exam: HMMT

problem:
Farmer Tim is lost in the densely-forested Cartesian plane. Starting from the origin he walks a sinusoidal path in search of home; that is, after $t$ minutes he is at position $(t, \sin t)$.

Five minutes after he sets out, Alex enters the forest at the origin and sets out in search of Tim. He walks in such a way that after he has been in the forest for $m$ minutes, his position is $(m, \cos t)$.

What is the greatest distance between Alex and Farmer Tim while they are walking in these paths?

solution:
At arbitrary time $t$, Farmer Tim is at position $(t, \sin t)$ and Alex is at position $(t-5, \cos t)$. Hence at time $t$, the distance, $d$, between Tim and Alex is $d=\sqrt{(\sin t-\cos t)^{2}+25}$. To find the maximum value of $d$, we solve for $t$ such that $\frac{d d}{d t}=0$.
$\frac{d d}{d t}=\frac{(\sin t-\cos t)(\cos t+\sin t)}{\sqrt{(\sin t-\cos t)^{2}+25}}$. Then $\frac{d d}{d t}=0 \Rightarrow \sin ^{2} t-\cos ^{2} t=0 \Rightarrow \sin ^{2} t=\cos ^{2} t$. Equality happens if $t$ is any constant multiple of $\frac{\pi}{4}$.

Notice that to maximize $d$, we need to maximize $(\sin t-\cos t)^{2}$. This is achieved when $\cos t=-\sin t$. Because we determined earlier that $t$ is a constant multiple of $\frac{\pi}{4}$, then with this new condition, we see that $t$ must be a constant multiple of $\frac{3 \pi}{4}$.

Then $(\sin t-\cos t)^{2}=2 \Rightarrow d=\sqrt{29}$.

problem_match: 1. Problem:

solution_match: Solution:


problem: A cube with sides 1 m in length is filled with water, and has a tiny hole through which the water drains into a cylinder of radius 1 m . If the water level in the cube is falling at a rate of $1 \mathrm{~cm} / \mathrm{s}$, at what rate is the water level in the cylinder rising?

solution:
The magnitude of the change in volume per unit time of the two solids is the same. The change in volume per unit time of the cube is $1 \mathrm{~cm} \cdot \mathrm{~m}^{2} / \mathrm{s}$. The change in volume per unit time of the cylinder is $\pi \cdot \frac{d h}{d t} \cdot m^{2}$, where $\frac{d h}{d t}$ is the rate at which the water level in the cylinder is rising.

Solving the equation $\pi \cdot \frac{d h}{d t} \cdot m^{2}=1 \mathrm{~cm} \cdot m^{2} / \mathrm{s}$ yields $\frac{1}{\pi} \mathrm{~cm} / \mathrm{s}$.

problem_match: 2. Problem:


problem: Find the area of the region bounded by the graphs of $y=x^{2}, y=x$, and $x=2$.

solution:
There are two regions to consider. First, there is the region bounded by $y=x^{2}$ and $y=x$, in the interval $[0,1]$. In this interval, the values of $y=x$ are greater than the values of $y=x^{2}$, thus the area is calculated by $\int_{0}^{1}\left(x-x^{2}\right) d x$.

Second, there is the region bounded by $y=x^{2}$ and $y=x$ and $x=2$, in the interval [1,2]. In this interval, the values of $y=x^{2}$ are greater than the values of $y=x$, thus the area is calculated by $\int_{1}^{2}\left(x^{2}-x\right) d x$. Then the total area of the region bounded by the three graphs is $\int_{0}^{1}\left(x-x^{2}\right) d x+\int_{1}^{2}\left(x^{2}-x\right) d x=1$.

problem_match: 3. Problem:


solution: Observe that $f(x)$ is merely an infinite geometric series. Thus $f(x)=\frac{1}{1-\frac{x}{2}}=\frac{2}{2-x}$. Then $\int_{0}^{1} \frac{2}{2-x}=2 \ln 2$. Then $\sqrt{e^{2 \ln 2}}=\sqrt{2^{2}}=2$.

problem_match: 4. Problem:


problem: Edward, the author of this test, had to escape from prison to work in the grading room today. He stopped to rest at a place 1,875 feet from the prison and was spotted by a guard with a crossbow. The guard fired an arrow with an initial velocity of $100 \mathrm{ft} / \mathrm{s}$. At the same time, Edward started running away with an acceleration of $1 \mathrm{ft} / \mathrm{s}^{2}$. Assuming that air resistance causes the arrow to decelerate at $1 \mathrm{ft} / \mathrm{s}^{2}$ and that it does hit Edward, how fast was the arrow moving at the moment of impact (in $\mathrm{ft} / \mathrm{s}$ )?

solution:
We use the formula for distance, $d=\frac{1}{2} a t^{2}+v t+d_{0}$. Then after $t$ seconds, Edward is at location $1875+\frac{1}{2}(1)\left(t^{2}\right)$ from the prison. After $t$ seconds, the arrow is at location $\frac{1}{2}(-1)\left(t^{2}\right)+100 t$ from the prison. When the arrow hits Edward, both objects are at the same distance away from the tower. Hence $1875+\frac{1}{2}(1)\left(t^{2}\right)=\frac{1}{2}(-1)\left(t^{2}\right)+100 t$. Solving for $t$ yields $t^{2}-100 t+1875=0 \Rightarrow t=25$ or $t=75$. Then it must be $t=25$, because after the arrow hits Edward, he will stop running.

