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!


1

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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