The Pointe Moving Check List: Fill & Download for Free

Are you alone? If yes, proceed to step 2. If not, immediately kill your companion and leave his body between you and the lion as you back away.A lion is bigger, stronger, has better natural weapons than you and is an experienced natural killing machine. You are not going to defeat it in any form of combat.A lion is faster than you, is a better climber, is superbly adapted to its home terrain and can go anywhere you can go. So, you aren't going to get away from it.You, however, are smarter than the lion. Much smarter. Now, You have already given up a huge part of that natural advantage by being a dumb ass and getting lost in lion country without the proper weapons and training to defend yourself.Lions are cats. They aren't that different than your friendly little domestic house cat you know so well. Cats are naturally curious creatures. They are also very cautious when given time to think. If you just run or scream hysterically like a panicky mouse, the lion won't have any time to think and will attack and kill you immediately.If you happen to have gotten lost with a giant cardboard box, place it on the ground between you and the lion. The lion will find the inner space of the box irresistible and will jump into the box and lose interest in you. If there are other lions around, they too will be drawn to the giant box and take turns sitting in the box.If you happen to have a laser pointer, go ahead and get it out. Point it at the tree or rock nearest the lion such that he can see the laser point. Move it in tiny circles up the tree or rock. The lion will chase the laser point up the tree ignoring you. If you are confronted by multiple lions, shine the laser point on one of the lions enticing the other lions to attack him thus starting a fight. Escape.If you happen to have a long rope, back away slowly trailing the rope behind you. The lion will attack the rope. Whatever you do, don't let the lion have the rope. When he grabs it, stop moving. The lion will forget what he was doing and let go of the rope. Repeat this step until you reach civilization.If you have tried everything listed here and the lion is still following you, take your shirt off and throw it at the lion. Then continue to back away. He will stop to smell the shirt. He has to. He is compelled to. Maybe you will get lucky and he will attempt to eat the shirt. If he does, he will discover that it tastes horrible and has little to no nutritional value. Since you smell like your shirt, hopefully he will assume the same about you. Repeat this with your pants, underwear, shoes and socks as necessary.Now you are ready for a truly desperate act. Attempt to make friends with the lion. Calmly walk towards the lion saying "here kitty kitty. Nice kitty" while making a scratching motion with your hand. Scratch his head. Check to see if he has a thorn in any of his feet. If so, remove it.

If you can’t drive yourself to the date, with your car, and pay for the date with your money.. you’re probably not ready to date.Dating is serious.. You date for an end goal. If you aren’t ready for an end goal involving someone- you aren’t ready to date.If you decide you’re not interested, move on..If you dress cheap you’ll be treated that way. Someone can have the most expensive outfit in the room and still act/look cheap.You should be treated the same whether your date buys you a meal at McDonald’s or the nicest restaurant in town.Now is the time to be picky.Be honest with your flaws. Know what won’t work well with your bad and what will.Be very cautious of how much you drink.No means no. Know your limits, stick to them.If you invite a guy in to stay on the first night, he may assume you do the same with everyone. It’s only natural..If you don’t want a man to have a dating check list, you shouldn’t either.There is no rule on who needs to pay. A relationship is a give and take. Don’t just be a taker.It’s okay to let the man know you’re interested!He doesn’t need to know about your food allergies, your family tree, your parent’s medical history! (you get the point) Save something for later!There were other things happening in your life before you started dating and it should stay that way. Continue living your life, hanging out with friends!You do you girl! Or boy!! Just keep smiling!Dating is fun.. but nothing beats a happy marriage!