After 25 seconds, the arrow is moving at a velocity of $100-25(1)=75 \mathrm{ft} / \mathrm{s}$.

problem_match: 6. Problem:


problem: A parabola is inscribed in equilateral triangle $A B C$ of side length 1 in the sense that $A C$ and $B C$ are tangent to the parabola at $A$ and $B$, respectively. Find the area between $A B$ and the parabola.

solution:
Suppose $A=(0,0), B=(1,0)$, and $C=\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$. Then the parabola in question goes through $(0,0)$ and $(1,0)$ and has tangents with slopes of $\sqrt{3}$ and $-\sqrt{3}$, respectively, at these points. Suppose the parabola has equation $y=a x^{2}+b x+c$. Then $\frac{d y}{d x}=2 a x+b$.

At point $(0,0), \frac{d y}{d x}=b$. Also the slope at $(0,0)$, as we determined earlier, is $\sqrt{3}$. Hence $b=\sqrt{3}$. Similarly, at point $(1,0), \frac{d y}{d x}=2 a+b$. The slope at $(1,0)$, as we determined earlier, is $-\sqrt{3}$. Then $a=-\sqrt{3}$.

Since the parabola goes through $(0,0), c=0$. Hence the equation of the parabola is $y=-\sqrt{3} x^{2}+\sqrt{3} x$. The desired area is simply the area under the parabolic curve in the interval $[0,1]$.

Hence $\int_{0}^{1}\left(-\sqrt{3} x^{2}+\sqrt{3} x\right) d x=\frac{\sqrt{3}}{6}$.

problem_match: 7. Problem:


problem: Find the slopes of all lines passing through the origin and tangent to the curve $y^{2}=x^{3}+39 x-35$.

solution:
Any line passing throug the origin has equation $y=m x$, where $m$ is the slope of the line. If a line is tangent to the given curve, then at the point of tangency, $(x, y), \frac{d y}{d x}=m$.

First, we calculate $\frac{d y}{d x}$ of the curve: $2 y d y=3 x^{2} d x+39 d x \Rightarrow \frac{d y}{d x}=\frac{3 x^{2}+39}{2 y}$. Substituting $m x$ for $y$, we get the following system of equations:

$$
\begin{aligned}
m^{2} x^{2} & =x^{3}+39 x-35 \\
m & =\frac{3 x^{2}+39}{2 m x}
\end{aligned}
$$

Solving for $x$ yields the equation $x^{3}-39 x+70=0 \Rightarrow(x-2)(x+7)(x-5)=0 \Rightarrow x=2$ or $x=-7$ or $x=5$. These solutions indicate the $x$-coordinate of the points at which the desired lines are tangent to the curve. Solving for the slopes of these lines, we get $m= \pm \frac{\sqrt{51}}{2}$ for $x=2$, no real solutions for $x=-7$, and $m= \pm \frac{\sqrt{285}}{5}$ for $x=5$. Thus $m= \pm \frac{\sqrt{51}}{2}, \pm \frac{\sqrt{285}}{5}$.

problem_match: 8. Problem:


problem: Evaluate $\sum_{n=1}^{\infty} \frac{1}{n \cdot 2^{n-1}}$.

solution:
Note that if we take the integral of $f(x)$ in problem 4, we get the function $F(x)=x+\frac{x^{2}}{2 \cdot 2}+$ $\frac{x^{3}}{3 \cdot 2^{2}}+\ldots$. Evaluating this integral in the interval $[0,1]$, we get $1+\frac{1}{2 \cdot 2}+\frac{1}{3 \cdot 2^{2}}+\ldots$, which is the desired sum.

Hence $\int_{0}^{1} \frac{2}{2-x} d x=2 \ln 2$.

problem_match: 9. Problem:


problem: Let $S$ be the locus of all points $(x, y)$ in the first quadrant such that $\frac{x}{t}+\frac{y}{1-t}=1$ for some $t$ with $0<t<1$. Find the area of $S$.

solution:
Solving for $t$ in the given equation, we get $t^{2}+(y-x-1) t+x=0$. Using the quadratic equation, we get $t=\frac{(x+1-y) \pm \sqrt{(y-x-1)^{2}-4 x}}{2}$. For all valid combinations of $(x, y), t$ is positive and less than 1 (this is easy to see by inspection). All valid combinations of $(x, y)$ are those that make $(y-x-1)^{2}-4 x \geq 0$.

Solving for $y$ in the equation $(y-x-1)^{2}-4 x=0$ yields $y^{2}-(2 x+2) y+(x-1)^{2} \geq 0 \Rightarrow y=(x+1) \pm 2 \sqrt{x}$. In the original equation, it is given that $\frac{x}{t}+\frac{y}{1-t}=1$, and $0<t<1$. This implies that $x, y<1$. Then the only possible $y<1$ that satisfies $(y-x-1)^{2}-4 x=0$ is $y=x+1-2 \sqrt{x}$.

Then to satisfy the inequality $(y-x-1)^{2}-4 x \geq 0$, we must have $y \leq x+1-2 \sqrt{x}$. Recall that this is when $0<y<1$. Hence we integrate in the interval $[0,1]: \in_{0}^{1} x+1-2 \sqrt{x}=\frac{1}{6}$.

problem_match: 10. Problem: