Hello,

First, thank you to all of the mathematicians here who answer problems.

You have been an enormous help to me I am very grateful.

Second, I have a question similar to this one: https://web2.0calc.com/questions/help_54234 and this one: https://web2.0calc.com/questions/math-help-thank-you_4 but I am still stuck when following Melody's work.

Here's the question:

Let G be the center of equilateral triangle XYZ. A dilation centered at G with scale factor -3/4 is applied to triangle XYZ to obtain triangle X'Y'Z'. Let A be the area of the region that is contained in both triangles XYZ and X'Y'Z'. Find A/[XYZ]

I have triangle XGY=9sqrt3/4

then multiply by 3/4

= 27sqrt3/16,

but I don't know what to do next.

Images are in the links above but the fraction in my problem is -3/4 instead of -2/3.

Thanks in advance for the help!!

Digscaves Apr 27, 2023

#1**0 **

Let $s$ be the side length of equilateral triangle $XYZ$, and let $s'$ be the side length of equilateral triangle $X'Y'Z'$. Since the dilation centered at $G$ with scale factor $-3/4$ sends $X$, $Y$, and $Z$ to $X'$, $Y'$, and $Z'$, respectively, we have $GX' = -3/4 \cdot GX$, $GY' = -3/4 \cdot GY$, and $GZ' = -3/4 \cdot GZ$. Since $G$ is the center of equilateral triangle $XYZ$, we have $GX = GY = GZ = s/\sqrt{3}$.

Let $P$ be the intersection of lines $XZ$ and $Y'Z'$. Since $XZ$ and $Y'Z'$ are parallel, we have $\triangle GPZ' \sim \triangle GPX$, and it follows that $GPZ' = (3/4) GPX$. Similarly, we can show that $GPY' = (3/4) GPX$. Therefore, we have

A=[GPY′XZ′]=[GPY′]+[GPX]+[PZ′Y′X]=3/4[GPX]+[PZ′Y′X]

To find $[GPX]$, note that $\triangle GPX$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle with hypotenuse $GP = s/\sqrt{3}$. Therefore, we have $[GPX] = (1/2) \cdot (s/\sqrt{3}) \cdot (s/2) = s^2/(4\sqrt{3})$.

To find $[PZ'Y'X]$, note that $PZ' = 3/4 \cdot GX = -3s/(4\sqrt{3})$, and $Y'X = XZ = s$. Therefore, we have $[PZ'Y'X] = (1/2) \cdot (-3s/(4\sqrt{3})) \cdot s = -3s^2/(8\sqrt{3})$.

Substituting these values, we get

A= 3/4*s^2/(4*sqrt(3)) - 3s^2/(8*sqrt(3)) = s^2/(8*sqrt(3))

Therefore, we have $A/[XYZ] = A/(s^2\sqrt{3}/4) = 4/(8\sqrt{3}) = sqrt(3)/6.

Guest Apr 27, 2023

#2**+1 **

I will unlock the original posts for you. I only answered one of them.

Please explain what you do not understand about my answer on the original post

https://web2.0calc.com/questions/math-help-thank-you_4

then send me a private message asking me to take a look.

It is an old thread so if you do not send me a message I probably will not see your new post.

Melody May 1, 2023

#3**+1 **

23/48

Step-by step explantion.

Let each side of triangle XYZ have length s. We therefore have:

s = YZ = YQ + QR + RZ = x + 2y

Since each side of triangle X1 Y2 Z2 is 3/4 the length of the corresponding side of triangle XYZ, we have:

3/4s = X1 Z1 = X1 Q + QP + P Z1 = 2x +y.

The system of equations:

x + 2y = s

2x + y = 3/4s

Multiply the second equation by 2, then subtract the first equation.

3x = 1/2s, so x = 1/6s.

Now we have y = 3/4s-2x = 5/12s.

A/[XYZ] = [XYZ] - 3[YPQ]/[XYZ]

= {[XYZ] - 3[(25)/144] * [XYZ]} / [XYZ]

={23/48* [XYZ]} / [XYZ]

= 23/48

**Cannot answer " url question "**

acyclics May 1, 2023