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i have tried so many times on this problem:

 

Let $G$ be the center of equilateral triangle $XYZ.$ A dilation centered at $G$ with scale factor $-\frac{4}{3}$ 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 $\frac{A}{[XYZ]}.$
 

i have so many wrong answers:

 

4/9
11/27
20/27
73/200
73/100
5/9
17/27

 

please help T^T

tysm!!

 Aug 1, 2021
 #1
avatar+1487 
+2

The correct answer is this:

 

A / [XYZ] = 851 / 999  laugh

 Aug 1, 2021
 #2
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HOLY C**P--- THANK YOU SO MUCH!! I was able to understand the explanation that the website gave me ^^

tysm!! you're a lifesaver :))

 Aug 1, 2021
 #3
avatar+114511 
+2

 

 

I let GF=1   (just for ease)

CX=2    (30,60,90 triangle)

This means that the sides of the original triangle are   \(2\sqrt3\;units\)

So the area of the original triangle is  

\(Area\;XYZ = \frac{1}{2}*2\sqrt3*2\sqrt3*sin60\\ Area \;XYZ= 6*\frac{\sqrt3}{2}\\Area \;XYZ= 3\sqrt3\;\;u^2\)

 

CG=4/3

So GX=2 - 4/3 = 2/3

 

\(tan60=\frac{XG}{GT}\\ \sqrt3=\frac{2}{3*GT}\\ GT=\frac{2}{3\sqrt3}\\ \)

 

\(\text{Area of }\triangle STX = GT*GX = \frac{2}{3\sqrt3}*\frac{2}{3}\\~\\ \text{Area of }\triangle STX = \frac{4}{9\sqrt3}\\~\\ \text{Area of intersection}=\triangle XYZ-3*\triangle XST\)

 

 

 \(Intersection\; area = 3\sqrt3-3*\frac{4\sqrt3}{27}=\frac{27\sqrt3-4\sqrt3}{9}=\frac{23\sqrt3}{9}\\~\\\)

 

\(\displaystyle \frac{\text{intersection area}}{\text{area of original triangle XYZ}}=\frac{23\sqrt3}{9}\times \frac{1}{3\sqrt3}=\frac{23}{27} \)

 Aug 2, 2021
edited by Melody  Aug 2, 2021
 #4
avatar+114511 
0

So guest, are you going to show some interest or was I only amusing myself?

Melody  Aug 3, 2021

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