AMC12

\section{多项式}

\begin{ltbox}\begin{example}
(2018 AMC 12A 21) Which of the following polynomials has the greatest real root?
\begin{tasks}(5)
\task $x^{19}+2018x^{11}+1$

\task $x^{17}+2018x^{11}+1$

\task $x^{19}+2018x^{13}+1$

\task $x^{17}+2018x^{13}+1$

\task $2019x+2018$
\end{tasks}
\end{example}\end{ltbox}
\begin{solution}
Solution 1

We can see that our real solution has to lie in the open interval $(-1,0)$. From there, note that $x^a < x^b$ if $a$, $b$ are odd positive integers if $a<b$, so hence it can only either be B or E(as all of the other polynomials will be larger than the polynomial B). E gives the solution $x=-\frac{2018}{2019}$. We can approximate the root for B by using $x=-\frac 12$. $$(- \frac {1}{2}) ^{17} - \frac{2018}{2048} + 1 \approx 0$$ therefore the root for B is approximately $-\frac 12$. The answer is $\fbox{B}$. (cpma213)

Solution 2 (Calculus version of solution 1)

Note that $a(-1)=b(-1)=c(-1)=d(-1) < 0$ and $a(0)=b(0)=c(0)=d(0) > 0$. Calculating the definite integral for each function on the interval $[-1,0]$, we see that $B(x)\rvert^{0}_{-1}$ gives the most negative value. To maximize our real root, we want to maximize the area of the curve under the x-axis, which means we want our integral to be as negative as possible and thus the answer is $\fbox{B}$.

Solution 3 (Alternate Calculus Version)

Newton's Method is used to approximate the zero $x_{1}$ of any real valued function given an estimation for the root $x_{0}$: $x_{1}=x_{0}-{\frac {f(x_{0})}{f'(x_{0})}}\,.$ After looking at all the options, $x_{0}=-1$ gives a reasonable estimate. For options A to D, $f(-1) = -2018$ and the estimation becomes $x_{1}=-1+{\frac {2018}{f'(-1)}}\,.$ Thus we need to minimize the derivative, giving us B. Now after comparing B and E through Newton's method, we see that B has the higher root, so the answer is $\fbox{B}$. (Qcumber)

Solution 4

Let the real solution to $B$ be $a.$ It is easy to see that when $a$ is plugged in to $A,$ since $-1 < a < 0,$ $a^{19} < a^{17}$ thus making the real solution to $A$ more "negative", or smaller than $B.$ Similarly we can assert that $D > C.$ Now to compare $B$ and $D,$ we can use the same method to what we used before to compare $B$ to $A,$ in which it is easy to see that the smaller exponent $(11)$ "wins". Now, the only thing left is for us to compare $B$ and $E.$ Plugging $\frac{-2018}{2019}$ (or the solution to $E$) into $B$ we obtain $\frac{(-2018)^{17}}{2019^{17}} + 2018\frac{(-2018)^{11}}{2019^{11}} + 1,$ which is intuitively close to $-1 - 2018 + 1 = -2018,$ much smaller than the solution the required $0.$ (For a more rigorous proof, one can note that $(\frac{2018}{2019})^{17}$ and $(\frac{2018}{2019}^{11})$ are both much greater than $(\frac{2018}{2019})^{2019} \approx \frac{1}{e},$ by the limit definition of $e.$ Since $- \frac{1}{e} - 2018 \cdot \frac{1}{e} + 1$ is still much smaller the required $0$ for the solution to $B$ to be a solution, our answer is $\boxed{B}.$
\end{solution}

\begin{ltbox}\begin{example}
(2019 AMC 10A 24) Let $p$, $q$, and $r$ be the distinct roots of the polynomial $x^3 - 22x^2 + 80x - 67$. It is given that there exist real numbers $A$, $B$, and $C$ such that $$\dfrac{1}{s^3 - 22s^2 + 80s - 67} = \dfrac{A}{s-p} + \dfrac{B}{s-q} + \frac{C}{s-r}$$ for all $s\not\in\{p,q,r\}$. What is $\tfrac1A+\tfrac1B+\tfrac1C$?
\begin{tasks}(6)
\task $243$

