Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: 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?
$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$

Solution: This problem is going to require some geometric probability, so let's dive right in.
Take three integers $x$ $y$ , and $z$ . Then for the triangle inequality to hold, the following $3$ inequalities must be true
\[x + y > z\] \[y + z > x\] \[x + z > y\]
Now, it would be really easy if these equations only had two variables instead of $3$ , because then we could graph it in a $2$ -dimensional plane instead of a $3$ -dimensional cube. So, we assume $z$ is a constant. We will deal with it later.
Now, since we are graphing, we should probably write these equations in terms of $y$ so they are in slope-intercept form and are easier to graph.
\[y > -x + z\] \[y > x - z\] \[y < x + z\]
Now, note that all solutions $(x,y)$ are in a $1$ $x$ $1$ square in the $xy$ -plane because $x$ $y \in [0, 1]$
I recommend drawing the following figure to get an idea of what is going on.
The first line is a line with a negative slope that cuts off a $45-45-90$ triangle with side length $z$ of the bottom left corner of the square. The second line is a line with a positive slope that cuts off a $45-45-90$ triangle with side length $1-z$ off the top left corner of the square. The third line also has a positive slope and cuts a $45-45-90$ triangle with side length $1 - z$ off the bottom right corner of the square.
Note: All triangles are $45-45-90$ because the lines have slopes of $1$ or $-1$
Using the $>$ and $<$ signs in the lines, we see that the area that satisfies all three inequalities is the area not enclosed in the three triangles. So, our plan of attack will be to
Find the area of the triangles -> Subtract that from the area of the square -> Use probability to get the answer.
Except, now, we have one problem. $z$ is still a variable. But, we want $z$ to be a constant. Well, what if we just took the area over every possible value of $z$ ? Well, that would be a bit hard, if not impossible to do by hand, but there is a handy math tool that will let us do that: the integral!
To find the area of the triangles, our plan of attack will be
Find the area in terms of $z$ -> take the integral from $0$ to $1$ of the expression for the area (this will cover every possible value of $z$
The area of the triangles is
\[\frac{z^2}{2} + (1-z)^2 = \frac{1}{2}\left(-3x^2 + 4x\right)\]
The integral from $0$ to $1$ is
\[\frac{1}{2}\int_0^1\left(-3x^2 + 4x\right)dx = \frac{1}{2}\]
The total area of all the possible unit squares is quite obviously
\[\int_0^11dx = 1\]
Thus, the area not enclosed by the triangles is $1 - \dfrac{1}{2} = \dfrac{1}{2}$ , and the total area of the square is $1$ . Thus, the desired probability is
\[\frac{\frac{1}{2}}{1} = \boxed{12}\]

Answer: C 12
Problem: 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?
$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$

Solution: Consider a stick of length $1$ . Cutting the stick at two random points gives a triangle from the three new segments. These two random points must be on opposite sides of the halfway mark. Thus, after the first cut is made, there is $\boxed{12}$ probability that the second cut is on the opposite side.

Answer: C 12
Problem: There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$

Solution: The acceleration must be zero at the $x$ -intercept; this intercept must be an inflection point for the minimum $a$ value.
Derive $f(x)$ so that the acceleration $f''(x)=0$ . Using the power rule, \begin{align*} f(x) &= x^3-ax^2+bx-a \\ f’(x) &= 3x^2-2ax+b \\  f’’(x) &= 6x-2a \end{align*} So $x=\frac{a}{3}$ for the inflection point/root. Furthermore, the slope of the function must be zero - maximum - at the intercept, thus having a triple root at $x=a/3$ (if the slope is greater than zero, there will be two complex roots and we do not want that).
The function with the minimum $a$
\[f(x)=\left(x-\frac{a}{3}\right)^3\] \[x^3-ax^2+\left(\frac{a^2}{3}\right)x-\frac{a^3}{27}\] Since this is equal to the original equation $x^3-ax^2+bx-a$ , equating the coefficients, we get that
\[\frac{a^3}{27}=a\rightarrow a^2=27\rightarrow a=3\sqrt{3}\] \[b=\frac{a^2}{3}=\frac{27}{3}=\boxed{9}\]

Answer: B 9
Problem: There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$

Solution: Let the roots of the polynomial be $r, s, t$ . By Vieta's formulas we have $r+s+t=a$ $rs+st+rt=b$ , and $rst=a$ . Since both $a$ and $b$ are positive, it follows that all 3 roots $r, s, t$ are positive as well, and so we can apply AM-GM to get \[\tfrac 13 (r+s+t) \ge \sqrt[3]{rst} \quad \Rightarrow \quad a \ge 3\sqrt[3]{a}.\] Cubing both sides and then dividing by $a$ (since $a$ is positive we can divide by $a$ and not change the sign of the inequality) yields \[a^2 \ge 27 \quad \Rightarrow \quad a \ge  3\sqrt{3}.\] Thus, the smallest possible value of $a$ is $3\sqrt{3}$ which is achieved when all the roots are equal to $\sqrt{3}$ . For this value of $a$ , we can use Vieta's to get $b=\boxed{9}$

Answer: B 9
Problem: There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$

Solution: We see that with cubics, the number $3$ comes up a lot, and as $9=3\cdot3$ has the most relation to $3$ , we can assume $b=\boxed{9}$

Answer: B 9
Problem: What is the value of $\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}$ when $a= \frac{1}{2}$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20$

Solution: By: Dragonfly
We find that $a^{-1}$ is the same as $2$ , since a number to the power of $-1$ is just the reciprocal of that number. We then get the equation to be
\[\frac{2\times2+\frac{2}{2}}{\frac{1}{2}}\]
We can then simplify the equation to get $\boxed{10}$

Answer: D 10
Problem: The harmonic mean of two numbers can be calculated as twice their product divided by their sum. The harmonic mean of $1$ and $2016$ is closest to which integer?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 45 \qquad \textbf{(C)}\ 504 \qquad \textbf{(D)}\ 1008 \qquad \textbf{(E)}\ 2015$

Solution: Since the harmonic mean is $2$ times their product divided by their sum, we get the equation
$\frac{2\times1\times2016}{1+2016}$
which is then
$\frac{4032}{2017}$
which is finally closest to $\boxed{2}$ .
-dragonfly

Answer: A 2
Problem: Let $x=-2016$ . What is the value of $\bigg|$ $||x|-x|-|x|$ $\bigg|$ $-x$
$\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048$

Solution: By: dragonfly
First of all, lets plug in all of the $x$ 's into the equation.
$\bigg|$ $||-2016|-(-2016)|-|-2016|$ $\bigg|$ $-(-2016)$
Then we simplify to get
$\bigg|$ $|2016+2016|-2016$ $\bigg|$ $+2016$
which simplifies into
$\bigg|$ $2016$ $\bigg|$ $+2016$
and finally we get $\boxed{4032}$

Answer: D 4032
Problem: Let $x=-2016$ . What is the value of $\bigg|$ $||x|-x|-|x|$ $\bigg|$ $-x$
$\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048$

Solution: Consider $x$ is negative.
We replace all instances of $x$ with $|x|$
$\bigg|$ $||x|+|x||-|x|$ $\bigg|$ $+|x|$
$=$ $\bigg|$ $|2x|-|x|$ $\bigg|$ $+|x|$
$=$ $|x|$ $+|x|$
$=|2x|$
$=4032 \implies \boxed{4032}$

Answer: D 4032
Problem: The ratio of the measures of two acute angles is $5:4$ , and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?
$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$

Solution: By: dragonfly
We set up equations to find each angle. The larger angle will be represented as $x$ and the smaller angle will be represented as $y$ , in degrees. This implies that
$4x=5y$
and
$2\times(90-x)=90-y$
since the larger the original angle, the smaller the complement.
We then find that $x=75$ and $y=60$ , and their sum is $\boxed{135}$

Answer: C 135
Problem: The ratio of the measures of two acute angles is $5:4$ , and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?
$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$

Solution: We can visualize the problem like so: [asy] path b = brace((0,10),(90,10),5); draw(b); label("$90^\circ$",b,N);  draw("$5x$",(0,0)--(75,0),N); draw((75,2.5)--(75,-2.5)); draw("$1y$",(75,0)--(90,0),N);  draw("$4x$",(0,-10)--(60,-10),S); draw((60,-7.5)--(60,-12.5)); draw("$2y$",(60,-10)--(90,-10),S);  draw((0,5)--(0,-15)); draw((90,5)--(90,-15)); [/asy] \[5x+1y = 90^\circ = 4x+2y\] Moving like terms to the same side gets $x = y$ , and substituting this back gets $6x = 90^\circ \implies x = \frac{90^\circ}{6} = 15^\circ$ , so the sum of the degree measures is $5x + 4x = 9x = 9(15) = \boxed{135}$ . ~ emerald_block

Answer: C 135
Problem: All three vertices of $\bigtriangleup ABC$ lie on the parabola defined by $y=x^2$ , with $A$ at the origin and $\overline{BC}$ parallel to the $x$ -axis. The area of the triangle is $64$ . What is the length of $BC$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16$

Solution: By: Albert471
Plotting points $B$ and $C$ on the graph shows that they are at $\left( -x,x^2\right)$ and $\left( x,x^2\right)$ , which is isosceles. By setting up the triangle area formula you get: $64=\frac{1}{2}*2x*x^2  =   64=x^3$ Making x=4, and the length of $BC$ is $2x$ , so the answer is $\boxed{8}$

Answer: C 8
Problem: Josh writes the numbers $1,2,3,\dots,99,100$ . He marks out $1$ , skips the next number $(2)$ , marks out $3$ , and continues skipping and marking out the next number to the end of the list. Then he goes back to the start of his list, marks out the first remaining number $(2)$ , skips the next number $(4)$ , marks out $6$ , skips $8$ , marks out $10$ , and so on to the end. Josh continues in this manner until only one number remains. What is that number?
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 56 \qquad \textbf{(D)}\ 64 \qquad \textbf{(E)}\ 96$

Solution: Following the pattern, you are crossing out...
Time 1: Every non-multiple of $2$
Time 2: Every non-multiple of $4$
Time 3: Every non-multiple of $8$
Following this pattern, you are left with every multiple of $64$ which is only $\boxed{64}$

Answer: D 64
Problem: A thin piece of wood of uniform density in the shape of an equilateral triangle with side length $3$ inches weighs $12$ ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of $5$ inches. Which of the following is closest to the weight, in ounces, of the second piece?
$\textbf{(A)}\ 14.0\qquad\textbf{(B)}\ 16.0\qquad\textbf{(C)}\ 20.0\qquad\textbf{(D)}\ 33.3\qquad\textbf{(E)}\ 55.6$

Solution: By: dragonfly
We can solve this problem by using similar triangles, since two equilateral triangles are always similar. We can then use
$\left(\frac{3}{5}\right)^2=\frac{12}{x}$
We can then solve the equation to get $x=\frac{100}{3}$ which is closest to $\boxed{33.3}$

Answer: D 33.3
Problem: A thin piece of wood of uniform density in the shape of an equilateral triangle with side length $3$ inches weighs $12$ ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of $5$ inches. Which of the following is closest to the weight, in ounces, of the second piece?
$\textbf{(A)}\ 14.0\qquad\textbf{(B)}\ 16.0\qquad\textbf{(C)}\ 20.0\qquad\textbf{(D)}\ 33.3\qquad\textbf{(E)}\ 55.6$

Solution: Another approach to this problem, very similar to the previous one but perhaps explained more thoroughly, is to use proportions. First, since the thickness and density are the same, we can set up a proportion based on the principle that $d=\frac{m}{V}$ , thus $dV=m$
However, since density and thickness are the same and $A$ is proportional to $s^2$ (recognizing that the area of an equilateral triangle is $\frac{(s)^2\sqrt{3}}{4}$ ), we can say that $m$ is proportional to $s^2$
Then, by increasing s by a factor of $\frac{5}{3}$ $s^2$ is increased by a factor of $\frac{25}{9}$ , thus $m=12*\frac{25}{9}$ or $\boxed{33.3}$

Answer: D 33.3
Problem: Carl decided to fence in his rectangular garden. He bought $20$ fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly $4$ yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl’s garden?
$\textbf{(A)}\ 256\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 384\qquad\textbf{(D)}\ 448\qquad\textbf{(E)}\ 512$

Solution: By Albert471
To start, use algebra to determine the number of posts on each side. You have (the long sides count for $2$ because there are twice as many) $6x = 20 + 4$ (each corner is double counted so you must add $4$ ) Making the shorter end have $4$ , and the longer end have $8$ $((8-1)*4)*((4-1)*4) = 28*12 = 336$ . Therefore, the answer is $\boxed{336}$

Answer: B 336
Problem: A quadrilateral has vertices $P(a,b)$ $Q(b,a)$ $R(-a, -b)$ , and $S(-b, -a)$ , where $a$ and $b$ are integers with $a>b>0$ . The area of $PQRS$ is $16$ . What is $a+b$
$\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12  \qquad\textbf{(E)}\ 13$

Solution: Note that the slope of $PQ$ is $\frac{a-b}{b-a}=-1$ and the slope of $PS$ is $\frac{b+a}{a+b}=1$ .  Hence, $PQ\perp PS$ and we can similarly prove that the other angles are right angles.  This means that $PQRS$ is a rectangle. By distance formula we have $(a-b)^2+(b-a)^2*2*(a+b)^2 = 256$ . Simplifying we get $(a-b)(a+b) = 8$ . Thus $a+b$ and $a-b$ have to be a factor of 8. The only way for them to be factors of $8$ and remain integers is if $a+b = 4$ and $a-b = 2$ . So the answer is $\boxed{4}$

Answer: A 4
Problem: A quadrilateral has vertices $P(a,b)$ $Q(b,a)$ $R(-a, -b)$ , and $S(-b, -a)$ , where $a$ and $b$ are integers with $a>b>0$ . The area of $PQRS$ is $16$ . What is $a+b$
$\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12  \qquad\textbf{(E)}\ 13$

Solution: Solution by e_power_pi_times_i
By the Shoelace Theorem, the area of the quadrilateral is $2a^2 - 2b^2$ , so $a^2 - b^2 = 8$ . Since $a$ and $b$ are integers, $a = 3$ and $b = 1$ , so $a + b = \boxed{4}$

Answer: A 4
Problem: How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$ , the line $y=-0.1$ and the line $x=5.1?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 57$

Solution: Solution by e_power_pi_times_i
Revised by Kinglogic and RJ5303707
[asy] Label l;  l.p=fontsize(8);  xaxis(-1,8,Ticks(l, 1.0));  yaxis(-1,16,Ticks(l, 1.0)); real f(real x)  {  return x * pi; }  D(graph(f,-1/pi,5.1)); D((5.1,-1)--(5.1,16)); D((-1,-0.1)--(8,-0.1)); for(int x = 0; x < 5.1; ++x) { 	for(int y = 0; y < 16; ++y) {     	if(x * pi > y) {         	D((x,y));         }     } } [/asy] (red shows lattice points within the triangle)
If we draw a picture showing the triangle, we see that it would be easier to count the squares vertically and not horizontally. The upper bound is $16$ squares $(y=5.1*\pi)$ , and the limit for the $x$ -value is $5$ squares. First we count the $1*1$ squares. In the rightmost column, there are $12$ squares with length $1$ because $y=4*\pi$ generates squares from $(4,0)$ to $(4,4\pi)$ , and continuing on we have $9$ $6$ , and $3$ for $x$ -values for $1$ $2$ , and $3$ in the equation $y=\pi x$ . So there are $12+9+6+3 = 30$ squares with length $1$ in the figure.
For $2*2$ squares, each square takes up $2$ units left and $2$ units up. Squares can also overlap. For $2*2$ squares, the rightmost column stretches from $(3,0)$ to $(3,3\pi)$ , so there are $8$ squares with length $2$ in a $2$ by $9$ box. Repeating the process, the next column stretches from $(2,0)$ to $(2,2\pi)$ , so there are $5$ squares. Continuing and adding up in the end, there are $8+5+2=15$ squares with length $2$ in the figure.
Squares with length $3$ in the rightmost column start at $(2,0)$ and end at $(2,2\pi)$ , so there are $4$ such squares in the right column. As the left row starts at $(1,0)$ and ends at $(1,\pi)$ there are $4+1=5$ squares with length $3$ . As squares with length $4$ would not fit in the triangle, the answer is $30+15+5$ which is $\boxed{50}$

Answer: D 50
Problem: Alice and Bob live $10$ miles apart. One day Alice looks due north from her house and sees an airplane. At the same time Bob looks due west from his house and sees the same airplane. The angle of elevation of the airplane is $30^\circ$ from Alice's position and $60^\circ$ from Bob's position. Which of the following is closest to the airplane's altitude, in miles?
$\textbf{(A)}\ 3.5 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 4.5 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 5.5$

Solution: We have two 30-60-90 triangles $ABC$ and $DBC$ that are perpendicular and share leg $BC$ (the altitude of the plane $p$ ). $AD=10$ The shared leg is the shortest leg of one triangle and the longest leg of the other. $A$ and $B$ are Bob and Alice respectively.
Find $AC$ and $DC$ in terms of $p$ . Use Pythagorean Theorem on triangle $ADC$ to produce $p=\sqrt{30}\implies\boxed{5.5}$

