. For obtuse triangles the sum of tangents is negative, and for right triangles the sum is ...

6

A hint: $3^\circ=45^\circ+30^\circ-72^\circ$. For $72^\circ$ however you have to know how to construct a regular pentagon or some golden ratio triangle.

5

It is quite clear that the bell-shaped curve is formed by an arc of the ellipse of parametric equation $(r\cos\theta,t\sin\theta)$, traced by point $B$, and by an arc of the envelope of the family of lines $BC$ and $AB$. In the following I'll assume $t\ge r$ and consider only the curve for $x\ge0$, as it is symmetrical around the $y$ axis. It is then enough ...

4

From basic trigonometry we get that that the $x$ coordinates tends to... $$a=\lim_{N \to \infty} \sum_{n=0}^{N} \frac{d}{2^n}\cos (45)$$ Here $\frac{d}{2^n}$ denotes the magnitude of the $n+1$th hypotenuse (we start the sum at $0$ that is why it's a bit weird with $n+1$). How did I get this? You implied the distances (hypotenuses) went $d,d/2,d/4$ and I ...

3

By Heron's formula $$16\Delta^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c) \tag{1}$$ and by the AM-GM inequality $$(-a+b+c)(a-b+c)(a+b-c) \leq \left(\frac{a+b+c}{3}\right)^3 \tag{2}$$ Equality is attained only at $a=b=c$.

3

Flip the whole image through the symmetry median of the $20-80-80$ triangle. The outer part forms a large equilateral triangle. Construct a smaller equilateral triangle outside the $80-80$ edge. Find the reflective and rotational symmetry, and check that $\alpha$ is half of $20^\circ$. Edit: changed diagram to name points

3

Following E.Girgin suggestion plus Richmond's method: Steps: Construct a regular hexagon $ABCDEF$ with centre $O$; Construct a square in such a way that $\widehat{AOG}=75^\circ$; let $B'=BC\cap OG$ and take $A'\in FA$ such that $OA'\perp OB'$; Let $\Gamma$ be the circle with centre $O$ through $A'$ and $M$ be the midpoint of $OA'$; let $N$ be the ...

3

X must be a right angle because it faces a $180°$ arc. Therefore, $z=180-90-25=65°$. What you have is correct. You know that y must be $90°$ because it also faces an $180°$ arc, and because triangle ACD is isosceles, W must be $\frac{180-90}{2}$ or $45°$.

2

As noted, we have $$a+b=12$$$$a^2+b^2=100$$ As you mentioned, this can be turned into a polynomial. However, notice that $$(a+b)^2-a^2-b^2=2ab=44$$Thus $$ab=22$$ and by taking the half of this we can find the desired area. Sure, there are two solutions. Namely that $$a=6-\sqrt{14}, b=6+\sqrt{14}$$ or $$a=6+\sqrt{14}, b=6-\sqrt{14}$$But notice that in both ...

2

First of all, the angle of view of a line segment are two arcs and Z are on these arc and also it is on a line, that is the $y=0$ line. Your task is to figure out the formula for the arcs and then find the intersections of $y=0$ on it. Edit: hint, you need to use the Central Angle Theorem to find the centre of the arc/circle. I do not understand why $ABZ$ ...

2

First, it should be obvious that if $A$ is a fixed vertex and $B$ moves in a straight line, then $AB$ cannot be constant. Equivalently stated, if $A$ is fixed and $AB$ is constant, then the locus of $B$ is a circle with $AB$ as radius. We conclude that the only reasonable interpretation of the question is to suppose $A$ is fixed and $B$ moves in a straight ...

2

The triangles $PAM,PLB$ are similar. Both are half an equilateral triangle. So $\frac{PM}{PA}=\frac{PL}{PB}=\frac{\sqrt3}{2}$. Note that if you reflect $PAM$ in the line $PM$ you get an equilateral triangle side $PA$. So $AM$ is half $PA$ and you can then get $PM$ by Pythagoras' Theorem. Hence the $\frac{\sqrt3}{2}$. So ...

2

First we write a formula for the sum of the angle measures of triangle PAB. I'm going to just use "APB" as "angle APB". We note that angle ABP is "O": $$O + \frac{A}{2} + APB = 180$$ Next we write a formula for the sum of the angle measures of triangle ABC: $$2\cdot O + A = 180$$ This gives you that $$O = \frac{180-A}{2}.$$ We substitue this value for "O" in ...

1

Taking @Frank000's suggestion to give a full answer for those that wish to see one possible method. Using the fact that $$\displaystyle \frac{c}{2} - \frac{ab}{c} = 7$$ we rearrange to find $$c^2 -14c = 2ab.$$ Note also that $a+b=72-c$, squaring this gives $$a^2 + 2ab + b^2 = 72^2 -144c +c^2,$$ since $a^2 + b^2 =c^2$ and we have an expression for $2ab$, ...

1

Check out tetration: https://en.wikipedia.org/wiki/Tetration#Iterated_powers_vs._iterated_exponentials The actual function you are describing is on that page as well, and is called a nested exponential. Tetration is a special case of a nested exponential. For the record, pentation is the next operation up (it's tetration nested like tetration is ...