\task $244$

\task $245$

\task $246$

\task $247$
\end{tasks}
\end{example}\end{ltbox}
\begin{solution}
Multiplying both sides by $(s-p)(s-q)(s-r)$ yields $$1 = A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q)$$ As this is a polynomial identity, and it is true for infinitely many $s$, it must be true for all $s$ (since a polynomial with infinitely many roots must in fact be the constant polynomial $0$). This means we can plug in $s = p$ to find that $\frac1A = (p-q)(p-r)$. Similarly, we can find $\frac1B = (q-p)(q-r)$ and $\frac1C = (r-p)(r-q)$. Summing them up, we get that $$\frac1A + \frac1B + \frac1C = p^2 + q^2 + r^2 - pq - qr - pr$$ By Vieta's Formulas, we know that $p^2 + q^2 + r^2 = (p+q+r)^2 - 2(pq + qr + pr) = 324$ and $pq + qr + pr = 80$. Thus the answer is $324 -80 = \boxed{\textbf{(B) } 244}$.

Note: this process of substituting in the 'forbidden' values in the original identity is a standard technique for partial fraction decomposition, as taught in calculus classes.
\end{solution}

 

\begin{ltbox}\begin{example}
(2015 Grade 6)
\begin{tasks}(5)
\task A

\task B

\task C

\task D
\end{tasks}
\end{example}\end{ltbox}
\begin{solution}

\end{solution}

 

\begin{ltbox}\begin{example}
(2015 Grade 6)
\begin{tasks}(5)
\task A

\task B

\task C

\task D
\end{tasks}
\end{example}\end{ltbox}
\begin{solution}

\end{solution}

 

\begin{ltbox}\begin{example}
(2015 Grade 6)
\begin{tasks}(5)
\task A

\task B

\task C

\task D
\end{tasks}
\end{example}\end{ltbox}
\begin{solution}

\end{solution}

\section{代数综合问题}

\begin{ltbox}\begin{example}
(2019 AMC 12B 8) Let $f(x) = x^{2}(1-x)^{2}$. What is the value of the sum $$f \left(\frac{1}{2019} \right)-f \left(\frac{2}{2019} \right)+f \left(\frac{3}{2019} \right)-f \left(\frac{4}{2019} \right)+\cdots + f \left(\frac{2017}{2019} \right) - f \left(\frac{2018}{2019} \right)?$$
\begin{tasks}(6)
\task $0$

\task $\frac{1}{2019^{4}}$

\task $\frac{2018^{2}}{2019^{4}}$

\task $\frac{2020^{2}}{2019^{4}}$

\task $1$
\end{tasks}
\end{example}\end{ltbox}
$$\begin{solution} First, note that $f(x) = f(1-x)$. We can see this since\[f(x) = x^2(1-x)^2 = (1-x)^2x^2 = (1-x)^{2}\left(1-\left(1-x\right)\right)^{2} = f(1-x)\]Using this result, we regroup the terms accordingly:\[\left( f \left(\frac{1}{2019} \right) - f \left(\frac{2018}{2019} \right) \right) + \left( f \left(\frac{2}{2019} \right) - f \left(\frac{2017}{2019} \right) \right) + \cdots + \left( f \left(\frac{1009}{2019} \right) - f \left(\frac{1010}{2019} \right) \right)\]\[= \left( f \left(\frac{1}{2019} \right) - f \left(\frac{1}{2019} \right) \right) + \left( f \left(\frac{2}{2019} \right) - f \left(\frac{2}{2019} \right) \right) + \cdots + \left( f \left(\frac{1009}{2019} \right) - f \left(\frac{1009}{2019} \right) \right)\]Now it is clear that all the terms will cancel out (the series telescopes), so the answer is $\boxed{\textbf{(A) }0}$. \end{solution}$$

\begin{ltbox}\begin{example}
(2018 AMC 12A 19) Let $A$ be the set of positive integers that have no prime factors other than $2$, $3$, or $5$. The infinite sum $$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots$$ of the reciprocals of the elements of $A$ can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?
\begin{tasks}(6)
\task $16$

