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Punkte1218
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 #4
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+1
24.09.2024
 #1
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0

To solve this problem, we need to understand the relationships within the triangle STU. Here are the steps to find the length of \( SX \):

 

### Step 1: Analyzing the Triangle


Given:


- \( S \), \( T \), and \( U \) are the vertices of the triangle.


- \( M \) is the midpoint of \( ST \).


- \( N \) is a point on \( TU \) such that \( SN \) is the altitude of the triangle.


- \( ST = SU = 13 \), \( TU = 8 \).


- \( UM \) and \( SN \) intersect at \( X \).

 

### Step 2: Applying the Median and Altitude Properties


Since \( M \) is the midpoint of \( ST \), \( SM = MT = \frac{13}{2} = 6.5 \).

 

Also, \( SN \) is an altitude, so it is perpendicular to \( TU \).

 

### Step 3: Use the Property of the Centroid


In any triangle, the centroid (intersection of the medians) divides each median in a 2:1 ratio. Since \( X \) is the intersection of the medians \( SN \) and \( UM \), it is the centroid of triangle \( STU \).

 

This implies:


\[
SX = \frac{2}{3} \times SN
\]


where \( SN \) is the altitude from \( S \) to \( TU \).

 

### Step 4: Calculate SN Using the Area of the Triangle


We use the fact that the area of the triangle can be calculated in two ways:


1. Using base \( TU \) and height \( SN \).


2. Using Heron's formula.

 

#### Heron's Formula:


First, calculate the semi-perimeter \( s \):


\[
s = \frac{ST + SU + TU}{2} = \frac{13 + 13 + 8}{2} = 17
\]


Then, calculate the area \( \Delta \):


\[
\Delta = \sqrt{s(s - ST)(s - SU)(s - TU)} = \sqrt{17(17 - 13)(17 - 13)(17 - 8)} = \sqrt{17 \times 4 \times 4 \times 9} = \sqrt{2448} = 24
\]

 

#### Area Using Altitude \( SN \):


The area can also be written as:


\[
\Delta = \frac{1}{2} \times TU \times SN = \frac{1}{2} \times 8 \times SN = 4 \times SN
\]


Equating the two expressions for the area:


\[
24 = 4 \times SN \implies SN = 6
\]

 

### Step 5: Calculate SX


Now that we know \( SN = 6 \), the length of \( SX \) is:


\[
SX = \frac{2}{3} \times 6 = 4
\]

 

Thus, the length of \( SX \) is \( \boxed{4} \).

14.08.2024
 #2
avatar+1218 
0

Given points \( A = (4, -4) \), \( B = (3, 8) \), and \( C = (-1, 2) \), and a point \( Q \) such that for any point \( P \) in the plane, the following equation holds:

 

\[
PA^2 + PB^2 + PC^2 = 3PQ^2 + k
\]

 

We need to find the constant \( k \).

 

### Step 1: Write down the expression for the sum of squares of distances


Let \( P = (x, y) \) and \( Q = (x_Q, y_Q) \). The squared distances from \( P \) to the points \( A \), \( B \), and \( C \) are:

 

\[
PA^2 = (x - 4)^2 + (y + 4)^2
\]


\[
PB^2 = (x - 3)^2 + (y - 8)^2
\]


\[
PC^2 = (x + 1)^2 + (y - 2)^2
\]

 

The sum of these squared distances is:

 

\[
PA^2 + PB^2 + PC^2 = \left[(x - 4)^2 + (y + 4)^2\right] + \left[(x - 3)^2 + (y - 8)^2\right] + \left[(x + 1)^2 + (y - 2)^2\right]
\]

 

Expanding these expressions:

 

\[
PA^2 = (x^2 - 8x + 16) + (y^2 + 8y + 16)
\]


\[
PB^2 = (x^2 - 6x + 9) + (y^2 - 16y + 64)
\]


\[
PC^2 = (x^2 + 2x + 1) + (y^2 - 4y + 4)
\]

 

Adding them together:

 

\[
PA^2 + PB^2 + PC^2 = \left[3x^2 + (-8x - 6x + 2x) + (16 + 9 + 1)\right] + \left[3y^2 + (8y - 16y - 4y) + (16 + 64 + 4)\right]
\]

 

Simplifying:

 

\[
PA^2 + PB^2 + PC^2 = 3x^2 - 12x + 26 + 3y^2 - 12y + 84
\]

