RedDragonl

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Questions 81
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 #3
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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.

Jun 4, 2024
 #1
avatar+352 
+1

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, \(z = 10\).

Jun 1, 2024
 #1
avatar+352 
+1

To find the probability that intervals \(I\) and \(J\) overlap, we need to consider the possible positions of the four points \(x_1, x_2, x_3,\) and \(x_4\) within the interval \([0, 1]\).

 

Without loss of generality, let's assume that \(x_1 < x_2\) and \(x_3 < x_4\).

 

For \(I\) and \(J\) to overlap, one of the following conditions must be true:

 

1.  \(x_1 < x_3 < x_2 < x_4\)


2.  \(x_3 < x_1 < x_4 < x_2\)

 

Let's calculate the probability for each condition:

 

1.  Probability of Condition 1:


   - The probability that \(x_1\) falls in the interval \([0, 1]\) is \(1\).


   - The probability that \(x_3\) falls in the interval \([x_1, 1]\) is \(1 - x_1\).


   - Given \(x_1\) and \(x_3\), the probability that \(x_2\) falls in the interval \((x_1, 1]\) is \(1 - x_1\).


   - Given \(x_1\), \(x_3\), and \(x_2\), the probability that \(x_4\) falls in the interval \((x_3, 1]\) is \(1 - x_3\).


   - So, the probability of Condition 1 is \(1 \times (1 - x_1) \times (1 - x_1) \times (1 - x_3) = (1 - x_1)^2 (1 - x_3)\).

 

2.  Probability of Condition 2:


   - The probability that \(x_3\) falls in the interval \([0, 1]\) is \(1\).


   - The probability that \(x_1\) falls in the interval \([0, x_3]\) is \(x_3\).


   - Given \(x_3\) and \(x_1\), the probability that \(x_4\) falls in the interval \((x_3, 1]\) is \(1 - x_3\).


   - Given \(x_3\), \(x_1\), and \(x_4\), the probability that \(x_2\) falls in the interval \((x_1, 1]\) is \(1 - x_1\).


   - So, the probability of Condition 2 is \(1 \times x_3 \times (1 - x_1) \times (1 - x_3) = x_3 (1 - x_1) (1 - x_3)\).

 

Now, since \(x_1, x_2, x_3,\) and \(x_4\) are chosen independently and uniformly at random in the interval \([0, 1]\), we can find the probability that intervals \(I\) and \(J\) overlap by summing the probabilities of Condition 1 and Condition 2:

 

\[ \text{Total probability} = (1 - x_1)^2 (1 - x_3) + x_3 (1 - x_1) (1 - x_3) \]

 

\[ = (1 - x_1) (1 - x_3) (1 - x_1 + x_3) \]

 

\[ = (1 - x_1) (1 - x_3) \]

 

Now, since \(x_1\) and \(x_3\) are chosen uniformly at random in the interval \([0, 1]\), we can find the expected value of the probability by integrating over the joint distribution of \(x_1\) and \(x_3\):

 

\[ \text{Expected probability} = \int_{0}^{1} \int_{0}^{1} (1 - x_1) (1 - x_3) \, dx_1 \, dx_3 \]

 

\[ = \int_{0}^{1} \left( \int_{0}^{1} (1 - x_1) (1 - x_3) \, dx_3 \right) dx_1 \]

 

\[ = \int_{0}^{1} \left( 1 - x_1 - \frac{1}{2} + \frac{x_1^2}{2} \right) dx_1 \]

 

\[ = \left[ x_1 - \frac{x_1^2}{2} - x_1 + \frac{x_1^3}{6} \right]_{0}^{1} \]

 

\[ = \left( 1 - \frac{1}{2} - 1 + \frac{1}{6} \right) - \left( 0 - 0 - 0 + 0 \right) \]

 

\[ = \frac{1}{3} - \frac{1}{2} = \frac{1}{6} \]

 

Therefore, the expected probability that intervals \(I\) and \(J\) overlap is \( \frac{1}{6} \).

May 1, 2024