In either 1) a brute-force or 2) an inventive fashion.Theorem: on the sides of an arbitrary planar triangle [math]A_1A_2A_3[/math] there constructed equilateral triangles with the remaining vertices located at the points [math]B_{1,2,3}[/math] in such a way that no equilateral triangle has common interior points with the parent triangle [math]A_1A_2A_3[/math]. Let the points [math]C_{1,2,3}[/math] be the centers of the equilateral triangles (Fig. 1):Prove that the triangle [math]C_1C_2C_3[/math] is also equilateral.Method 1, Brute ForceEven though we qualify the upcoming proof as brute-force, it still has its charm as it showcases the tactical power of complex numbers.Before we begin, however, we strongly encourage the readers to work their way through this Quora answer where we generated a proof of a related theorem by using the idea of modeling masses with complex numbers.Proof: let [math]O[/math] be the origin of an arbitrarily chosen coordinate system in the complex plane. Then, as was explained in the answer linked to, to the point [math]A_1[/math] in the complex plane there will correspond the vector [math]\overrightarrow{OA_1}[/math] to which we correlate the corresponding complex number denoted as [math]a_1[/math].Likewise, to the point [math]C_3[/math] in the same complex plane there will correspond the vector [math]\overrightarrow{OC_3}[/math] to which we correlate the corresponding complex number denoted as [math]c_3[/math] and so on for all the remaining points.We, thus, see that one way to get to the point [math]C_3[/math] starting at the origin and traveling through the point [math]A_1[/math] would be:[math]\overrightarrow{OC_3} = \overrightarrow{OA_1} + \overrightarrow{A_1C_3} \tag{1}[/math]in vector notation or pictorially (Fig. 2):Using the side [math]A_3B_3[/math] of the equilateral triangle [math]A_1B_3A_3[/math], construct yet another equilateral triangle [math]B_3D_3A_3[/math] as shown in the figure above. Then, by the property of the centroid of an equilateral triangle, it should be evident that the length of the vector [math]A_1C_3[/math] is one third of the length of the vector [math]A_1D_3[/math]:[math]3\cdot \overrightarrow{A_1C_3} = \overrightarrow{A_1D_3} \tag{2}[/math]But the vector [math]A_1D_3[/math], playing the role of a diagonal of the rhombus [math]A_1B_3D_3A_3[/math], is the sum of two vectors of equal length - by the problem statement:[math]\overrightarrow{A_1D_3} = \overrightarrow{A_1B_3} + \overrightarrow{A_1A_3} \tag{3}[/math]The vector [math]A_1A_3[/math] is the sum of the vectors [math]-OA_1[/math] and [math]OA_3[/math]:[math]\overrightarrow{A_1A_3} = -\overrightarrow{OA_1} + \overrightarrow{OA_3} \tag{4}[/math]What about the vector [math]A_1B_3[/math]? This is where the tactical power of complex numbers manifests itself already: the vector [math]A_1B_3[/math] is the vector [math]A_1A_3[/math] rotated about the vertex [math]A_1[/math] counterclockwise by [math]\pi/3[/math] radians (Fig. 3):also by the problem statement:[math]\overrightarrow{A_1B_3} = \omega\cdot \overrightarrow{A_1A_3} \tag{5}[/math]where the number [math]\omega[/math] has the following form:[math]\omega = e^{i\pi/3}, \;\omega^3 = e^{i\pi} = -1 \tag{6}[/math]Switching to the corresponding complex numbers and combining our results (2–6), we find for the vector [math]OC_3[/math]:[math]c_3 = a_1 + \dfrac{1}{3}\cdot(-a_1 + a_3 - \omega a_1 + \omega a_3) \tag*{}[/math]or, bringing the RHS of the above relation under one denominator:[math]c_3 = \dfrac{2a_1 + a_3 - \omega a_1 + \omega a_3}{3} \tag{7}[/math]Cyclically permuting all the relevant vertices, in exactly the same fashion we obtain for the remaining centers [math]c_{1,2}[/math]:[math]c_2 = \dfrac{2a_3 + a_2 - \omega a_3 + \omega a_2}{3} \tag{8}[/math]and:[math]c_1 = \dfrac{2a_2 + a_1 - \omega a_2 + \omega a_1}{3} \tag{9}[/math]Lastly, in order to prove that the triangle [math]C_1C_2C_3[/math] is equilateral all we have to do is pick any vector represented by one of its sides, rotate it by [math]\pi/3[/math] radians and check to see if it coincides with the corresponding vector represented by the side of the said triangle emanating from the same vertex.Namely, remembering that a rotation of a plane corresponds to the multiplication by [math]\omega[/math], we are to check that, say:[math]\omega\cdot \overrightarrow{C_3C_1} = \overrightarrow{C_3C_2} \tag{10}[/math]In other words, we are to answer the following question: is the relationship shown in (10) an