 

Thus:

 

\[
PA^2 + PB^2 + PC^2 = 3(x^2 - 4x + \frac{26}{3}) + 3(y^2 - 4y + 28)
\]

 

### Step 2: Express the right-hand side


The expression for \( 3PQ^2 \) is:

 

\[
3PQ^2 = 3\left[(x - x_Q)^2 + (y - y_Q)^2\right] = 3\left[(x^2 - 2xx_Q + x_Q^2) + (y^2 - 2yy_Q + y_Q^2)\right]
\]

 

Expanding:

 

\[
3PQ^2 = 3x^2 - 6xx_Q + 3x_Q^2 + 3y^2 - 6yy_Q + 3y_Q^2
\]

 

### Step 3: Set up the equation


We equate \( PA^2 + PB^2 + PC^2 \) with \( 3PQ^2 + k \):

 

\[
3x^2 - 12x + 110 + 3y^2 - 12y + 84 = 3x^2 - 6xx_Q + 3x_Q^2 + 3y^2 - 6yy_Q + 3y_Q^2 + k
\]

 

By comparing coefficients, we get:

 

\[
-12x = -6xx_Q \quad \text{and} \quad -12y = -6yy_Q
\]

 

So:

 

\[
x_Q = 2, \quad y_Q = 2
\]

 

Now, matching the constant terms:

 

\[
110 + 84 = 3x_Q^2 + 3y_Q^2 + k
\]


\[
194 = 3(2^2) + 3(2^2) + k = 12 + 12 + k = 24 + k
\]

 

Thus:

\[
k = 194 - 24 = 170
\]

 

### Final Answer:


The constant \( k \) is \( \boxed{170} \).

10.08.2024
 #1
avatar+1218 
0

To find the probability that all three pieces of a stick, which is 6 units long, are shorter than 6 units after it is broken at two random points, we can use a geometric interpretation.

 

1. **Setting Up the Problem:**


- Let the stick start at 0 and end at 6 (it has a total length of \( L = 6 \) units).


- Denote the two break points as \( X_1 \) and \( X_2 \), where \( 0 < X_1 < X_2 < 6 \).


- The resulting pieces of the stick will have lengths:


- Piece 1: from 0 to \( X_1 \) (length \( X_1 \))


- Piece 2: from \( X_1 \) to \( X_2 \) (length \( X_2 - X_1 \))


- Piece 3: from \( X_2 \) to 6 (length \( 6 - X_2 \))

2. **Condition for the Pieces:**


- We need each piece to be less than 6 units in length:


- \( X_1 < 6 \)


- \( X_2 - X_1 < 6 \)


- \( 6 - X_2 < 6 \)


- The inequalities simplify to:


- \( X_1 < 6 \) (which will always be true since \( X_1 < X_2 < 6 \))


- \( X_2 < 6 + X_1 \) (or simply \( X_2 < 6 \) due to initial constraints)


- \( X_2 > 0 \) (will also hold since \( X_2 > X_1 > 0 \))

3. **Valid Conditions:**


- The only actual restriction is that \( X_2 < 6 \) (which is always satisfied because \( 0 < X_1 < X_2 < 6 \)).


- The lengths of the pieces will always be less than the length of the original stick (6 units).

4. **Geometric Approach:**


- We represent the breaking points \( (X_1, X_2) \) in the coordinate plane (in the region where \( 0 < X_1 < X_2 < 6 \)).


- The total area of the triangle formed by these restrictions in this coordinate system is represented by:


\[
\text{Area of valid region} = \frac{1}{2} \times 6 \times 6 = 18
\]


- However, the actual region that must be considered is bounded by the triangle formed by \( (0,0), (6,0), (6,6) \). The area of the square is \( 6 \times 6 = 36 \).

5. **Calculating the Probability:**


- The area of the successful outcomes where all pieces are shorter than 6 units is the same as the area where the inequalities hold. As we analyzed, if we consider the constraints, the condition is satisfied for all relative placements of the breaking points.


- Thus the probability \( P \) of obtaining such a division where each piece is shorter than 6 units is simply:
\[
P = \frac{\text{area of valid choices}}{\text{total area}} = \frac{18}{36} = \frac{1}{2}
\]

Thus, the probability that all three resulting pieces are shorter than 6 units is \(\frac{1}{2}\) or 50%.

10.08.2024