Points P,Q,R, and S are chosen on the sides of parallelogram ABCD, so that P is on line AB, Q is on line BC, R is on line CD, S is on line DA, and AP=BQ=CR=DS=1/3 AB. Compute the ratio of the area of PQRS to the area of ABCD.  The special case where ABCD is a square is solved here:
121292
Points $P$, $Q$, $R$ and $S$ are chosen on the sides of parallelogram $ABCD$, so that $P$ is on line $AB$, $Q$ is on line $BC$, $R$ is on line $CD$, $S$ is on line $DA$, and $AP=BQ=CR=DS=\frac{1}{3}(AB)$. Compute the ratio of the area of $PQRS$ to the area of $ABCD$.
4804
I’m sorry I couldn’t upload a photo, so I’ll try to explain it as best as I can.     The quadrilateral has a pair of opposite and equal sides, and has a pair of opposite equal angles (85 degrees in the question).   Can I say this is a parallelogram or not?    I tried drawing a shape with those conditions that is not a parallelogram and I failed. Thanks in advance, and again sorry for not having a photo and I hope I explained it well enough.
84719
Question :  Let $z_1,z_2,z_3,z_4$ be the position vectors of the vertices for quadrilateral $ABCD$. Prove that $ABCD$ is a parallelogram if only if $z_1-z_2-z_3+z_4=0$  What i've done so far :  Proof of $p\Rightarrow q$  Let $z_1=x_1+iy_1,\quad z_2=x_2+iy_2, \quad z_3=x_3+iy_3, \quad z_4=x_4+iy_4$  $ABCD$ is a parallelogram $\Leftrightarrow\,\overrightarrow{AB}=\overrightarrow{CD}\, \land\, \overrightarrow{AD}=\overrightarrow{BC} $   $\begin{aligned} \overrightarrow{AB}=\overrightarrow{CD} &amp;\Rightarrow |z_1-z_2|=|z_3-z_4|\\ &amp;\Leftrightarrow x_1^2-2x_1x_2+x_2^2+y_1^2-2y_1y_2+y_2^2=x_3^2-2x_3x_4+x_4^2+y_3^2-2y_3y_4+y_4^2\\ \overrightarrow{AD}=\overrightarrow{BC} &amp;\Rightarrow |z_1-z_4|=|z_2-z_3|\\ &amp;\Leftrightarrow x_1^2-2x_1x_4+x_4^2+y_1^2-2y_1y_4+y_4^2=x_2^2-2x_2x_3+x_3^2+y_2^2-2y_2y_3+y_3^2 \end{aligned}$  I'm stuck and see no conclusion here.  Proof $p\Leftarrow q$  I don't have an idea for this.  Please help me, thanks.
33433
Find the number $b$ such that the line $y = b$ divides the region bounded by the curves $y = x^2$  and $y = 4$ into two regions with equal area.  I could find the area of this region but i have no clue how to split it parallel to the $x$ axis perfectly... this is too abstract for me...  I found the whole area and got $\dfrac{32}{3}$... I tried plugging that back into the definite integral but that definitely wasn't right...
27964
Let $m_{a},m_{b},m_{c}$ be the lengths of the medians and $a,b,c$ be the lengths of the sides of a given triangle , Prove the inequality :   $$m_{a}m_{b}m_{c}\leq\frac{Rs^{2}}{2}$$  Where :   $s : \operatorname{Semiperimeter}$  $R : \operatorname{circumradius}$   I know the relation :   $$m_{a}^{2}=\frac{2(b^{2}+c^{2})-a^{2}}{4}$$  But when I multiple together I dont get simple formulas!  So, I need help finding a solution. Thanks!
