A geometrical puzzle involving calculus

Question

Some time ago I stumbled across a problem from the Putnam Mathematical Competition. I could not find it, but I remember the text quite well.

There are two vectors: a=(10, $y$) and b=($x$,10), where $0 ≤ x ≤ 10$, and $0 ≤ y ≤ 10$. We have to compute the probability of these two vectors forming a parallelogram with an area A ≥ 50. The hint in the text is that the probability can be expressed as $$\ln(\sqrt a) + \frac{b}{c}$$ (where $a, b, c \in \mathbb{N}$; $b$ and $c$ are coprime and $a$ is as smaller as possible).

Here is a visual representation for the problem :

First, I calculated the area of the parallelogram: $A = ||\vec{a} \times \vec{b}|| = xy - 100$, which implies that $0 ≤ A ≤ 100$. We can write the inequality $A ≥ 50$ as $xy ≤ 50$.

If we assume that $y$ is a given (we could repeat this reasoning for $x$), then $x ≤ \frac{50}{y}$. If $y ≤ 5$ then all $x$ values $≤ 10$ are acceptable; therefore the probability of $A ≥ 50$ is at least $\frac{1}{2}$. When $y ≥ 5, x ≤ 50$; so when $y$ increases the portion of acceptable $x$ values decreases. I calculated $$\int_{5}^{10} \frac{50}{y} \cdot \frac{1}{10} \ dy$$ to count these values. The idea is that for every $y$ value greater than 5 the probability of $x$ values being less than $\frac{50}{y}$ is $\frac{\frac{50}{y}}{10}$, where 10 is the segment of all possible $x$ values. The integration yields $5 \ln (2)$. Using the law of total probability I wrote:

$$P(A \geq 50) = P(y < 5) \cdot P(A \geq 50 | y < 5) + P(y \geq 5) \cdot P(A \geq 50 | y > 5)$$

$$P(A \geq 50) = \frac{1}{2} \cdot 1 + \frac{1}{2} \cdot (5 \ln(2)) \approx 2.23.$$

But why is the probability greater than 1?

Answer 1

I am not able to understand your method, but here is an alternative. Additionally, note that the area is given by $A = 100 - xy$, since $A > 0$.

$$A = 100 - xy \implies y = \dfrac{100 - A}{x}$$

Observe, $A$ is just a constant now that affects the dilation of the $y$; the hyperbola.

As a matter of fact, we could describe the above families of hyperbola created by $y = \dfrac{100 - A}{x}$ as our families of "the area of our paralellogram". These are hyperbolas are bounded within a $10 \times 10$ box, corresponding to the domain $0 \le x \le 10$ and $0 \le y \le 10$. The explanation here may be vague, so I have drawn a picture below:

Since, $100 \ge A \ge 50$ can take on infinite real values, the infinite sum of hyperbolas conditioned by $100 \ge A \ge 50$ would represent the desired area. This is equivalent to the orange coloured section in the picture above.

$y = \dfrac{100 - (A = 50)}{x} = \dfrac{50}{x}$, is bounded by the box at $(x, y) = (5, 10) \lor (10, 5)$.

Hence,

$$\text{Pr}(A > 50) = \dfrac{\sum{\{y \, := A > 50\}}}{\sum{\{y \, := A > 0\}} = \text{all possible A}} \\ \quad \\$$ $$= \dfrac{\overbrace{5 \cdot 10 + \large{\int^{10}_{5}{\frac{50}{x} dx}}}^{\large{\text{Our "infinite sum of hyperbolas conditioned by $100 \ge A \ge 50$"}}}}{10 \cdot 10} \\ \quad \\= \dfrac{\ln{2} + 1}{2} \\= \ln{\sqrt{2}} + \dfrac{1}{2} \\\approx 0.847$$

Answer 2

$A = 100 - xy$

If $A>50$ then $xy < 50$

$\frac {\int_0^{10}\int_0^{\min(10,\frac {50}{x})} \ dy\ dx}{\int_0^{10}\int_0^{10} \ dy\ dx}$

$\frac {1}{100} (\int_0^5\int_0^{10} \ dy\ dx + \int_5^{10}\int_0^{\frac {50}x} \ dy \ dx)$

$\frac {1}{2} (1 + \ln 2)$