blackpanther

avatar
Nombre de usuarioblackpanther
Puntuación1689
Membership
Stats
Preguntas 428
Respuestas 33

 #2
avatar+1689 
0

To solve the problem, let's break it down step by step:

 

Step 1: Understand the Dilation

 

The dilation is centered at GG, with a scale factor of −75-\frac{7}{5}.

 

A negative scale factor means the resulting triangle X′Y′Z′X'Y'Z' is flipped across the center GG.

 

The magnitude of the scale factor (75\frac{7}{5}) indicates the relative size of the dilated triangle compared to the original triangle.

 

Step 2: Ratio of Areas

 

The area of a dilated figure changes by the square of the scale factor. Thus:

 

Area of X′Y′Z′=(75)2⋅Area of XYZ=4925⋅Area of XYZ\text{Area of } X'Y'Z' = \left( \frac{7}{5} \right)^2 \cdot \text{Area of } XYZ = \frac{49}{25} \cdot \text{Area of } XYZ

 

Step 3: Overlap of Triangles

 

Since the dilation is centered at GG, the two triangles share the same center and are similar.

 

The region of overlap forms a smaller triangle that is geometrically similar to both XYZXYZ and X′Y′Z′X'Y'Z',

 

with sides proportionate to the smaller absolute scale factor 75−1=25\frac{7}{5} - 1 = \frac{2}{5}.

 

Step 4: Area of Overlap

 

The overlap triangle has sides scaled by 25\frac{2}{5}, so its area is proportional to the square of this scale:

 

Area of Overlap=(25)2⋅Area of XYZ=425⋅Area of XYZ\text{Area of Overlap} = \left( \frac{2}{5} \right)^2 \cdot \text{Area of } XYZ = \frac{4}{25} \cdot \text{Area of } XYZ

 

Step 5: Ratio A[XYZ]\frac{A}{[XYZ]}

 

The ratio of the area of the overlap to the area of triangle XYZXYZ is:

 

A[XYZ]=425⋅Area of XYZArea of XYZ=425\frac{A}{[XYZ]} = \frac{\frac{4}{25} \cdot \text{Area of } XYZ}{\text{Area of } XYZ} = \frac{4}{25}

 

Final Answer:

 

4/25

25 ene 2025
 #1
avatar+1689 
-5

Let's analyze the given relations:

 

f(x+1x​)=2f(x)​: This relation suggests a halving property.

 

If we substitute x=2n+12​, we get: f(2n+12​+12n+12​​)=2f(2n+12​)​

 

Simplifying the fraction inside the argument gives: f(2n+32​)=2f(2n+12​)​

 

f(1−x)=1−f(x): This relation suggests a complementary property.

 

If we substitute x=21​, we get: f(1−21​)=1−f(21​) Simplifying gives: f(21​)=21​

 

Now, let's consider the infinite series: S=f(32​)+f(52​)+f(72​)+⋯+f(2n+12​)+⋯

 

Using the halving property, we can rewrite this series as: S=2[f(52​)+f(72​)+⋯+f(2n+12​)+⋯]

 

Notice that the terms inside the brackets are almost the same as the original series, except for the first term. So, we can write:

 

S=2(S−f(32​))

 

Solving for S, we get: S=2f(32​)

 

To find f(32​), we can use the complementary property: f(1−32​)=1−f(32​) f(31​)=1−f(32​)

 

Now, using the halving property on f(31​): f(31​+131​​)=2f(31​)​ f(41​)=21−f(32​)​

 

We can continue this process to find a pattern: f(51​)=21−21−f(32​)​​ f(61​)=21−21−21−f(32​)​​​

 

We can see that this pattern converges to 21​ as we keep applying the halving property. Therefore, f(31​)=21​.

 

Substituting this back into the equation for f(32​), we get: 21​=1−f(32​) f(32​)=21​

 

Finally, substituting this value into the expression for S, we get: S=2⋅21​=1

 

So, the value of the given infinite series is 1.

20 dic 2024
 #2
avatar+1689 
0

To find the largest possible value of \( CF \), we'll use the relationship between the altitudes in a triangle and its area. The area \( \Delta \) of a triangle can be expressed using any of its altitudes:

 

\[
\Delta = \frac{1}{2} \times \text{base} \times \text{height}
\]

 

For triangle \( ABC \), we have the following relationships:

 

- \( \Delta = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times CA \times BE = \frac{1}{2} \times AB \times CF \)

 

Let the side lengths be \( a = BC \), \( b = CA \), and \( c = AB \). Then:

 

\[
\Delta = \frac{1}{2} \times a \times 12 = \frac{1}{2} \times b \times 16 = \frac{1}{2} \times c \times CF
\]

 

This simplifies to:

 

\[
\Delta = 6a = 8b = \frac{1}{2} c \times CF
\]

 

Equating these expressions:

 

\[
6a = 8b \quad \text{and} \quad 6a = \frac{1}{2} c \times CF
\]

 

### Step 1: Solve for \( a \) and \( b \)


From \( 6a = 8b \):

 

\[
\frac{a}{b} = \frac{8}{6} = \frac{4}{3}
\]

 

Thus, \( a = \frac{4}{3}b \).

 

 

### Step 2: Substitute into \( 6a = \frac{1}{2} c \times CF \)


Substitute \( a = \frac{4}{3}b \) into \( 6a = \frac{1}{2} c \times CF \):

 

\[
6 \times \frac{4}{3}b = \frac{1}{2} c \times CF
\]

 

Simplifying:

 

\[
8b = \frac{1}{2} c \times CF \quad \text{so} \quad 16b = c \times CF
\]

 

### Step 3: Find the Maximum Value of \( CF \)


We need to maximize \( CF \), which is a positive integer. Since \( 16b = c \times CF \), and \( c \) and \( CF \) are integers, \( CF \) is maximized when \( c \) is minimized.

 

Given the relationship \( 6a = 8b \), \( a = \frac{4}{3}b \), and \( c = \frac{16b}{CF} \), the smallest integer value of \( c \) occurs when \( CF \) is as large as possible.

 

If \( CF = 16 \), then:

 

\[
16b = c \times 16 \quad \text{so} \quad c = b
\]

 

Since \( c \) is minimized and \( CF \) is maximized at \( 16 \), this is the largest possible value for \( CF \).

 

Thus, the largest possible value of \( CF \) is \( \boxed{24} \).

14 ago 2024