11856
I get the graphic explanation, i.e. that the area $dA$ of the sector's increment can be looked upon as a polar "rectangle" as $dr$ and $d\theta$ are infinitesimal, but how do you prove this rigorously?    1. Geometrically, the exact area would be $$\frac{(r+dr)^2d\theta}{2} - \frac{r^2d\theta}{2}$$$$= (r + \frac{dr}{2}) dr d\theta  $$$$= r dr d\theta + \frac{dr^2 d\theta}{2}.$$  How do we get rid of $\frac{dr^2 d\theta}{2}$? Is it too insignificant in value compared to $r dr d\theta$ so that it vanishes? If so, is it the reason it gets ignored? And what's more, geometrically speaking, why would the two areas - $dx dy$ and $r dr d\theta$, be equal?  2. Approaching it algebraically, setting $$x = r\sin\theta$$$$y = r \cos\theta$$  gives $$\frac{dx}{d\theta} = r\cos\theta,  dx = r\cos{\theta} d\theta$$$$ \frac{dy}{dr} = \cos\theta,  dy = \cos\theta dr. $$  Multiplying both equations, side by side, gives $$dxdy = r\cos^2\theta dr d\theta.$$  Again I get an extra term, which is $\cos^2 \theta$. In both cases I am unable to derive that $dxdy = rdrd\theta$.  What do I do wrong in my reasoning? It all makes me think that I'm getting something essential terribly wrong.
24707
A square with edge length 2 cm has semicircles drawn on each side. Find the total area of the shaded region.   Here is an image of the diagram shown :       Please show your work in pictures, numbers, words, anything. (Try to keep it to a Grade 8 level too)
103690
Let $S$  be a square of unit area. Consider any quadrilateral whose $4$ vertices lie on each side square $S$. Let the length of the sides of this quadrilateral be $a,b,c,d$. Then prove that $$2 \leq a^2+b^2+c^2+d^2 \leq 4$$     This problem appeared in IIT JEE $1997$ (re-exam).I really do not know know how to approach this problem properly.I have managed to prove $a^2 + b^2 + c^2 +d^2  \leq 4$ but not the initial inequality.
19597
I need to find the area of a parallelogram with vertices $(-1,-1), (4,1), (5,3), (10,5)$.  If I denote $A=(-1,-1)$, $B=(4,1)$, $C=(5,3)$, $D=(10,5)$, then I see that $\overrightarrow{AB}=(5,2)=\overrightarrow{CD}$. Similarly $\overrightarrow{AC}=\overrightarrow{BD}$. So I see that these points indeed form a parallelogram.  It is assignment from linear algebra class. I wasn't sure if I had to like use a matrix or something.
107441
I had some troubles with this problem :     Let $ABCD$ be a convex quadrilateral. $M$ and $N$ are the midpoints of   the diagonals $AC$ and $BD$. The sides $AB$ and $CD$ are extended   until they intersect. The intersection point is $E$. The sides $AD$   and $BC$ are extended until they intersect. The intersection point is   $F$. Let $P$ be the midpoint of the segment $[EF]$. Prove that $M$, $N$, $P$ are    collinear.     First, I found that a quadrilateral in which the opposite sides interesect is also known as a complete quadrilateral. Then, the line $M-N-P$ is known as Newton-Gauss line and the problem above as Newton's Problem.  I've taught about solving it using areas. I've used the property that median divides the triangle in two echivalent triangles (with the same area). Many properties can be derived from it.  I haven't figure out, but I'm interested in a proof using areas. I would appreciate some suggestions.  Thanks!
12824
I've been having a problem with these problems for a week now so I thought I'd post it on MSE. Read the text below. . I've also typed it up:  Annabel made a shape by placing identical square tiles in a frame as shown in the diagram above. The tiles are arranged in columns. Each column touches the base but no columns touch the side or top. There are no empty gaps between columns. The frame can be enlarged as needed.  These are the questions. I've also typed them up.    A. Show that it is possible to arrange $7$ tiles so that the ant walks $8,9,10,11,12,13,14,15$ tile edges.   B. Show 6 ways of arranging 7 tiles so that the ant walks a total of 9 tile edges.  C. Show it is possible to arrange 49 tiles so that the ant walks a total of less than 21 tile edges.  D. Show four arrangements of 137 tiles, each arrangement with a different maximum height, so that the ant walks a total of 34 tile edges.  For question $a$, is there a formula relating the number of tile edges exposed to the number of tiles. These are my answers for $a$    For the above picture, it's the left hand side which has the $a$ answers.  However, I'm having a bit of trouble with $b$. I've already got four answers down but where are the other two. My four current answers are in the right hand side of the last image. Also, I'll also appreciate a formula to show the relationship.  I'm also having trouble with $c$. I found out that you get 21 tile edges exposes when you make a $7*7$ square but the answer is asking for less than 21. Can anyone help me. Also, as in the previous questions, a formula to show the relationships would be great.  And for $d$, I am totally lost. I have no idea where to start, and I certainly can't find out four arrangements. So can anyone help me? And as always, a formula is nice  P.S. Please explain this in an understandable formula, as I'm only a Year 7 with the ability to understand linear algebra and parts of quadratics and trigonometry. Also, sorry about the poor imaging.
