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5

The hint. By the affine transformation with determinant $\Delta$ write the equation of this ellipse in the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the needed area is equal to $\frac{\pi ab}{\Delta}$ I got the following. $$\left(2x+y+\frac{1}{2}\right)^2+\left(x+y+\frac{5}{2}\right)^2=\frac{1}{2}.$$ Can you end it now?

4

Triangles $ABE$ and $CDE$ are similar, with ratio $3:5$. Hence $$AE=\frac{3}{8}AC=\frac{3}{8}\cdot 5=\frac{15}{8}$$ By the same similarity, if $h$ is the height of triangle $CDE$ from $E$ to $CD$, then $$h=\frac{5}{8}BC=\frac{5}{8}\cdot 4=\frac{5}{2}$$ hence the area of triangle $CDE$ is $$\frac{1}{2}\cdot CD\cdot h=\frac{1}{2}\cdot 5\cdot \frac{5}{2}=\... 3 For first question you may use similar triangles - \triangle AEB and \triangle CED:$$\dfrac{3}{5} = \dfrac{x}{5-x}$$2 Take a unit sphere S_n. Construct a vector field \mathbf n=\mathbf x defined on S_n. Then you can easily show that this vector field is continuous, has unit length, and normal to S_n. So by definition you have defined orientation to your surface, and this the surface is orientable. 2 The complicated version is to write coordinates for each point, so point 1 would be (x_1,y_1). You can write differential equations for each point, so$$x_1'=v\frac {x_2-x_1}{\sqrt{ (x_2-x_1)^2+(y_2-y_1)^2}}$$and similarly for the others, then solve the coupled differential equations. This throws away the knowledge that the triangle stays equilateral,... 2 Let A(t), B(t), C(t) be your points (as functions of time t. For convenience, take the origin to be the centroid of the triangle at t=0, so A(0)+B(0)+C(0)=0. Then if R is a rotation by \pi/3 in the appropriate direction, B(0) = R A(0), C(0) = R B(0) and A(0) = R C(0). For some v > 0 the motion of your points is governed by the ... 1 You have to recall the definition of the topology on G_n(V). A metric h on V is a hermitian metric, i.e. a complex scalar product on V. Using this scalar product, each subspace W \subset V determines the orthogonal projection p_W^h : V \to V onto W. This is unique linear endomorphism such that p_W^h(V) = W, p_W^h(w) = w for all w \in W ... 1 For the first part use similarity of triangles to simplify computation. You have similar triangles ABE and CDE with ratio of sides being 3 to 5 You also know AC=5 so you find AE=5\times \frac {3}{8}=\frac {15}{8} which is choice D Your have solved the second part correctly. 1 There is no simple general formula for such problems. In the case at hand it is expected that you recognize the curve as an ellipse E whose axes are not parallel to the x- and the y-axes. Solving the given equation for y should therefore result in two functions x\mapsto y_+(x) and x\mapsto y_-(x) describing the "upper half" and the "lower half" ... 1 The way without trigonometry and the Pythagoras's theorem: Let \measuredangle PDA=\alpha. Thus, since PBCQ is cyclic, we obtain:$$\measuredangle BCQ=\measuredangle BCP+\measuredangle PCQ=\alpha+90^{\circ}-2\alpha=90^{\circ}-\alpha=\measuredangle BQC and we are done!

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Modifying a bit your sketch and splitting the speed into a - "height compressing" component ( red in the sketch), and - a "tangential speed" (blue), corresponding to a rotation then it is easy to express the process, specially using complex numbers in a Argand plane. And it comes out also clearly that, apart rotation, the height is compressed ...

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