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The formula of finding the volume of a tetrahedron is this: \(\dfrac{a^3}{6 \sqrt2}\) where a is the side length of the tetrahedron. So the volume of this tetrahedron is \(\dfrac{1}{6 \sqrt2}=\dfrac{\sqrt2}{12}\)

And now, we're done!  Hope this helped!

Let's let the final length of the triangle be equal to x. By the Triangle inequality, 

24+49>x

24+x>49

49+x>24.

The first inequality gives us the equation x<73. The second equation gives up x>25. The final equation gives up x>-25. From these equations, we know that 25 <73. So the smallest possible length of x is 26. So the perimeter of the triangle is 26+24+49=99. 

And now we're done!  Hope this helped!

Let the center of the smaller semi-circle be \(X\) and the center of the small circle be \(O\). 
\(\triangle BOX\) is a right angle triangle with right angle at \(B\). So:

\(BO^2+BX^2=OX^2\\ (2-r)^2+1^2=(1+r)^2\\ 6r=4\\ \boxed{r=\displaystyle\frac{2}{3}}\)

Let the center of the small circle be O, and the point where circle O is tangent to the circle with diameter AC be D, and let the radius of circle O be r. Lastly, let the midpoints of AB and BC be points M and N, respectively.

Since B is the center of a semicircle, BD must also be the same length as AB (because they are the radius) = 2.

MO = 1 + r, and MB = 1. OMB forms a right triangle with angel OBM = 90 degrees, since O lies on the perpendicular bisector of AC.

This means that OB, by the pythagorean theorem, has length \(\sqrt{(1+r)^2-1}=\sqrt{r^2+2r}\).

Additionally, DO = r, and lines on the same line as OB. Recalling that DB = 2, we can set an equation: DO + OB = 2:

\(r + \sqrt{r^2+2r}=2\)

\(\sqrt{r^2+2r}=2-r\)

\(r^2+2r=(2-r)^2=r^2-4r+4\)

\(2r = -4r+4\)

\(6r=4\)

\(r={2\over3}\)

Because triangle ABC is a right triangle with B = 90 degrees, then we know that the pythagorean theorem is a possible strategy.

AB = 3, and BC = 4, and they are both the legs of the right triangle because they include the right angle, and so by pythagorean theorem:

\(AC^2=AB^2+BC^2=3^2+4^2=25\)

So, AC = 5 units.

Thanks, I can confirm this is the correct answer, admist this confusion. 

We can verify proyaop's answer through similar triangles \(\triangle BEF \sim\triangle ECD\), so to check we make sure \(\frac{BF}{EF}=\frac{ED}{CD}\), by proyaop's answer, \(BF=3-\frac{3}{4}=\frac{9}{4}\), \(EF=\frac{3}{4}\) , \(ED=\frac{3}{4}\), \(CD=1-\frac{3}{4}=\frac{1}{4}\), so \(\frac{\frac{9}{4}}{\frac{3}{4}}=\frac{\frac{3}{4}}{\frac{1}{4}}=3\).

Asinus did nothing majorly wrong, i think they misread the side lengths as 1 and 2 .  

The probability that at most one of them shows tails = the probability that none of them show tails (all of them are heads) + the probability that one of them shows tails.

For any n, the probability that none of them shows tails is \({(\frac{1}{2})}^{n}\), because heads would needed to be landed on every single time.

The probability that one of these show tails is more difficult. Suppose the first one is tails, and the rest are heads, but we also note that any of these can be the one that lands on heads, so there are n ways to arrage these, so overall the probability is \(n*{(\frac{1}{2})}^{n}\).

Adding these together - \(n*{(\frac{1}{2})}^{n}+{(\frac{1}{2})}^{n}=(n+1){(\frac{1}{2})}^{n}=\frac{5}{16}\), \((n*1)*(\frac{1}{{2}^{n}})=\frac{n+1}{{2}^{n}}=\frac{5}{16}\). We see one such value that satisfies our equality is n = 4. But we know the rate of increase for exponential functions is exponetially increasing/decreasing, so after that the gap between n+1 and 2ⁿ, will only constantly increase, therefore our only positive solution is 4.

 Find AE.

Hello magenta!

