Improve Your Geometry Skills with Practice Problems

Given a circle with diameter $d$, is there a way to find the length of the chord f that cuts off an arc of length $l$?

I am trying to understand the relationships between chords and arcs etc, and this concept came up - is there a general formula for it?

(And before I get shut down for this being a homework question, it is not. The concept came up in independent study.)

I am trying to implement a fitness function for a genetic algorithm and have problem with understanding what function maximization means. In the paper which I use as the basis of the algorithm, the fitness function is defined as (1-RC) + (1-RO), where RC and RO are floating point values between 0 and 1, which I calculate using another formulas. However in the next paragraph of the article it is written - "This function (fitness function) has to be maximized"

I am struggling to understand what that means. Could you please explain it in easy-to-understand way for a Math noob? Thank you!

I have an optimization problem of the form
Max $x_1+x_2+x_3+\cdots <!--\; \cdots \; -->+x_n$
subject to $x_0^2+x_1^2+x_2^2+\cdots <!--\; \cdots \; -->+x_n^2+x_{12}^2+x_{13}^2+x_{14}^2+\cdots <!--\; \cdots \; -->+x_{1n}^2+x_{23}^2+\cdots <!--\; \cdots \; -->+x_{2n}^2+\cdots <!--\; \cdots \; -->+x_{nn-<!--\; -\; -->1}^2+x_{123}^2+\cdots <!--\; \cdots \; -->+x_{\mathrm{12..}n}^2=1$
$x_0+x_1+x_2+\cdots <!--\; \cdots \; -->+x_n+x_{12}+x_{13}$+$x_{14}+\cdots <!--\; \cdots \; -->+x_{1n}+x_{23}+\cdots <!--\; \cdots \; -->+x_{2n}+\cdots <!--\; \cdots \; -->+x_{nn-<!--\; -\; -->1}+x_{123}+\cdots <!--\; \cdots \; -->+x_{\mathrm{12..}n}=1$
where $x_0,x_1,x_2,\dots <!--\; \dots \; -->,x_{123\dots <!--\; \dots \; -->n}$ are unrestricted.

What is the upper bound for the cost function ..? Thank you.

Avery and Sasha were comparing parabola graphs on their calculators. Avery had drawn $y=0.001x^2$ in the window $-<!--\; -\; -->1000\le <!--\; \le \; -->x\le <!--\; \le \; -->1000$ and $0\le <!--\; \le \; -->y\le <!--\; \le \; -->1000$ , and Sasha had drawn $y=x^2$ in the window $-<!--\; -\; -->k\le <!--\; \le \; -->x\le <!--\; \le \; -->k$ and $0\le <!--\; \le \; -->y\le <!--\; \le \; -->k$ . Except for scale markings on the axes, the graphs looked the same! What was the value of k?

I noticed that I get the exact error, using midpoint rule error bound formula, but with $f^{″}(\frac{b-<!--\; -\; -->a}{2})$ for K, i.e. :
$\frac{K(b-<!--\; -\; -->a{)}^3}{24n^2}$
$\frac{f^{″}(\frac{b-<!--\; -\; -->a}{2})(b-<!--\; -\; -->a{)}^3}{24n^2}$

In his 5 postulates Euclid said:"All the right angles are equal". What he really meant by that ? Does he meant that we can impose any right angle on each other or the meaning there was that if you find this angle through the length of arc and radius of the circle then you get the same number ? I also heard that this postulate can be proven (and I remembered that I saw such a question there, but now I can't find it). So what is the proof (based on the definition of a right angle like right angle is an angle which is equal to its adjacent angle)

What is the formula to calculate the distance (arc length) between 2 points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ on the circumference of a circle of radius $r$ without knowing the angle $\mathrm{\theta <!--\; \theta \; -->}$ between them. I found that arc length can be calculated knowing $\mathrm{\theta <!--\; \theta \; -->}$. But I know only the $x,y,z$ co-ordinates of 2 points on the circumference of a circle. Please suggest.

Solve the trignometric equation $2\sin <!--\; \; -->(3x+\mathrm{\pi <!--\; \pi \; -->}/4)=\sqrt{1+8\sin <!--\; \; -->2x{\cos }^2<!--\; \; -->x}$

My question is that in general, is there a case where saddle point be the global max of a function? I am solving a game theory question which the optimal solution is the saddle point. Can I conclude that the optimal solution is at the boundaries?

Find the equation to the pair of angle bisectors of the pair of lines $(ax+by{)}^2=3(bx-<!--\; -\; -->ay{)}^2$.

Efforts:
$(ax+by{)}^2=3(bx-<!--\; -\; -->ay{)}^2$
After simplifying, I got:
$x^2(a^2-<!--\; -\; -->3b^2)+8abxy+y^2(b^2-<!--\; -\; -->3a^2)=0$
Now, what should I do next?

