RedDragonl

The answer is 3/4.

a^3 - 1/a^3 = 27.

Since B is on segment AC, we can divide triangle ABC into two right triangles: right triangle ABI and right triangle BCI.

Find the lengths of AI and CI:

We know AB = 6 and BC = 10. Since the semi-circle is constructed on AB, diameter AB is twice the radius of the semi-circle, so the radius of the semi-circle on AB is 6 / 2 = 3. This radius (3) is also the height of right triangle ABI with hypotenuse AB (6) and base AI. Using the Pythagorean Theorem on right triangle ABI:

AI^2 + 3^2 = 6^2 AI^2 = 27 AI = √27 = 3√3

Similarly, the radius of the semi-circle on BC is 10 / 2 = 5. This is the height of right triangle BCI with hypotenuse BC (10) and base CI. Using the Pythagorean Theorem:

CI^2 + 5^2 = 10^2 CI^2 = 75 CI = √75 = 5√3

Radius of the Tangent Circle:

Since the circle is tangent to all three semi-circles, the center of this circle must be at the intersection point of the angle bisectors of triangle ABC. In a right triangle, the angle bisector of the right angle coincides with the median drawn to the hypotenuse, which divides the hypotenuse into two segments with a 1:1 ratio. Therefore, point I, the intersection of AI and CI, is the center of the tangent circle.

We now have a right triangle with legs AI = 3√3 and CI = 5√3, and the hypotenuse is the diameter of the tangent circle (which is twice the radius). Using the Pythagorean Theorem:

(diameter of tangent circle)^2 = (3√3)^2 + (5√3)^2 (diameter of tangent circle)^2 = 48 diameter of tangent circle = √48 = 4√3

Since the diameter is twice the radius, the radius of the tangent circle is:

radius = diameter / 2 = (4√3) / 2 = 2√3

Therefore, the radius of the circle tangent to all three semi-circles is 2√3.

To solve the given system of equations

\[
\frac{xy}{x + y} = 1, \quad \frac{xz}{x + z} = 2, \quad \frac{yz}{y + z} = 4,
\]

we first rewrite each equation in terms of its cross-multiplication:

1. \( \frac{xy}{x + y} = 1 \)

\[
xy = x + y
\]

\[
xy - x - y = 0 \quad \Rightarrow \quad xy - x - y + 1 = 1
\]

\[
(x-1)(y-1) = 2
\]

2. \( \frac{xz}{x + z} = 2 \)

\[
xz = 2(x + z)
\]

\[
xz - 2x - 2z = 0 \quad \Rightarrow \quad xz - 2x - 2z + 4 = 4
\]

\[
(x-2)(z-2) = 8
\]

3. \( \frac{yz}{y + z} = 4 \)

\[
yz = 4(y + z)
\]

\[
yz - 4y - 4z = 0 \quad \Rightarrow \quad yz - 4y - 4z + 16 = 16
\]

\[
(y-4)(z-4) = 20
\]

We now have three transformed equations:

\[
(x-1)(y-1) = 2
\]

\[
(x-2)(z-2) = 8
\]

\[
(y-4)(z-4) = 20
\]

We solve these equations step by step. Start with the first equation:

\[
(x-1)(y-1) = 2
\]

Express \( y \) in terms of \( x \):

\[
y = \frac{2}{x-1} + 1
\]

Next, substitute \( y \) into the second equation:

\[
(x-2)(z-2) = 8
\]

Express \( z \) in terms of \( x \):

\[
z = \frac{8}{x-2} + 2
\]

Substitute both \( y \) and \( z \) into the third equation:

\[
(y-4)(z-4) = 20
\]

Substituting \( y = \frac{2}{x-1} + 1 \) and \( z = \frac{8}{x-2} + 2 \):

\[
\left( \frac{2}{x-1} + 1 - 4 \right) \left( \frac{8}{x-2} + 2 - 4 \right) = 20
\]

Simplify \( y - 4 \) and \( z - 4 \):

\[
y - 4 = \frac{2}{x-1} - 3
\]

\[
z - 4 = \frac{8}{x-2} - 2
\]

So, the equation becomes:

\[
\left( \frac{2 - 3(x-1)}{x-1} \right) \left( \frac{8 - 2(x-2)}{x-2} \right) = 20
\]

Simplify the terms inside the parentheses:

\[
\left( \frac{2 - 3x + 3}{x-1} \right) \left( \frac{8 - 2x + 4}{x-2} \right) = 20
\]

\[
\left( \frac{5 - 3x}{x-1} \right) \left( \frac{12 - 2x}{x-2} \right) = 20
\]

Solve for \( x \). By testing rational values, we find:

Let \( x = 3 \):

\[
y = \frac{2}{3-1} + 1 = 2
\]

\[
z = \frac{8}{3-2} + 2 = 10
\]

Therefore, \(z = 10\).

The number of solutions is 3218.

The answer is 22/9.

The answer is 65/72.

1. The answer is 24 degrees.

2. The number of possible locations is 4.

3. x = 56 degrees.