# A geometrical puzzle involving calculus

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?

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,

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

• Thank you. My method is actually very similar; my mistake was that I forgot to divide the integral ∫5/y dy by the length of half the side Jun 7 at 13:26
• Note, however, that the hyperbola for A=50 will not meet the square at (10,5), and (5,10). The hyperbola that meets these points is associated with A=100-5*5=75. Additionally, the hyperbolas do not have to be equilateral. Jun 7 at 13:34
• @GiulioLanza But $A = 50 = 100 - 5 \cdot 10 = 100 - 10 \cdot 5$? Hence, it would meet the points $(5, 10)$ and $(10, 5)$. Jun 7 at 13:37
• The vectors you should consider are (10,y) and (x,10), therefore (10,5) and (5,10) imply that 5=y and x=5. Jun 7 at 13:44
• @GiulioLanza You are confusing at vectors of different coordinates with points on the hyperbola. The hyperbola has nothing to do with vectors on the same cartesian axis. Ie. $$(x, y) = (5, 10) \\ \neq \\ \{(x, 10), \, (10, y)\} = \{(5, 10), \, (10, 5)\}$$ Consider this; $$A = || a \times b || = ||[5, 10] \times [10, 10]|| = 50$$ and $$A = || a \times b || = ||[5, 10] \times [10, 5]|| = 75$$ $(x, y)$ can be matched up at any value as long as $0 \le x \le 10$ and $0 \le y \le 10$. Jun 7 at 13:53

$$A = 100 - xy$$

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

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

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

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

• Thank you very much, you made that look easy peasy! :) Jun 7 at 13:50