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This is a problem from "Scuola Normale Superiore" admission test 2017. Can anybody help me to solve it?

Let $ \alpha, \beta, \gamma $ and $ \delta $ $ \in \mathbb{R}$. We call $ S $ the set of points $ (x, y, z) $ of the Euclidean space such that $ z \leq min(\alpha x + \beta y, \gamma x + \delta y) $. Suppose then that for some $ r, s \in \mathbb{R} $ happens that for every $ (x, y, z) \in S $ we have $ z \leq rx + sy $. Prove that in this case there is a number $ t \in \mathbb{R} $ with $ 0 \leq t \leq 1 $ such that $ r = t \alpha + (1-t) \gamma $ and $ s = t \beta + (1-t) \delta $.

Note: in the solution we can assume that $ (\alpha, \beta) \neq (0,0) $, $ (\gamma, \delta) \neq (0,0) $ and that the lines $ \alpha x + \beta y = 0 $ and $ \gamma x + \delta y = 0 $ are distinct

Jeff18
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2 Answers2

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Hint: you are looking at the intersection of two half-spaces, bounded by two planes $\pi: z=\alpha x + \beta y$ and $\sigma: z=\gamma x + \delta y$ both including the origin. If the two planes are distinct, their intersection must be a line, and you can imagine obtaining $\pi$ by slowly rotating $\sigma$ around that line and viceversa. The planes that you obtain as you go through that rotation (in the form $z=t\cdot(\alpha x + \beta y)+(1-t)\cdot(\gamma x + \delta y)$ for $0\leq t\leq 1$) are all the planes that bound half-spaces containing the intersection of your original half-spaces - why? Begin by answering why the latter planes must all contain the entire line $\pi\cap\sigma$.

Note that the hypothesis that $\pi\neq\sigma$ is not necessary (why?) it just leaves out an unnecessary complication.

Anonymous
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Second hint: if you can't "see" the solution from the hint above and/or find the problem too difficult, try a warm-up with the problem in 2D instead of 3D, as follows.

2D version of the problem: Consider, in the $y,z$ plane, the set of points $S=\{(y,z):z\leq \min(\beta y, \delta y)\}$. Suppose then that for some $s\in\mathbb{R}$ it happens that, for every $(y,z)\in S$, we have $z\leq sy$. Prove that in this case there is a number $t\in\mathbb{R}$ with $0\leq t\leq 1$ such that $s=t\beta + (1-t)\delta$.

N.B. do not just prove it. See it. In particular, draw the two lines $z(y)=\beta y$ and $z(y)=\delta y$, see that $S$ is the region of the plane "below" both lines, and see where the generic line $z_t(y)=\left(t\beta + (1-t)\delta\right)y$ lies. Then see how any line in the form $z(y)=sy$ with no portion of $S$ above it, must have $s=\left(t\beta + (1-t)\delta\right)y$, for $0\leq t\leq 1$.

From 2D to 3D: As in the first hint above, consider the two planes $\pi: z=\alpha x + \beta y$ and $\sigma: z=\gamma x + \delta y$, both including the origin. In this case, $S$ is the portion of the Euclidean space the falls "below" both planes. If the two planes are distinct, their intersection, which obviously contains the origin, must be a line. If you rotate the axes so that the new $x$ axis coincides with this line, and look at the $y,z$ plane, you are back to the the 2D version of the problem (and the proof of the 2D version yields a proof of the 3D version).

Bonus points! How can you extend the problem to $4$ dimensions? How can you extend it to $k$ dimensions for any $k\in\mathbb{N}$?

Anonymous
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