Strange proof of Schwarz Inequality with Pythagorean Theorem
Does anyone know what is going on in this proof of the Schwarz inequality? Most importantly: how can one assume that $c^2\leqq <!--\; \leqq \; -->\Vert <!--\; \Vert \; -->A{\Vert <!--\; \Vert \; -->}^2$, or later on, that $c^2\Vert <!--\; \Vert \; -->B\Vert <!--\; \Vert \; -->\leqq <!--\; \leqq \; -->\Vert <!--\; \Vert \; -->A{\Vert <!--\; \Vert \; -->}^2$? This would imply that $\Vert <!--\; \Vert \; -->A-<!--\; -\; -->cE{\Vert <!--\; \Vert \; -->}^2$, or in the latter case $\Vert <!--\; \Vert \; -->A-<!--\; -\; -->cB{\Vert <!--\; \Vert \; -->}^2$, could also be equal to zero (seeing as it's a smaller-than-or-equal-to sign). But since we're using the Pythagorean theorem to arrive at these relations, $\Vert <!--\; \Vert \; -->A-<!--\; -\; -->cE{\Vert <!--\; \Vert \; -->}^2$ or $\Vert <!--\; \Vert \; -->A-<!--\; -\; -->cB{\Vert <!--\; \Vert \; -->}^2$ could not possibly be zero, as then one of the sides of our 'triangle' would be zero, and we wouldn't be able to use the Pythagorean theorem to justify this conclusion. So why isn't it just a '<'-sign? As I said, I don't see how vector A−cE or A−cB can have length zero if we're using the Pythagorean theorem, that is, if we're presupposing the existence of a triangle. If either A−cE or A−cB equals zero, there wouldn't be a triangle anymore, and we wouldn't be able to apply the Pythagorean theorem.
Edit: fixed link.
Edit2: I must add that the author maintains an unusual definition for the scalar component; instead of the conventional $\frac{\mathbf{A}\cdot <!--\; \cdot \; -->\mathbf{B}}{\Vert <!--\; \Vert \; -->B\Vert <!--\; \Vert \; -->}$, i.e., $\mathbf{A}\cdot <!--\; \cdot \; -->\stackrel{\mathbf{^<!--\; ^\; -->}}{\mathbf{B}}$, the author defines $\frac{\mathbf{A}\cdot <!--\; \cdot \; -->\mathbf{B}}{\mathbf{B}\cdot <!--\; \cdot \; -->\mathbf{B}}$ to be the scalar component, i.e., $\frac{\mathbf{A}\cdot <!--\; \cdot \; -->\stackrel{\mathbf{^<!--\; ^\; -->}}{\mathbf{B}}}{\Vert <!--\; \Vert \; -->B\Vert <!--\; \Vert \; -->}$. In other words, the author divides the 'conventional' scalar component by the the norm of B, thus making 'his' scalar component a number which is in fact the proportion of the 'regular' scalar component to the entire length of the vector on which this component is projected.
Edit3: Since people don't seem to understand my confusion, let me phrase my question as explicitely as possible. My question is: why is the $\ge <!--\; \ge \; -->$ sign used? Why not just ">"? In what situation could $\Vert <!--\; \Vert \; -->A{\Vert <!--\; \Vert \; -->}^2$ equal $t^2\Vert <!--\; \Vert \; -->E{\Vert <!--\; \Vert \; -->}^2$ (the $\ge <!--\; \ge \; -->$ sign means greater than OR EQUAL TO, so in what situation could the left hand side ever equal the right hand side?) The problem is: since we're using the Pythagorean theorem for our proof, we're presupposing the existence of a triangle (or otherwise we wouldn't be able to use the Pythagorean theorem), and as such, $\Vert <!--\; \Vert \; -->A{\Vert <!--\; \Vert \; -->}^2$ can never equal $t^2\Vert <!--\; \Vert \; -->E{\Vert <!--\; \Vert \; -->}^2$, because this would mean that $\Vert <!--\; \Vert \; -->A-<!--\; -\; -->tE{\Vert <!--\; \Vert \; -->}^2$ is equal to zero, and that one of the sides of our triangle has length zero. Then we wouldn't have a triangle, and we wouldn't be justified in using the Pythagorean theorem in our particular situation. So let me ask it again: why is the $\ge <!--\; \ge \; -->$ sign used, instead of the ">"-sign?

Each angle of a rectangle is trisected. The intersections of the pairs of trisectors adjacent to the same side always form:
$$\begin{gathered} \mathbf{\text{(A)}}\text{a square}\mathbf{\text{(B)}}\text{a rectangle}\mathbf{\text{(C)}}\text{a parallelogram with unequal sides} \\ \mathbf{\text{(D)}}\text{a rhombus}\mathbf{\text{(E)}}\text{a quadrilateral with no special properties} \end{gathered}$$

Find k such that $P(0,2)$ is equidistant from $(3,k)$ and $(k,5)$ . Why can't we use the Midpoint Formula?