\task $17$

\task $19$

\task $23$

\task $36$
\end{tasks}
\end{example}\end{ltbox}
$$\begin{solution} It's just\[\sum_{a=0}^\infty\frac1{2^a}\sum_{b=0}^\infty\frac1{3^b}\sum_{c=0}^\infty\frac{1}{5^c} =\sum_{a=0}^\infty\sum_{b=0}^\infty\sum_{c=0}^\infty\frac1{2^a3^b5^c} = 2 \cdot \frac32 \cdot \frac54 = \frac{15}{4}\Rightarrow\textbf{(C)}.\]since this represents all the numbers in the denominator. (athens2016) \end{solution}$$

\begin{ltbox}\begin{example}
(2015 Grade 6)
\begin{tasks}(6)
\task $0$

\task $0$

\task $0$

\task $0$
\end{tasks}
\end{example}\end{ltbox}
\begin{solution}

\end{solution}

\begin{ltbox}\begin{example}
(2015 Grade 6)
\begin{tasks}(6)
\task $0$

\task $0$

\task $0$

\task $0$
\end{tasks}
\end{example}\end{ltbox}
\begin{solution}

\end{solution}

\section{数论}

\begin{ltbox}\begin{example}
(2018 AMC 12B 17) Let $p$ and $q$ be positive integers such that $$\frac{5}{9} < \frac{p}{q} < \frac{4}{7}$$ and $q$ is as small as possible. What is $q-p$?
\begin{tasks}(6)
\task $7$

\task $11$

\task $13$

\task $17$

\task $19$
\end{tasks}
\end{example}\end{ltbox}
\begin{solution}
Solution 1

We claim that, between any two fractions $a/b$ and $c/d$, if $bc-ad=1$, the fraction with smallest denominator between them is $\frac{a+c}{b+d}$. To prove this, we see that

$$\frac{1}{bd}=\frac{c}{d}-\frac{a}{b}=\left(\frac{c}{d}-\frac{p}{q}\right)+\left(\frac{p}{q}-\frac{a}{b}\right) \geq \frac{1}{dq}+\frac{1}{bq},$$ which reduces to $q\geq b+d$. We can easily find that $p=a+c$, giving an answer of $\boxed{\textbf{(A)}\ 7}$.

Solution 2 (requires justification)

Assume that the difference $\frac{p}{q} - \frac{5}{9}$ results in a fraction of the form $\frac{1}{9q}$. Then,

$9p - 5q = 1$

Also assume that the difference $\frac{4}{7} - \frac{p}{q}$ results in a fraction of the form $\frac{1}{7q}$. Then,

$4q - 7p = 1$

Solving the system of equations yields $q=16$ and $p=9$. Therefore, the answer is $\boxed{\textbf{(A)}\ 7}$

Solution 3

Cross-multiply the inequality to get $$35q < 63p < 36q.$$
Then, $$0 < 63p-35q < q,$$ $$0 < 7(9p-5q) < q.$$
Since $p$, $q$ are integers, $9p-5q$ is an integer. To minimize $q$, start from $9p-5q=1$, which gives $p=\frac{5q+1}{9}$. This limits $q$ to be greater than $7$, so test values of $q$ starting from $q=8$. However, $q=8$ to $q=14$ do not give integer values of $p$.

Once $q>14$, it is possible for $9p-5q$ to be equal to $2$, so $p$ could also be equal to $\frac{5q+2}{9}.$ The next value, $q=15$, is not a solution, but $q=16$ gives $p=\frac{5\cdot 16 + 1}{9} = 9$. Thus, the smallest possible value of $q$ is $16$, and the answer is $16-9= \boxed{\textbf{(A)}\ 7}$.