15334
Below the parallelogram is obtained from square by stretching the top side while fixing the bottom. Since area of parallelogram is base times height, both square and parallelogram have the same area.    This is true no matter how far I stretch the top side.    In below figure it is easy to see why both areas are same.   But it's not that obvious in first two figures. Any help seeing why the area doesn't change in first figure?
72954
There is a hint to show that if there were points in $\mathbb{Q}$, then there would exist $r,s, t \in \mathbb{Z}$ with $\gcd(r, t) = 1$ such that $2s^2 = t^4 - 17r^4$ , and then show that any prime dividing $s$ is a quadratic residue modulo $17$.  I'm stuck on the first bit - if you suppose $\left(\frac{a}{b}, \frac{c}{d}\right)$ is a point on the curve in $\mathbb{Q}$ (so $a, b, c, d \in \mathbb{Z}$) then you get to $2a^2d^4 = b^2c^4 - 17b^2d^4$, but I'm not sure where to go from here since you don't know that $b$ is a square?
101930
I have this algorithm and Im searching p and q:  $n=p^2 * q $  $l=(p-1)*(q-1) / \gcd (p-1,q-1)$  $d\equiv l^{-1} \pmod n$  And the values are:  For $n$:  1043247381320041410146374847683905608886067988802815759655490071506964811427733496668055364700822779908166748193534341903285387332658081562041534228961538451830025041762537639985150344917584064827615356535763872539925989984117068207555498700585376330731428359489791600319636079777025344217248475485835298689437926003958578141784379857951264973109214803916445702964047899447002191548965239884095802091798787988745415178578597206815122915438351636672681355517982410123569961242563194480399347961753678159210217993526738306796503660563163029287419732210247579793266891085178817047536758465239950861266592524206390869822972469137566045176690277813352563392838847417674529251627419479815097868890993153215515488877097760272399483111013602618554846096893805108827211344467797072352072430591318485091677011261912992157762106672216780766138057271434398274272110062768324672899209610001835423246735718435995420720133657793744810610657   for $d$:  105717877672224295784291675385546750129601472163379527375367128238003471793097341763391999503913659542087520808364287167645349555897774428176288829031477703594776341268160433668976237198330124904372546897953955241778475322074160469767162566234358907740412805211247598698481640207956827199048419949848619731050860757808082106844953602176503247786535423423104239566910713528691298692332149783689032722772009974404716783312200885555343877611094998117307970399241640213553820043460140493803404190774171912962651075100983834071259021038322886114660028788229094872545516054757700068264628024524394995795135464746570071201   How can I get $p$ and $q$?   Thanks!
60506
Given:   The total length of ad + dc The lengths of each ab, bc and ca That angle adb = angle cdb   How can I work out lengths ad and cd
103809
How am I to "[u]se induction to show that $n$ straight lines in the plane divide the plane into $\frac{n^2+n+2}{2}$ regions"? It is assumed here that no two lines are parallel and that no three lines have a common point. Further, this is not a non-Euclidean problem, but I wouldn't mind a discussion on the non-Euclidean nature of the problem.  I was thinking it might be easier to show using the unit sphere.    A GOOD REFERENCE I FOUND THAT WILL BE USEFUL TO ANYONE COMING ACROSS THIS PROBLEM IS Concrete Mathematics, Graham, Knuth, Patashnik pp. 4–8.
62773
$ABC$ is an acute triangle. The angle bisector $AD$, the median $BE$ and the altitude $CF$ are concurrent. Prove that angle $A$ is more than $45$ degrees. Here $D,E,F$ are  points on $BC,CA,AB$ respectively.  This question was asked in 4th All-Soviet Union Mathematics Competition  in 1970 and the wording of the problem is same.