Cosine law:\(c^2=a^2+b^2-2ab\ cos\ \gamma\)

\(\overline{AE}\ ^2=3^2+5^2-2\cdot 3\cdot 5\cdot cos\ 60^{\circ}\\ \overline{AE}=\sqrt{9+25-30\cdot \frac{1}{2}}=\sqrt{19}\\ \color{blue}\overline{AE}=4.359\)

 !

Through construction, AB = AF + FB = 3.

AF = EF = x. Since ADEF is a square, DE is parallel to AB, and thus angle CED is congruent to angle CBA by corresponding angles, and ADE = angle EFB because they are both 90 degrees. Thus, triangle CDE is similar to EFB is similar to CAB, and CA : AB = 1 : 3, likewise, FE : FB = 1 : 3. Substituting, we have x : FB = 1 : 3, and therefore FB = 3x. Substituting this into the original equation, AB = x + 3x = 4x = 3. And thus, x = 3/4. The area of the square would be x^2 = 9/16.

If you construct a unit circle at a point, say (1,1), you will see that it satisfies the given constructions. 

The radius of a unit circle is just 1 unit.

If angle BAC = angle ABC = 60 degrees, then angle BCA = 60 degrees by triangle interior angle sum, and thus triangle ABC is equilateral. The angle bisector of the apex of an isosceles triangle (which includes the equilateral triangle) is also the height of the triangle. Since AD = 24, and angle A is being bisected, AD is also the height with length 24. By 30 - 60 - 90, the base of triangle ABC has length BC = 8sqrt(3), and the area is b*h/2 = 96sqrt(3) units^2.

Find the area of the square.

Hello magenta!

\(1:3=x:(3-x)\\ 3x=3-x\\ 4x=3\\ x=\frac{3}{4}\\ \color{blue}A=\frac{3}{4}\cdot \frac{3}{4}=\frac{9}{16}\)

  !

The answer is \(16 \sqrt 3\)

\(\angle CAX=\angle CBX\) (isoceles triangle)
In an isoceles triangle, the angle bisector is also the altitude and median so:
\(AX=BX=7.5\)
\(\triangle CAX=(AX)(CX)/2\)
\(\triangle CBX=(BX)(CX)/2\)
Since \(AX=BX\):

\(\boxed{\frac{\triangle BCX}{\triangle ACX}=1}\)

As follows:

Hi Juriemagic,

A smart little kid could use trial and error.

We can rewrite the given equation as:

ab−c2a+b−c​+bc−a2b+c−a​+ca−b2c+a−b​=1.

Notice that the left-hand side of the equation resembles the sum of the reciprocals of three numbers, but with some extra terms. Let's manipulate the equation to exploit this similarity.

First, we can factor the quadratic terms in the denominators:

(a−c)(b)a+b−c​+(b−a)(c)b+c−a​+(c−b)(a)c+a−b​=1.

Then, we can multiply both sides of the equation by the common denominator, which is the product of all the denominators:

(a−c)(b)⋅(a−c)(b)a+b−c​+(b−a)(c)⋅(b−a)(c)b+c−a​+(c−b)(a)⋅(c−b)(a)c+a−b​=(a−c)(b)⋅(b−a)(c).

Expanding both sides, we get:

a(b+c−a)+b(c+a−b)+c(a+b−c)=(a−c)(b)(c−a).

This simplifies to:

3abc=abc−a2b−ab2−a2c−ac2−b2c.

Since a+b+c=1, we can substitute:

3abc=abc−(a2+b2+c2)−(ab+ac+bc).

We know that (a+b+c)2=a2+b2+c2+2(ab+ac+bc), so:

3abc=abc−(a+b+c)2+2(ab+ac+bc).

Substituting a+b+c=1, we get:

3abc=abc−1+2(ab+ac+bc).

Finally, we can factor out abc to find:

2abc=abc−1.

Solving for abc, we get:

abc = 1/3.

Therefore, the answer is abc = 1/3.

First, we expand the product in the given equation: k = (6x+12)(x−8) = 6x2−36x+96.

We see that a=6, and b=−36, so the least possible value of k occurs when x = 2a−b ​ = 2⋅6−(−36) ​= 3​.

Setting x = 3 into the given equation, we see that this value does indeed result in the least possible value of k:

k = 6(3)+(−36)+96 = 150−108 = 42​.

Therefore, the answer is 42.