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Ksju [112]
4 years ago
8

Pigeon Hole Principle :

Mathematics
1 answer:
yaroslaw [1]4 years ago
5 0

Answer:

A proof can be as follows:

Step-by-step explanation:

Let S=\{a_{1},a_{2},...,a_{n},a_{n+1}\} be a set of n+1 integers. By the division algorithm the possible remainders when we divide by n are 0,1,2,....,n-1. Then, each integer a_{i}\in S can be written as:

a_{i}=np_{i}+r_{i},\,\,0\leq r_{i}

Observe that the set of remainders \{r_{1},r_{2},...,r_{n+1}\} has n+1 elements and each element has n possible values. By the Pigenhole principle at least two remainders have the same value. Suppose that this two elements are a_{i}, a_{j}. Then,

\begin{array}{c}a_{i}=np_{i}+r\\a_{j}=np_{j}+r\end{array}

Where r_{i}=r_{j}=r. Then,  a_{i}-a_{j}=np_{i}-np_{i}=n(p_{i}-p_{j}). Then we have that n divides a_{i}-a_{j}.

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First of all, let <span>θθ</span> be some angle in <span><span>(0,π)</span><span>(0,π)</span></span>. Then

<span><span><span>θθ</span> is acute <span>⟺⟺</span> <span><span>θ<<span>π2</span></span><span>θ<<span>π2</span></span></span> <span>⟺⟺</span> <span><span>cosθ>0</span><span>cos⁡θ>0</span></span>.</span><span><span>θθ</span> is right <span>⟺⟺</span> <span><span>θ=<span>π2</span></span><span>θ=<span>π2</span></span></span> <span>⟺⟺</span> <span><span>cosθ=0</span><span>cos⁡θ=0</span></span>.</span><span><span>θθ</span> is obtuse <span>⟺⟺</span> <span><span>θ><span>π2</span></span><span>θ><span>π2</span></span></span> <span>⟺⟺</span> <span><span>cosθ<0</span><span>cos⁡θ<0</span></span>.</span></span>

Now, to see if (say) angle <span>AA</span> of the triangle <span><span>ABC</span><span>ABC</span></span> is acute/right/obtuse, we need to check whether <span><span>cos∠BAC</span><span>cos⁡∠BAC</span></span> is positive/zero/negative. But what is <span><span>cos∠BAC</span><span>cos⁡∠BAC</span></span>? It is the angle made by the vectors <span><span><span>AB</span><span>−→−</span></span><span><span>AB</span>→</span></span> and <span><span><span>AC</span><span>−→−</span></span><span><span>AC</span>→</span></span>. (When you are computing the angle at a particular vertex <span>vv</span>, you should make sure that both the vectors corresponding to the two adjacent sides have that vertex <span>vv</span> as the initial point.) We will first compute these two vectors:

<span><span><span><span>AB</span><span>−→−</span></span>=(0,0,0)−(1,2,0)=(−1,−2,0)</span><span><span><span>AB</span>→</span>=(0,0,0)−(1,2,0)=(−1,−2,0)</span></span><span><span><span><span>AC</span><span>−→−</span></span>=(−2,1,0)−(1,2,0)=(−3,−1,0)</span><span><span><span>AC</span>→</span>=(−2,1,0)−(1,2,0)=(−3,−1,0)</span></span>Therefore, the angle between these vectors is given by:<span><span><span>cos∠BAC=<span><span><span><span>AB</span><span>−→−</span></span>⋅<span><span>AC</span><span>−→−</span></span></span><span>|<span><span>AB</span><span>−→−</span></span>||<span><span>AC</span><span>−→−</span></span>|</span></span>=…</span>(1)</span><span>(1)<span>cos⁡∠BAC=<span><span><span><span>AB</span>→</span>⋅<span><span>AC</span>→</span></span><span>|<span><span>AB</span>→</span>||<span><span>AC</span>→</span>|</span></span>=…</span></span></span>Can you take it from here? From the sign of this value, you should be able to decide if angle <span>AA</span> is acute/right/obtuse.

Now, do the same procedure for the remaining two angles <span>BB</span> and <span>CC</span> as well. That should help you solve the problem.

A shortcut. Since you are not interested in the actual values of the angles, but you need only whether they are acute, obtuse or right, it is enough to compute only the sign of the numerator (the dot product between the vectors) in formula (1). The denominator is always positive.

6 0
4 years ago
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