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Difference between revisions of "Random Problem"

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==Solution==
 
==Solution==
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???
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'''Submitted by BinouTheGuineaPig''' | ''A step-by-step solution''
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We create a cross-section of the cylinder in the cone, "slicing down" from the apex of the cone as follows.
 +
 
 +
[png]
 +
 
 +
Volume of the cylinder
 +
 
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<math>V=\pi{r^2}h</math>
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<math>V=\pi{r^2}(14-2r)</math>
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 +
<math>V=14\pi{r^2}-2\pi{r^3}</math>
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 +
Now, we find all inflection points in the graph of this equation, by finding the values of <math>r</math> where the gradient is <math>0</math>, in other words where <math>\frac{dV}{dr}=0</math>.
 +
 
 +
<math>\frac{dV}{dr}=28\pi{r}-6\pi{r^2}</math>
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<math>28\pi{r}-6\pi{r^2}=0</math>
 +
 
 +
<math>3r^2-14r=0</math>
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<math>r(3r-14)=0</math>
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 +
<math>r=0</math> or <math>r=\frac{14}{3}</math>
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 +
We will obviously use the larger value of <math>r</math> to find the maximum volume of the cylinder, as follows.
 +
 
 +
<math>V_{max}=14\pi(\frac{14}{3})^2-2\pi(\frac{14}{3})^3</math>
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 +
<math>V_{max}=\frac{2744}{27}\pi \equiv \frac{a}{b}\pi</math>
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Therefore, the maximum cylinder volume can be expressed as <math>\boxed{\frac{2744}{27}\pi}</math>.
  
 
== Hard Problem ==
 
== Hard Problem ==

Revision as of 16:19, 24 March 2023

Easy Problem

The sum\[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\]can be expressed as $a-\frac{1}{b!}$, where $a$ and $b$ are positive integers. What is $a+b$?

$\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$

Solution

Submitted by BinouTheGuineaPig | A step-by-step solution

We see that the general form for each term can be expressed in terms of $n$ as follows.

$\frac{n}{(n+1)!} =\frac{(n+1)-1}{(n+1)!}$

$\qquad\qquad = \frac{n+1}{(n+1)(n!)} - \frac{1}{(n+1)!}$

$\qquad\qquad = \frac{1}{n!} - \frac{1}{(n+1)!}$

Now, to find the entire sum, it can be expressed as follows.

$\sum_{n=1}^{2022} \frac{n}{(n+1)!}$

$=\sum_{n=1}^{2022} \frac{1}{n!} - \frac{1}{(n+1)!}$

$=(\frac{1}{1!}+\cancel{\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots+\frac{1}{2022!}}) - (\cancel{\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots+\frac{1}{2022!}}+\frac{1}{2023!})$

Here, we see that a whole chunk of terms cancel each other out, leaving us with

$=1-\frac{1}{2023!} \equiv a-\frac{1}{b!}$

This means $a+b=1+2023=2024$.

Therefore, the answer is $\boxed{\textbf{(E)}}$.

Medium Problem

Show that there exist no finite decimals $a = 0.\overline{a_1a_2a_3\ldots a_n}$ such that when its digits are rearranged to a different decimal $b = 0.\overline{a_{b_1}a_{b_2}a_{b_3}\ldots a_{b_n}}$, $a + b = 1$.

Solution

???

Hardish Problem

A cylinder is inscribed in a circular cone with base radius of $7$ and height of $14$. What is the maximum possible volume of this cylinder is $\frac{a}{b}\pi$?

Solution

Submitted by BinouTheGuineaPig | A step-by-step solution

We create a cross-section of the cylinder in the cone, "slicing down" from the apex of the cone as follows.

[png]

Volume of the cylinder

$V=\pi{r^2}h$

$V=\pi{r^2}(14-2r)$

$V=14\pi{r^2}-2\pi{r^3}$

Now, we find all inflection points in the graph of this equation, by finding the values of $r$ where the gradient is $0$, in other words where $\frac{dV}{dr}=0$.

$\frac{dV}{dr}=28\pi{r}-6\pi{r^2}$

$28\pi{r}-6\pi{r^2}=0$

$3r^2-14r=0$

$r(3r-14)=0$

$r=0$ or $r=\frac{14}{3}$

We will obviously use the larger value of $r$ to find the maximum volume of the cylinder, as follows.

$V_{max}=14\pi(\frac{14}{3})^2-2\pi(\frac{14}{3})^3$

$V_{max}=\frac{2744}{27}\pi \equiv \frac{a}{b}\pi$

Therefore, the maximum cylinder volume can be expressed as $\boxed{\frac{2744}{27}\pi}$.

Hard Problem

A regular $48$-gon is inscribed in a circle with radius $1$. Let $X$ be the set of distances (not necessarily distinct) from the center of the circle to each side of the $48$-gon, and $Y$ be the set of distances (not necessarily distinct) from the center of the circle to each diagonal of the $48$-gon. Let $S$ be the union of $X$ and $Y$. What is the sum of the squares of all of the elements in $S$?

Solution

???

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