Todai Entrance Exam: Math 2019 – Problem 3

(1)

    \[ \mathbb{P}(Q \in AB) = \mathbb{P} \left (\Theta \in \left [0, \frac{\pi}{2}\right ] \right ) = \frac{\frac{\pi}{2}}{2\pi} = \frac{1}{4} \]

(2)

The easy way:

The projection of AB on Ox is OA. So \mathbb{E}(X | Q \in AB) = \frac{1}{2}|OA| = \frac{1}{2}.

The hard way:

Let M = (M_{x}, M{y}) be the intersection between AB and y = x. As the triangle ABC is symmetric with respect to the line y = x, every point (X, Y) on MA has a corresponding point (2M_{X} - X, 2M_{Y} - Y) on MB. So:

    \[ \mathbb{E}(X | Q \in AB) = \frac{1}{2}\left ( \sum_{Q \in MA} X + \sum_{Q \in MB} X \right ) = \frac{1}{2}\left ( \sum_{Q \in MA} X + \sum_{Q \in MA} (2M_{X} - X) \right ) = M_{X} = \frac{1}{2} \]

Note that M = \left ( \frac{1}{2}, \frac{1}{2} \right ) as d_{AB} : y = 1 - x.

(3)

We need to find g and h. First, we find g:

    \[ g(\Theta) = \mathbb{P}(\Theta | Q \in BC) = \frac{1}{\hat{BOC}} = \frac{1}{\cos^{-1}\left ( \frac{\vec{OB} \cdot \vec{OC}}{||\vec{OB}||||\vec{OC}||} \right )} = \frac{1}{\cos^{-1} \frac{-1}{\sqrt{2}}} = \frac{1}{135} \]

Now, we find h. We can represent Q coordination in terms of \theta.

    \[ Q(X, Y) = Q(\sqrt{X^{2} + Y^{2}}\cos\theta, \sqrt{X^{2} + Y^{2}}\sin\theta) \]

Note that:

    \[ Q \in BC, d_{BC} : y = 2x + 1 \Rightarrow Q(X, Y) = Q(X, 2X + 1) \]

So:

    \[ Q(X, Y) = Q(\sqrt{5X^{2} + 4X + 1}\cos\theta, \sqrt{5X^{2} + 4X + 1}\sin\theta) \]

Therefore:

    \[ X = \sqrt{5X^{2} + 4X + 1}\cos\theta \Leftrightarrow \theta = h(X) = \cos^{-1} \frac{X}{\sqrt{5X^{2} + 4X + 1}} \]

Now compute f(x):

    \[ \frac{\mathrm{d} h}{\mathrm{d} x}(x) = -\frac{1}{\sqrt{1 - \frac{x^{2}}{5x^{2} + 4x + 1}}}\frac{\sqrt{5x^{2} + 4x + 1} - x\frac{10x + 4}{2\sqrt{5x^{2} + 4x + 1}}}{5x^{2} + 4x + 1} = -\frac{2x + 1}{(5x^{2} + 4x + 1)|2x + 1|} \]

    \[ f(x) = g(h(x))\left | \frac{\mathrm{d} h}{\mathrm{d} x}(x) \right | = \frac{1}{135} \frac{1}{5x^{2} + 4x + 1} \]

Note that \forall x \in \mathbb{R}, 5x^{2} + 4x + 1 > 0.

(4)

    \[ \alpha = \int_{C}^{B}xf(x)dx = \int_{-1}^{0}xf(x)dx = \frac{1}{135}  \int_{-1}^{0}\frac{x}{5x^{2} + 4x + 1}dx \]

    \[ \alpha = \frac{1}{1350}\int_{-1}^{0} \frac{10x - 4}{5x^{2} + 4x + 1}dx - \frac{2}{135}\int_{-1}^{0} \frac{1}{(5x + 2)^{2} + 1}dx \]

Let u = 5x^{2} + 4x + 1 \Rightarrow du = (10x - 4)dx:

    \[ \int_{-1}^{0} \frac{10x - 4}{5x^{2} + 4x + 1}dx = \int_{2}^{1} \frac{1}{u}du = \ln{u} \big |^{1}_{2} = -\ln 2 \]

Let v = 5x + 2 \Rightarrow dv = 5dx, and thanks to math.com:

    \[ \int_{-1}^{0} \frac{1}{(5x + 2)^{2} + 1}dx = \frac{1}{5}\int_{9}^{2} \frac{1}{v^{2} + 1}dv = \frac{1}{5}\arctan{v} \big |^{2}_{-3} = \frac{1}{5}(\arctan(2) + \arctan(3)) \]

    \[ \alpha = -\frac{\ln{2}}{1350} - \frac{2}{675}\arctan(2) - \frac{2}{675}\arctan(3) \]

(5) The same procedure can be applied to AC. Let \beta be the expectation of X given Q is located on AC, then:

    \[ \beta = \frac{1}{675}\ln{\frac{2}{3}} + \frac{1}{270}(\arctan(2) - \arctan(3)) \]

    \[ \mu = \frac{1}{|AB| + |BC| + |AC|}(|AB|\mathbb{E}(X | Q \in AB) + |BC|\alpha + |AC|\beta) \]

    \[ \mu = \frac{1}{\sqrt{2} + 2\sqrt{5}}\left ( \frac{\sqrt{2}}{2} - \frac{\sqrt{5}\ln{2}}{1350} - \frac{2\sqrt{5}}{675}\arctan(2) - \frac{2\sqrt{5}}{675}\arctan(3) + \frac{\sqrt{5}}{675}\ln{\frac{2}{3}} + \frac{\sqrt{5}}{270}\arctan(2) - \frac{\sqrt{5}}{270}\arctan(3) \right ) \]


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