Solution 4

Graph the regions $y > \frac{5}{9}x$ and $y < \frac{4}{7}x$. Note that the lattice point $(16,9)$ is the smallest magnitude one which appears within the region bounded by the two graphs. Thus, our fraction is $\frac{9}{16}$ and the answer is $16-9= \boxed{\textbf{(A)}\ 7}$.

Remark: This also gives an intuitive geometric proof of the mediant using vectors.

Solution 5 (Using answer choices to prove mediant)

As the other solutions do, the mediant $=\frac{9}{16}$ is between the two fractions, with a difference of $\boxed{\textbf{(A)}\ 7}$. Suppose that the answer was not $A$, then the answer must be $B$ or $C$ as otherwise $p$ would be negative. Then, the possible fractions with lower denominator would be $\frac{k-11}{k}$ for $k=12,13,14,15$ and $\frac{k-13}{k}$ for $k=14,15,$ which are clearly not anywhere close to $\frac{4}{7}\approx 0.6$

Solution 6

Inverting the given inequality we get $$\frac{7}{4} < \frac{q}{p} < \frac{9}{5}$$
which simplifies to $$35p < 20q < 36p$$
We can now substitute $q = p + k$. Note we need to find $k$.

$$35p < 20p + 20k < 36p$$
which simplifies to $$15p < 20k < 16p$$
Cleary $p$ is greater than $k$. We will now substitute $p = k + x$ to get

$$15k + 15x < 20k < 16k + 16x$$
The inequality $15k + 15x < 20k$ simplifies to $3x < k$. The inequality $20k < 16k + 16x$ simplifies to $k < 4x$. Combining the two we get $$3x < k < 4x$$
Since $x$ and $k$ are integers, the smallest values of $x$ and $k$ that satisfy the above equation are $2$ and $7$ respectively. Substituting these back in, we arrive with an answer of $\boxed{\textbf{(A)}\ 7}$.

Solution 7

Start with $\frac{5}{9}$. Repeat the following process until you arrive at the answer: if the fraction is less than or equal to $\frac{5}{9}$, add $1$ to the numerator; otherwise, if it is greater than or equal to $\frac{4}{7}$, add one $1$ to the denominator. We have:

$$\frac{5}{9}, \frac{6}{9}, \frac{6}{10}, \frac{6}{11}, \frac{7}{11}, \frac{7}{12}, \frac{7}{13}, \frac{8}{13}, \frac{8}{14}, \frac{8}{15}, \frac{9}{15}, \frac{9}{16}$$
$16 - 9 = \boxed{\textbf{(A)}\ 7}$.

Solution 8

Because q and p are positive integers with $p<q$, we can let $q=p+k$ where $k\in{\mathbb{Z}}$. Now, the problem condition reduces to

$\frac{5}{9}<\frac{p}{p+k}<\frac{4}{7}$

Our first inequality is $\frac{5}{9}<\frac{p}{p+k}$ which gives us $5p+5k<9p\implies \frac{5}{4}k<p$.

Our second inequality is $\frac{p}{p+k}<\frac{4}{7}$ which gives us $7p<4p+4k\implies p<\frac{4}{3}k$.

Hence, $\frac{5}{4}k<p<\frac{4}{3}k\implies 15k<12p<16k$.

It is clear that we are aiming to find the least positive integer value of k such that there is at least one value of p that satisfies the inequality.

Now, simple casework through the answer choices of the problem reveals that $q-p=p+k-p=k\implies k\ge{\boxed{7}}$.

Solution 9 (Quick inspection)
Checking possible fractions within the interval can get us to the answer, but only if we do it with more skill. The interval can also be written as $0.5556<x<0.5714$. This represents fraction with the numerator a little bit more than half the denominator. Every fraction we consider must not exceed this range.

The denominators to be considered are $9,10,11,12...$. We check $\frac{6}{10}, \frac{6}{11}, \frac{7}{12}, \frac{7}{13}, \frac{8}{15}, \frac{9}{16}$. At this point we know that we've got our fraction and our answer is $16-9=\boxed{\textbf{A } 7}$

The inspection was made faster by considering the fact that $\frac{a+1}{b+1}>\frac{a}{b}$.

So, once a fraction was gotten which was greater than $\frac{4}{7}$ we jump to the next denominator.

We then make sure we consider fractions with higher positive difference between the denominator and numerator. And we also do not forget that the numerator must be greater than half of the denominator.

($\frac{8}{14}$ was obviously skipped because it is equal to $\frac{4}{7}$.)
\end{solution}

\begin{ltbox}\begin{example}
(1973 USAMO 2) Let $\{X_n\}$ and $\{Y_n\}$ denote two sequences of integers defined as follows:

$X_0=1$, $X_1=1$, $X_{n+1}=X_n+2X_{n-1}$ $(n=1,2,3,\dots),$
$Y_0=1$, $Y_1=7$, $Y_{n+1}=2Y_n+3Y_{n-1}$ $(n=1,2,3,\dots)$.
Thus, the first few terms of the sequences are:

$X:1, 1, 3, 5, 11, 21, \dots$,
$Y:1, 7, 17, 55, 161, 487, \dots$.
Prove that, except for the "1", there is no term which occurs in both sequences.
\end{example}\end{ltbox}
\begin{solution}
We can look at each sequence $\bmod{8}$:

$X$: $1$, $1$, $3$, $5$, $3$, $5$, $\dots$,
$Y$: $1$, $7$, $1$, $7$, $1$, $7$, $\dots$.
Proof that $X$ repeats $\bmod{8}$:
The third and fourth terms are $3$ and $5$ $\bmod{8}$. Plugging into the formula, we see that the next term is $11\equiv 3\bmod{8}$, and plugging $5$ and $3$, we get that the next term is $13\equiv 5\bmod{8}$. Thus the sequence $X$ repeats, and the pattern is $3,5,3,5,\dots$.

Proof that $Y$ repeats $\bmod{8}$:
The first and second terms are $1$ and $7$ $\bmod{8}$. Plugging into the formula, we see that the next term is $17\equiv 1\bmod{8}$, and plugging $7$ and $1$, we get that the next term is $23\equiv 7\bmod{8}$. Thus the sequence $Y$ repeats, and the pattern is $1,7,1,7,\dots$.

Combining both results, we see that $X_i$ and $Y_j$ are not congruent $\bmod{8}$ when $i\geq 3$ and $j\geq 2$. Thus after the "1", the terms of each sequence are not equal.
\end{solution}

$$\begin{ltbox}\begin{example} (2020CHKMO Q1) Given that ${a_n}$ and ${b_n}$ are two sequences of integers defined by \begin{align*} a_1=1, a_2=10, a_{n+1}=2a_n+3a_{n-1} & ~~~\text{for }n=2,3,4,\ldots, \\ b_1=1, b_2=8, b_{n+1}=3b_n+4b_{n-1} & ~~~\text{for }n=2,3,4,\ldots. \end{align*} Prove that, besides the number $1$, no two numbers in the sequences are identical. \end{example}\end{ltbox}$$
\begin{solution}
In the standard way, we compute the explicit formulas for each sequence:
$$\begin{align*} a_n &= \frac{11 \cdot 3^n + 21(-1)^n}{12} \\ b_n &= \frac{9 \cdot 4^n + 16 (-1)^n }{20}\\ \end{align*}$$ Thus it is enough to show that the equation
$$55 \cdot 3^n + 105(-1)^n = 27 \cdot 4^m + 48(-1)^m$$ has no solutions over the integers. Assume otherwise for the sake of contradiction. One can easily check that there is no $b_i=10=a_2$. Henceforth we assume $n > 2$. Then taken mod 27, we obtain
$$105 \equiv \pm 48 \pmod{27}$$ which is false.

