okok so I'm going to go insane, as you have been informed from the title. these problems are overdue-- and you know why-? bEcauSE iVE tRiEd 5 aNswers fOR eACH oF TheM aND tHey'Re aLL wROnG aaaAAAAAa hALP PLS


1. Let $A_1 A_2 A_3 \dotsb A_{10}$ be a regular polygon. A rotation centered at $A_1$ with an angle of $\alpha$ takes $A_4$ to $A_8$. Given that $\alpha < 180^\circ$, find $\alpha,$ in degrees.


2. 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 $\frac{A}{[XYZ]}.$


bye- anyone who helps is legit a lifesaver ūüė©


tysm whoever has the guts to even attempt this ;-;

also im sorry, I don't normally come to webcalc unless something like this happens :(

again, thank y'all <3


and uh sorry to all the people who think im cheating bc im really not. this is my last resort, and if anyone wants to see my work from the wrong answers I have tried before, I will happily give them away -n-


anyway, ty :)

 Jun 20, 2021

1.  For this question, we are provided that each of the sides of the polygon is the same, and that it has 10 sides.  Because of this, the specified figure is a regular decagon, with angle measures of 144 degrees.  



To start, we go online and google regular decagon, to find a suitable image/graph (visualization is key).  This is the one I found:













Let the topmost point be \(A_1\), and we label all points from \(A_1\) counterclockwise, getting \(A_4\) and \(A_8\).  Now, we connect \(A_1\)\(A_4\), and \(A_8\), forming an isoceles triangle.  Taking the Trapizoid \(A_1\)\(A_2\), \(A_3\), \(A_4\) into consideration, we draw the perpendicular from \(A_3\) to segment \(A_1 A_4.  \), point of intersection being X.  From this, we find that the angle \(A_3 A_4 X\) is 90 degrees, so the angle \(A_4 A_3 X\) has measure of 144-90, or 54 degrees.  


This means that angle \(A_3 A_4 X\) is 36 degrees, and that angle \(A_1, A_4 A_8\) is 90 degrees-36 degrees, or 54 degrees.  By the same reasoning, \(A_4 A_8 A_1\) is also 54 degrees.  This means that the angle measure of \(A_4 A_1 A_8\) is 180-54*2, or 180-108, 72 degrees.  Thus, to move point \(A_4\) to \(A_8\), we rotate around\( A_1\) by 72 degrees.  



Answer is 72 degrees.


2.  Unfortunately, I do not know how to do this question because I am very unfamiliar with dilations.  Good Luck though!.  

 Jun 20, 2021

wow I didn't expect someone to reply that fast!!! thank you so much!! :))) i really appreciate it

 Jun 20, 2021

No problem, most of the time was spent formating LaTeX and the picture anyways.  :)

EnchantedLava68  Jun 20, 2021

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