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2021/09/15  阅读：43  主题：默认主题

# Mathematics Interview Questions (XV)

### 76. What are the conditions for which a cubic equation (with real coefficients) has two, one, or no solutions?

Solution. It cannot have two or one roots since (i) it must cross the -axis at some point due to the intermediate value theorem; and (ii) if it has a complex solution , then its conjugate must appear as well. Consider a general cubic curve with roots and . Then its discriminant . If only one root, say , is real, then and are complex conjugates, which implies that is a purely imaginary number, and thus that is real and negative. On the other hand, and are also complex conjugates, and the product is real and positive. Thus, the discriminant is the product of a single negative number and several positive ones. That gives . Now expand the discriminant using Viète's formulae for cubic equations, giving as the final condition.

### 77. What is the area between two circles, with radii one, that go through each other's centres?

Solution. Consider two circles such as and which satisfy the conditions in the question. They intersect at the point where , i.e., , at which . The area is twice the sector formed by the circle with endpoints the two points of intersection minus the area of the central rhombus. (The reader is advised, at this point, to draw a diagram.) The angle formed by the origin and the two pints of intersection is , so the area of the sector is . The area of the rhombus, on the other hand, is given by . The total area is therefore .

### 78. If every term in a sequence is defined by the sum of every term before it, give a general formula for the th term.

Solution. from the question, so and . is therefore a geometric series with initial term and common ratio , giving .

### 79. Is ?

Solution. To answer this question, we need to define first what a decimal representation is.

Definition. A decimal expression of a real number is the infinite series , written in the form .

Using this definition, . (Actually, we can prove that (i) there exists a decimal expression for each real number; and (ii) if a real number has two different decimal expressions, then one ends in and the other in . The proofs are left to the reader.)

### 80. Why are there no Pythagorean triples in which both and are odd?

Solution. [Almost identical to question 75] The term Pythagorean triples refer to the triples for which and . Suppose and are both odd, then (taking as an example) we can write them in the form , i.e., , and . As a result, . However, can only be , leading to a contradiction. Hence, no such Pythagorean triples exist.

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2021/09/15  阅读：43  主题：默认主题