identity?Such a check amounts to a finite and lengthy computation and, thus, the qualification of this proof as brute-force.For the vector [math]C_3C_1[/math] we have:[math]\overrightarrow{C_3C_1} = -\overrightarrow{OC_3} + \overrightarrow{OC_1} \tag*{}[/math]or, in our short-hand notation:[math]\overrightarrow{C_3C_1} = -c_3 + c_1 \tag*{}[/math]Since the important property of [math]\omega[/math] is known from (6), we roll up our sleeves and mindlessly shuffle symbols, ignoring the [math]3s[/math] in the denominators across the relations (7, 8, 9) since in (10) they will cancel out:[math]\omega\cdot(-c_3 + c_1 ) = \tag*{}[/math][math]= -2\omega a_1 - \omega a_3 + \omega^2a_1 - \omega^2a_3 + \tag*{}[/math][math]+ 2\omega a_2 + \omega a_1 - \omega^2a_2 + \omega^2 a_1 \tag*{}[/math]For the vector [math]C_3C_2[/math] in (10) we have:[math]\overrightarrow{C_3C_2} = -\overrightarrow{OC_3} + \overrightarrow{OC_2} \tag*{}[/math]in the vector notation or in our short-hand notation:[math]\overrightarrow{C_3C_2} = - c_3 + c_2 = \tag*{}[/math][math]= -2a_1-a_3+\omega a_1 - \omega a_3 + 2a_3+a_2 - \omega a_3 + \omega a_2 \tag*{}[/math]We hope that the readers now really see why this is a brute-force proof.Equating the coefficients across all the [math]a_1[/math] terms on both sides of the above two relations, we mechanically have:[math]2\omega^2 - \omega = -2 + \omega \tag*{}[/math]which is to say that:[math]\omega^2 - \omega + 1 = 0 \tag*{}[/math]But we already know that [math]\omega^3 = -1[/math], meaning that [math]\omega^3 + 1 = 0[/math]. Decomposing the sum of two cubes into the following factors:[math]x^3 + 1 = (x+1)(x^2 - x + 1) \tag*{}[/math]we see that since [math]\omega \neq - 1[/math], it follows that it necessarily must be the case that:[math]\omega^2 - \omega + 1 = 0 \tag*{}[/math]Check for the complex number [math]a_1[/math] representing our vertex [math]A_1[/math].Do the same for the number [math]a_2[/math]:[math]2\omega-\omega^2 = 1 + \omega \tag*{}[/math]which is to say that:[math]\omega^2 - \omega + 1 = 0 \tag*{}[/math]Check for the complex number [math]a_2[/math] representing our vertex [math]A_2[/math].Do the same for the number [math]a_3[/math]:[math]-\omega-\omega^2 = -1 - \omega + 2 - \omega \tag*{}[/math]which is again to say that:[math]\omega^2 - \omega + 1 = 0 \tag*{}[/math]Check for the complex number [math]a_3[/math] representing our vertex [math]A_3[/math]. Thus, we have verified that the relationship shown in (10) is indeed an identity. Hence, the triangle [math]C_1C_2C_3[/math] is equilateral. [math]\blacksquare[/math]Method 2, InvenitveAs we already noted, in this Quora answer, where we modeled point masses with complex numbers, we solved a related problem except that in that problem the triangles that we constructed on the sides of the parent triangle were isosceles and not equilateral.Don’t matter - the idea and the blueprint for the proof of the Napoleon’s theorem using complex numbers in an inventive way is still the same.Proof: by the problem statement our vector [math]C_3A_3[/math] is actually the vector [math]C_3A_1[/math] rotated by [math]2\pi/3[/math] radians about the point [math]A_1[/math] counterclockwise (Fig. 4):since the homothety in this case reduces to the identity transformation where the coefficient of homothety [math]k[/math] is an identical unity, [math]k=1[/math], and in plain English it simply means that after we rotate the vector [math]C_3A_1[/math] about the point [math]C_3[/math], we do not do any non trivial scaling, up or down:[math]a_3 - c_3 = \omega\cdot(a_1 - c_3) \tag{11}[/math]where the complex number [math]\omega[/math] is slightly different now:[math]\omega = e^{2i\pi/3},\;\omega^3 = e^{2i\pi} = 1 \tag{12}[/math]Bending the relation in (11) to the will of the Law of the Lever due to Archimedes, we re-write it as follows:[math]1\cdot(a_3 - c_3) + (-\omega)\cdot(a_1 - c_3) = 0 \tag{13}[/math]which, interpreted in the sense of centers of mass, means that the point [math]C_3[/math] is the (complex) center of mass of these two points:the point [math]A_3[/math] loaded with the mass of [math]1[/math] and denoted as [math]1A_3[/math] andthe point [math]A_1[/math] loaded with the mass [math](-\omega)[/math] and denoted as [math](-\omega)A_1[/math]If it is not clear why the above statement is true then we suggest that you work your way through the answer linked to, since here we are referring to the relation (2) of the theorem T2 in that answer.Likewise, moving