2. $a_n \ (\text{mod}\ 9)$ is $(1,10,5,4,5,4,5,4,...)$
$b_n\ (\text{mod}\ 9)$ is $(1,8,1,8,1,8,1,8,...)$
Now you have to observe that $n\ge 3\implies b_n\ge a_1$
\end{solution}

\begin{ltbox}\begin{example}
(2019 AMC 10A 25) For how many integers $n$ between $1$ and $50$, inclusive, is $$\frac{(n^2-1)!}{(n!)^n}$$ an integer? (Recall that $0! = 1$.)
\begin{tasks}(5)
\task $31$

\task $32$

\task $33$

\task $34$

\task $35$
\end{tasks}
\end{example}\end{ltbox}
\begin{solution}
Solution 1

The main insight is that

$$\frac{(n^2)!}{(n!)^{n+1}}$$
is always an integer. This is true because it is precisely the number of ways to split up $n^2$ objects into $n$ unordered groups of size $n$. Thus,

$$\frac{(n^2-1)!}{(n!)^n}=\frac{(n^2)!}{(n!)^{n+1}}\cdot\frac{n!}{n^2}$$
is an integer if $n^2 \mid n!$, or in other words, if $n \mid (n-1)!$. This condition is false precisely when $n=4$ or $n$ is prime, by Wilson's Theorem. There are $15$ primes between $1$ and $50$, inclusive, so there are $15 + 1 = 16$ terms for which

$$\frac{(n^2-1)!}{(n!)^{n}}$$
is potentially not an integer. It can be easily verified that the above expression is not an integer for $n=4$ as there are more factors of $2$ in the denominator than the numerator. Similarly, it can be verified that the above expression is not an integer for any prime $n=p$, as there are more factors of p in the denominator than the numerator. Thus all $16$ values of n make the expression not an integer and the answer is $50-16=\boxed{\textbf{(D)}\ 34}$.

Solution 2

We can use the P-Adic Valuation (more info could be found here: Mathematicial notation) of n to solve this problem (recall the P-Adic Valuation of 'n' is denoted by $v_p (n)$ and is defined as the greatest power of some prime 'p' that divides n. For example, $v_2 (6)=1$ or $v_7 (245)=2$ .) Using Legendre's formula, we know that :

$$v_p (n!)= \sum_{i=1}^\infty \lfloor \frac {n}{p^i} \rfloor$$
Seeing factorials involved in the problem, this prompts us to use Legendre's formula where n is a power of a prime.

We also know that , $v_p (m^n) = n \cdot v_p (m)$ . Knowing that $a\mid b$ if $v_p (a) \le v_p (b)$ , we have that :

$$n \cdot v_p (n!) \le v_p ((n^2 -1 )!)$$ and we must find all n for which this is true.

If we plug in $n=p$, by Legendre's we get two equations:

$$v_p ((n^2 -1)!) = \sum_{i=1}^\infty \lfloor \frac {n^2 -1}{p^i} \rfloor = (p-1)+0+...+0 = p-1$$
And we also get :

$$v_p ((n!)^n) = n \cdot v_p (n!)= n \cdot \sum_{i=1}^\infty \lfloor \frac {n}{p^i} \rfloor = p \cdot ( 1+0+...0) = p$$
But we are asked to prove that $n \cdot v_p (n!) \le v_p ((n^2 -1 )!) \Longrightarrow p \le p-1$ which is false for all 'n' where n is prime.

Now we try the same for $n=p^2$ , where p is a prime. By Legendre we arrive at:

$$v_p ((p^4 -1)!) = p^3 + p^2 + p -3$$ and $$p^2 \cdot v_p (p^2 !) = p^3 + p^2$$
Then we get:

$$p^2 \cdot v_p (p!) \le v_p ((n^4 -1)!) \Longrightarrow p^3 + p^2 \le p^3 + p^2 + p -3$$ Which is true for all primes except for 2, so $2^2 = 4$ doesn't work. It can easily be verified that for all $n=p^i$ where $i$ is an integer greater than 2, satisfies the inequality : $$n \cdot v_p (n!) \le v_p ((n^2 -1 )!).$$
Therefore, there are 16 values that don't work and $50-16 = \boxed{\mathbf{(D)}\ 34}$ values that work.