1

$$4\Delta = \sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$$ by Heron's formula, hence by dividing both sides by $4\Delta^2$ we get (since $\frac{a}{2\Delta}=\frac{1}{h_a}$): $$\frac{1}{\Delta} = ... 1 It is incorrect First of all, Z(x, 0) Now for every possible value of x, a line can be made that is intersecting line AB at 45  degree. i think you mean point A and B is seen at 45 degrees. i got the answer this way. m_1 = \frac{7+1}{3+3} m_1 = \frac{4}{3} tan 45 = \frac{-m_1 + m_2}{1 + m_1*m_2} 1 + m_1*m_2 = { + m_2 -m_1} ... 1 Let E be a point inside the triangle such that EB=EC=BC. \qquad\qquad\qquad Since \angle{ABE}=\frac{180^\circ-20^\circ}{2}-60^\circ=20^\circ, we can see that \triangle{ABD} and \triangle{BAE} are congruent. Hence,$$\angle{DBC}=\angle{ABC}-\angle{ABD}=\angle{ABC}-\angle{BAE}=80^\circ-10^\circ=\color{red}{70^\circ}.$$1 If H is the intersection between AF and CD, then GF=(1/2)FH so that GF=(1/3)GH, that is: GF=(1/3)AG. If AM and FN are the altitudes of triangles AEO and FEO with respect to EO as common base, then AGM and FGN are similar triangle and FN=(1/3)AM. It follows that Area_{EOF}={1\over3}Area_{EOA}. On the other hand triangles AEG ... 1 One can always look for a smarter proof, but unfortunately a geometric proof is unlikely to be available because with the help of more Galois theory than I want to go into one can show that no geometric construction (with ruler and compasses) will allow you to construct the triangle given only the lengths of the angle bisectors. That strongly suggests that ... 1 Apologies about the lack of a diagram. It is important that you supply it. Let ABC be a triangle, and let M_a be the midpoint of side BC. Let \theta=\angle AM_aB. Then \angle AM_aC=180^\circ -\theta. Let m_a be the length of the median AM_a. By the Cosine Law applied to \triangle AM_aB we have$$c^2=(a/2)^2+m_a^2-2(a/2)(m_a)\cos\theta.$$... 1 your method is correct. Good job sorry it took so long, it's hard to review things not in math Jax I highly recommend you learn 1 I trust you know that \angle BDC=2 \angle A  Construct an isosceles \triangle BDC . Drop perpendicular from D to BC. See that a=2R \cdot \sin (\frac {1}{2} \angle BDC)  Substitute the first equality I mentioned 1$$A+B+C=180°B=90°A>=2C+30$$Hence,$$A+C=90°A>=2C+30°$$Let us set A=2C+30°, getting 2C+30°+C=90°, so C=20° and A=70°. If C is decreased then A, will still meet the equals below "Hence," so A>=70°. 1 You're off to a good start! Some people learn better geometrically, however I'll do this algebraically. Like you said, we have two angles we're trying to find the "size" of. Let's call them \angle a and \angle b. You've already pointed out that$$\angle a+\angle b=90^\circ Now, one of these angles must be smaller than the other (or the same size if ...

1

Here’s a somewhat different way to define these functions that might make the relationships to sides of triangle clearer. Consider the unit circle with center $O$ and two rays from $O$ that intersect the circle at points $A$ and $B$, with an acute angle $\theta=\angle AOB$ between them, as shown below. Instead of defining the trigonometric functions as ...

1

Every right angle with have distinct sides a, b, and hypotenuse h, always with the condition $a^2 + b^2 = h^2$. Each triangle with an (a,b,h) sides will have a distinct angle with with a one-to-one coorespondence between angle $\theta$ and the set of $(a/h, b/h, 1)$ where $(a/h, b/h, 1)$ represents a class of similar right triangles-- similar up to a ...

1

On the straight line $AB$, let $D$ and $E$ be two distinct points such that $\frac{AD}{DB}=\frac{1}{2}=\frac{AE}{EB}$. On the straight line $BC$, let $F$ and $G$ be two distinct points such that $\frac{BF}{FC}=\frac{2}{3}=\frac{BG}{GC}$. Consider the intersection points $Q$ and $R$ of the circle with diameter $DE$ with the circle with diameter $FG$. Then, ...

1

If you assume that the sides have been labelled so that $a^2 + b^2 = c^2$, then $2c^2 = 200$ becomes $c = 10$ and $a + b = 12$ So $144 = (a+b)^2 = a^2 + b^2 + 2ab = 100 + 2ab$. So $ab = 22$ the area of the triangle is $\dfrac 12 ab = 11$.

1

The parametrization that you found requires $u\ge0$, $v\ge0$, and $u+v\le1$. To handle the case $u+v\ge1$, you can apply the same parameterization but replacing coefficient $u$ with $1-u$ and coefficient $v$ with $1-v$ (since in that case $1-u+1-v\le1$, while $1-u$ and $1-v$ continue to be nonnegative). Then the coefficient $1-u-v$ becomes $u+v-1$. This ...

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