cyclically around the vertices of the parent triangle, we find that the point [math]C_2[/math] is the (complex) center of mass of these two points:the point [math]A_2[/math] loaded with the mass of [math]1[/math] and denoted as [math]1A_2[/math] andthe point [math]A_3[/math] loaded with the mass [math](-\omega)[/math] and denoted as [math](-\omega)A_3[/math]and that the point [math]C_1[/math] is the (complex) center of mass of these two points:the point [math]A_1[/math] loaded with the mass of [math]1[/math] and denoted as [math]1A_1[/math] andthe point [math]A_2[/math] loaded with the mass [math](-\omega)[/math] and denoted as [math](-\omega)A_2[/math]Next, the location of the center of mass of two massive points will not change if we uniformly scale the masses of both points up or down.We decide to scale the masses of the points [math]A_{1,2}[/math] uniformly down by [math]\omega[/math]. Hence, the point [math]C_1[/math] will also be the center of mass of the points [math](1/\omega)A_1[/math] and [math](-1)A_2[/math].We now consider the following four massive points loaded with masses modeled with complex numbers:[math]\{(1/\omega)A_1, 1A_2, (-1)A_2, (-\omega)A_3 \} \tag*{}[/math]As required by the theorem T1 in the answer linked to, the total sum of this discrete number of points is not zero and is well-defined since [math]\omega\neq 0[/math] and a positive unity is delivered by the third power of [math]\omega[/math] and not the second:[math]\dfrac{1}{\omega} - \omega = \dfrac{1-\omega^2}{\omega}\neq 0 \tag*{}[/math]By the formula (1) from the answer linked to, for the center of mass [math]Z[/math] of this system we have on the one hand:[math]\left(\dfrac{1}{\omega} - \omega\right)\cdot z = (1/\omega)a_1 + (-\omega)a_3 \tag{14}[/math]since the [math]A_2[/math]-bound terms cancel out.Now we uniformly scale the masses in the above points [math]A_1[/math] and [math]A_3[/math] down by [math](-\omega)[/math]. For the point [math]A_3[/math] we trivially have:[math]\dfrac{-\omega}{-\omega} = 1 \tag*{}[/math]For the point [math]A_1[/math] we have via (12):[math]\dfrac{1}{-\omega^2} = \dfrac{\omega^3}{-\omega^2} = -\omega \tag*{}[/math]Thus, we see that the center of mass of our carefully chosen system of four points is one and the same as the center of mass of these two points: [math]1A_3[/math] and [math](-\omega)A_1[/math].If we go back to our bulleted lists then we will quickly find that it is the point [math]C_3[/math] that is the center of mass of the points [math]1A_3[/math] and [math](-\omega)A_1[/math]. Hence, the center of mass of our four points is located at the point [math]C_3[/math].On the other hand, in our relation in (14) we can very much not cancel out the like terms and regroup them as follows:[math]\left(\dfrac{1}{\omega} - \omega\right)\cdot z = \left(1a_2+(-\omega)a_3\right) + \left((-1)a_2 + (1/\omega)a_1\right) \tag{15}[/math]But, still according to our bulleted lists above, the first parenthesis on the RHS of (15) can be replaced with:[math](1 - \omega)c_2 \tag*{}[/math]which follows from the analog of (13).The second parenthesis on the RHS of (15) can be replaced with:[math](-1 + 1/\omega)c_1 \tag*{}[/math]and the relation in (15), then, becomes:[math]\left(\dfrac{1}{\omega} - \omega\right)\cdot z = (1 - \omega)c_2 + (-1 + 1/\omega)c_1 \tag*{}[/math]which is to say that:[math]\dfrac{(1-\omega)(1+\omega)}{\omega}\, z = (1-\omega)c_2 + \dfrac{1-\omega}{\omega}\,c_1\tag*{}[/math]Multiply both sodes of the above relation by [math]\omega/(1-\omega)[/math]:[math](1+\omega)z = \omega c_2 + c_1 \tag*{}[/math]But we already established earlier that [math]z = c_3[/math]:[math](1+\omega)c_3 = \omega c_2 + c_1 \tag*{}[/math]The above relation can be re-written as follows:[math](-\omega)(c_2 - c_3) = c_1 - c_3 \tag{16}[/math]Now, according to the calculus of complex numbers, when we multiply a number by [math](-1)[/math] then geometrically we rotate the point in the complex plane corresponding to that number about origin counterclockwise by [math]\pi[/math] radians (Fig. 5):As such, what the relation in (16) is telling us is that the vector [math]C_3C_1[/math], that corresponds to the difference [math]c_1 - c_3[/math], is obtained by rotating the vector [math]C_3C_2[/math], that corresponds to the difference [math]c_2 - c_3[/math], by [math]-\pi/3[/math] radians or clockwise, in the negative direction (Fig. 6):which proves that the triangle [math]C_1C_2C_3[/math] is indeed equilateral. [math]\blacksquare[/math]