For $1\le <!--\; \le \; -->i\le <!--\; \le \; -->215$ let $a_i={\displaystyle \frac{1}{2^i}}$ and $a_{216}={\displaystyle \frac{1}{2^{215}}}$. Let $x_1,x_2,\dots <!--\; \dots \; -->,x_{216}$ be positive real numbers such that ${\displaystyle \underset{i=1}{\overset{216}{\sum <!--\; \sum \; -->}}{x}_i=1}$ and
$\underset{1\le <!--\; \le \; -->i<j\le <!--\; \le \; -->216}{\sum <!--\; \sum \; -->}{x}_i{x}_j={\displaystyle \frac{107}{215}}+\underset{i=1}{\overset{216}{\sum <!--\; \sum \; -->}}{\displaystyle \frac{a_i{x}_i^2}{2(1-<!--\; -\; -->a_i)}}.$
Find the maximum possible value of $x_2.$
I simplified the condition to ${\displaystyle \underset{i=1}{\overset{216}{\sum <!--\; \sum \; -->}}{\displaystyle \frac{x_i^2}{1-<!--\; -\; -->a_i}}={\displaystyle \frac{1}{215}},}$ but I'm not sure what to do next.

Solving trigonometric equation $\sin <!--\; \; -->(x)+\tan <!--\; \; -->(x)=0$

Write the equation of the ellipse, knowing it’s center $C(2,1)$ and the ends of two conjugate diameters $A(5,1)$, $B(0,3)$ .

The hyperbolic sine function, $\sinh <!--\; \; -->(x)$ , is defined by the equation:
$\sinh <!--\; \; -->(x)=\frac{e^x-<!--\; -\; -->e^{-<!--\; -\; -->x}}{2}$
Find a formula for its inverse,
${\sinh }^{-<!--\; -\; -->1}<!--\; \; -->(x)$

For fixed $g$, I want to find maximum $b$ with
$-<!--\; -\; -->2b(3t^2(s+1)+6t(s+1)+3s+2)-<!--\; -\; -->2g(6ts+3t+6s+2)-<!--\; -\; -->3t{s}^2+6ts+3t-<!--\; -\; -->3s^2+3s+1>0$
for some nonnegative reals $t$,$s$. Here $g$,$b$ are also $\ge <!--\; \ge \; -->0$. Can it be possible to get a function $f$ such that we get the upper bound on $b$ as $f(g)$ i.e, $b<f(g)$? If one can find $t$,$s$ for which $b$ is maximum, one will get $f(g)$.

Using expansion of $\sin <!--\; \; -->8\mathrm{\theta <!--\; \theta \; -->}$ to find equation with given roots

Let $f(x_1,\dots <!--\; \dots \; -->,x_n)=x_1{x}_2\cdot <!--\; \cdot \; -->\cdot <!--\; \cdot \; -->\cdot <!--\; \cdot \; -->x_n$. Let $A:=\{x\in <!--\; \in \; -->{\mathbb{R}}^n:x_1+x_2+\cdots <!--\; \cdots \; -->+x_n=n,x_i\ge <!--\; \ge \; -->0\mathrm{\forall <!--\; \forall \; -->}i\}$. I want to find the global max of $f$ under the constrain given by $A$. MY approach:
Put $F(x_1,\dots <!--\; \dots \; -->,x_n)=\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{x}_i-<!--\; -\; -->n=0$ is the constrain. We use lagrange multipliers:
$\mathrm{\nabla <!--\; \nabla \; -->}f=\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{\nabla <!--\; \nabla \; -->}F⟹<!--\; ⟹\; -->(x_2\cdot <!--\; \cdot \; -->\cdot <!--\; \cdot \; -->\cdot <!--\; \cdot \; -->x_n,\dots <!--\; \dots \; -->,x_1\cdots <!--\; \cdots \; -->x_{n-<!--\; -\; -->1})=(\mathrm{\lambda <!--\; \lambda \; -->},\mathrm{\lambda <!--\; \lambda \; -->},\dots <!--\; \dots \; -->,\mathrm{\lambda <!--\; \lambda \; -->}).$
Hence we have
$x_2\cdot <!--\; \cdot \; -->\cdot <!--\; \cdot \; -->\cdot <!--\; \cdot \; -->x_n=x_1{x}_3\cdot <!--\; \cdot \; -->\cdot <!--\; \cdot \; -->\cdot <!--\; \cdot \; -->x_n=\cdots <!--\; \cdots \; -->=x_1\cdot <!--\; \cdot \; -->\cdot <!--\; \cdot \; -->\cdot <!--\; \cdot \; -->x_{n-<!--\; -\; -->1}$
this implies that
$x_1=x_2=\cdots <!--\; \cdots \; -->=x_n$
We also know $\sum <!--\; \sum \; -->x_i=n⟹<!--\; ⟹\; -->nX=n⟹<!--\; ⟹\; -->X=1$ if we let $X=x_i$
So, max occurs at $(1,1,\dots <!--\; \dots \; -->,1)$ which is $f(1,1,\dots <!--\; \dots \; -->,1)=1$ Is this correct? I feel im missing something. thanks for any feedback.