Solution 3 (Guessing)

First, we see that $n=1, 6, 8, 9, 10, 12, 14$ work. This leads us to the conclusion of $50-16 = \boxed{\textbf{(D)}\ 34}$
\end{solution}

\begin{ltbox}\begin{example}
(2015 Grade 6)
\begin{tasks}(5)
\task A

\task B

\task C

\task D
\end{tasks}
\end{example}\end{ltbox}
\begin{solution}

\end{solution}

 

\begin{ltbox}\begin{example}
(2015 Grade 6)
\begin{tasks}(5)
\task A

\task B

\task C

\task D
\end{tasks}
\end{example}\end{ltbox}
\begin{solution}

\end{solution}

 

 

\begin{ltbox}\begin{example}
(2015 Grade 6)
\begin{tasks}(5)
\task A

\task B

\task C

\task D
\end{tasks}
\end{example}\end{ltbox}
\begin{solution}

\end{solution}

\begin{ltbox}\begin{example}
(2015 Grade 6)
\begin{tasks}(6)
\task A

\task B

\task C

\task D

\task D
\end{tasks}
\end{example}\end{ltbox}
\begin{solution}

\end{solution}

\section{概率}

\begin{ltbox}\begin{example}
(2016 AMC 12A 23) Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
\begin{tasks}(6)
\task $\frac16$

\task $\frac13$

\task $\frac12$

\task $\frac23$

\task $\frac56$
\end{tasks}
\end{example}\end{ltbox}
\begin{solution}
Solution 1: Super WLOG

WLOG assume $a$ is the largest. Scale the triangle to $1,{b}/{a},{c}/{a}$ or $1,x,y$ with $x,y \in [0,1]$ and equally likely to be any pair of numbers within the interval. Then, $x+y>1$, meaning the solution is $\boxed{\textbf{(C)}\;1/2}$, as shown in the graph below.

Solution 2: Conditional Probability

WLOG, let the largest of the three numbers drawn be $a>0$. Then the other two numbers are drawn uniformly and independently from the interval $[0,a]$. The probability that their sum is greater than $a$ is $\boxed{\textbf{(C)}\;1/2.}$

Solution 8: 3D geometry

We can draw a 3D space where each coordinate is in the range [0,1]. Drawing the lines $x+y>z, x+z>y,$ and $y+z>x,$ We have a 3D space that consists of two tetrahedrons. One is a regular tetrahedron with side length $\sqrt2$ and the other has 3 sides of length $\sqrt2$ and 3 sides of length $1.$ The volume of this region is $\frac 1 2$. Hence our solution is $C.$
\end{solution}

\begin{ltbox}\begin{example}
(2015 AMC 12A 23) Let $S$ be a square of side length 1. Two points are chosen independently at random on the sides of $S$. The probability that the straight-line distance between the points is at least $\frac12$ is $\frac{a-b\pi}{c}$, where $a,b,$ and $c$ are positive integers and $\text{gcd}(a,b,c) = 1$. What is $a+b+c$?
\begin{tasks}(6)
\task $59$

\task $60$

\task $61$

\task $62$

\task $63$
\end{tasks}
\end{example}\end{ltbox}
\begin{solution}
Divide the boundary of the square into halves, thereby forming 8 segments. Without loss of generality, let the first point $A$ be in the bottom-left segment. Then, it is easy to see that any point in the 5 segments not bordering the bottom-left segment will be distance at least $\dfrac{1}{2}$ apart from $A$. Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least 0.5 apart from $A$ is $\dfrac{0 + 1}{2} = \dfrac{1}{2}$ because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.)