Answer: E 5.5
Problem: Alice and Bob live $10$ miles apart. One day Alice looks due north from her house and sees an airplane. At the same time Bob looks due west from his house and sees the same airplane. The angle of elevation of the airplane is $30^\circ$ from Alice's position and $60^\circ$ from Bob's position. Which of the following is closest to the airplane's altitude, in miles?
$\textbf{(A)}\ 3.5 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 4.5 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 5.5$

Solution: Non-trig solution by e_power_pi_times_i
Set the distance from Alice's and Bob's position to the point directly below the airplane to be $x$ and $y$ , respectively. From the Pythagorean Theorem, $x^2 + y^2 = 100$ . As both are $30-60-90$ triangles, the altitude of the airplane can be expressed as $\dfrac{x\sqrt{3}}{3}$ or $y\sqrt{3}$ . Solving the equation $\dfrac{x\sqrt{3}}{3} = y\sqrt{3}$ , we get $x = 3y$ . Plugging this into the equation $x^2 + y^2 = 100$ , we get $10y^2 = 100$ , or $y = \sqrt{10}$ $y$ cannot be negative), so the altitude is $\sqrt{3*10} = \sqrt{30}$ , which is closest to $\boxed{5.5}$

Answer: E 5.5
Problem: Alice and Bob live $10$ miles apart. One day Alice looks due north from her house and sees an airplane. At the same time Bob looks due west from his house and sees the same airplane. The angle of elevation of the airplane is $30^\circ$ from Alice's position and $60^\circ$ from Bob's position. Which of the following is closest to the airplane's altitude, in miles?
$\textbf{(A)}\ 3.5 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 4.5 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 5.5$

Solution: Let Alice be at point $A$ , Bob be at point $B$ . Let the plane be at point $P$ and $X$ be the projection of $P$ onto the ground (the plane with contains $A$ and $B$ ). Let the height of the plane, or $PX$ , be $h$ . So, because of the $30-60-90$ triangles, \[AX = \dfrac{h}{\sqrt{3}}, BX = h\sqrt{3}\] By Pythagorean Theorem on $\triangle ABX$ \[\dfrac{h^2}{3} + 3h^2 = 100 \implies \dfrac{10h^2}{3} = 100 \implies h = \sqrt{30},\] which is clossest to $\boxed{5.5}.$

Answer: E 5.5
Problem: The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution: The sum of the geometric sequence is $\frac{a}{1 - r}$ where $a$ is the first term and $r$ is the common ratio. We know the second term, $ar,$ is equal to $1.$ Thus $ar = 1 \Rightarrow a = \frac{1}{r}.$ This means, \[S = \frac{a}{1 - r} = \frac{1/r}{1 - r} = \frac{1}{r(1 - r)}.\] In order to minimize $S,$ we maximize the denominator. By AM-GM, \[\frac{(r) + (1 - r)}{2} \ge \sqrt{r(1-r)} \Rightarrow \frac{1}{4} \ge r(1-r).\] Equality occurs at $r = 1-r \Rightarrow r = \frac{1}{2}.$ This gives the minimum value of $S$ as $\frac{\frac{1}{1/2}}{1 - \frac{1}{2}} = \boxed{4}.$

Answer: E 4
Problem: The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution: A geometric sequence always looks like
\[a,ar,ar^2,ar^3,\dots\]
and they say that the second term $ar=1$ . You should know that the sum of an infinite geometric series (denoted by $S$ here) is $\frac{a}{1-r}$ . We now have a system of equations which allows us to find $S$ in one variable.
\begin{align*} ar&=1 \\ S&=\frac{a}{1-r} \end{align*}
$\textbf{Solving in terms of \textit{a} then graphing}$
\[S=\frac{a^2}{a-1}\]
We seek the smallest positive value of $S$ . We proceed by graphing in the $aS$ plane (if you want to be rigorous, only stop graphing when you know all the rest you didn't graph is just the approaching of asymptotes and infinities) and find the answer is $\boxed{4}.$

Answer: E 4
Problem: The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution: \[\textbf{Completing the Square and Quadratics}\] Let $r$ be the common ratio. If the second term is $1$ , the first must be $\frac{1}{r}$ . By the infinite geometric series formula, the sum must be \[S=\frac{\frac{1}{r}}{1-r}\] This equals $\frac{1}{r(1-r)}$ . To find the minimum value of S, we must find the maximum value of the denominator, $r(1-r)$ , which is $\frac{1}{4}$ , completing the square. Thus, the minimum value of $S$ is $\boxed{4}$

Answer: E 4