If the second point $B$ is on the left-bottom segment, then if $A$ is distance $x$ away from the left-bottom vertex, then $B$ must be at least $\dfrac{1}{2} - \sqrt{0.25 - x^2}$ away from that same vertex. Thus, using an averaging argument we find that the probability in this case is $$\frac{1}{\frac{1}{2}^2} \int_0^{\frac{1}{2}} \dfrac{1}{2} - \sqrt{0.25 - x^2} dx = 4\left(\frac{1}{4} - \frac{\pi}{16}\right) = 1 - \frac{\pi}{4}.$$
(Alternatively, one can equate the problem to finding all valid $(x, y)$ with $0 < x, y < \dfrac{1}{2}$ such that $x^2 + y^2 \ge \dfrac{1}{4}$, i.e. (x, y) is outside the unit circle with radius 0.5.)

Thus, averaging the probabilities gives $$P = \frac{1}{8} \left(5 + \frac{1}{2} + 1 - \frac{\pi}{4}\right) = \frac{1}{32} (26 - \pi).$$
Our answer is $\textbf{(A)}$.

Case Solution
Fix one of the points on a SIDE. There are three cases: the other point is on the same side, a peripheral side, or the opposite side, with probability $0.25, 0.5, 0.25$, respectively.

Opposite side: Probability is obviously $1$, no matter what.

Same side: Pretend the points are on a line with coordinates $x$ and $y$. If $|a-b| \le \frac{1}{2}$, drawing a graph will give probability $\frac{1}{4}$.

Peripheral side: superimpose a coordinate system over the points; call them $(x, 0)$ and $(0, y)$. WLOG set $x, y >= 0$ and $x, y <= 1$. We need $x^2+y^2>0.25$, and drawing the coordinate system with bounds $(0, 0), (1, 0), (0, 1), (1, 1)$ gives probability $1-\frac{\pi}{16}$ that the distance between the points is $>0.5$.

Adding these up and finding the fraction gives us $\frac{1}{32} (26 - \pi)$ for an answer of $\textbf{(A)}\ 59$.

Solution 5 (Area)

Choose a certain side for one of the points to be on. Let the distance from the point to the vertex on its left be $x$

We split this into two cases:

Case 1: $0\leq x\leq \frac12$:

The total length of the segments for which the other point can be on such that the straight-line distance between the points is less than $\frac12$ is $$\sqrt{\frac14-x^2}+x+\frac12.$$ We can graph this in the Cartesian plane and find the area of the region below the curve and above the line $y=0$.

Case 2: $\frac12< x\leq 1$:

This is basically Case 1 but flipped over the line $x=\frac12$.

So our total probability is 1 minus the area of the graph over the total area (4, perimeter of square). Notice that the desired area of the region below the curve we found earlier is the sum of a quarter circle with radius $\frac12$ and centered at $(0,0)$ and a trapezoid with height $\frac12$ and bases of length $\frac12$ and $\frac32$. Adding this all up then multiplying by 2, we have $$\frac{\pi}{8}+\frac34$$ and then the probability of the desired result would be $$1-\frac{\pi+6}{32}=\frac{26-\pi}{32}$$ and our answer is $26+1+32=\boxed{59}$.
\end{solution}

\begin{ltbox}\begin{example}
(2015 Grade 6)
\begin{tasks}(6)
\task A

\task B

\task C

\task D

\task D
\end{tasks}
\end{example}\end{ltbox}
\begin{solution}

\end{solution}

\begin{ltbox}\begin{example}
(2015 Grade 6)
\begin{tasks}(6)
\task A

\task B

\task C

\task D

\task D
\end{tasks}
\end{example}\end{ltbox}
\begin{solution}

\end{solution}

 

\begin{ltbox}\begin{example}
(2015 Grade 6)
\begin{tasks}(6)
\task A

\task B

\task C

\task D

\task D
\end{tasks}
\end{example}\end{ltbox}
\begin{solution}

\end{solution}

\begin{ltbox}\begin{example}
(2015 Grade 6)
\begin{tasks}(6)
\task A

\task B

\task C

\task D

\task D
\end{tasks}
\end{example}\end{ltbox}
\begin{solution}

\end{solution}