Week 13

Example 2. Triangle ABC is a right triangle with ∠ACB as its right angle, ∠ABC = 60° and AB = 10. Point P is randomly chosen inside ABC, and BP is extended to meet AC at D. What is the probability that BD > 5√2?

Example 3. Find all real polynomials f(x) such that f(x − 1)f(x + 1) = f(f(x)) for all xR.

Of course, you can find the solution just below, but it is highly recommended that, you first try to solve it on your own.

Just remember the words of Paul Halmos, who says “the only way to learn mathematics is to do mathematics”.

Photo by Simon Berger on Pexels.com

Solution 1. From the given relation, we find that

Since (b + c) and (d + a) are ≠ 0 (otherwise, the given relation becomes undefined), only possibilities are a = c or a + b + c + d = 0.

Solution 2. Since AB = 10 and ∠ABC = 60°, we have

It is common in inequality problems to first consider the boundary case. Suppose that E is the point on AC with BE = 5√2. Then

and ∆ECB is an isosceles right triangle.

So BD < 5√2 when D is between E and C, that is, when P is inside ∆BEC. The probability that this will occur is

Hence the probability that BD > 5√2 is

Solution 3. Let d = deg f. We know that (fg)(x)= f(x)g(x) and (f g) = f(g(x)). Hence deg (fg) = deg f + deg g and deg (f g) = deg f × deg g.

Then LHS of the equation is of degree 2d, whereas the RHS is of degree d2. Then 2d = d2, implying that d = 0 or d = 2.

A polynomial of zero degree is constant, and the only constant f(x) = c satisfying this equation is c = 0 or c = 1. Let us choose a degree 2 polynomial as f(x) = ax2 + bx + c, where a ≠ 0. Then

f(x − 1) f(x + 1) = f(f(x)) = a{f(x)}2 + bf(x) + c

⇒ {a(x − 1)2 + b(x − 1) + c}{a(x + 1)2 + b(x + 1) + c}  = a(ax2 + bx + c)2 + b(ax2 + bx + c) + c

⇒ {a(x2 – 2x + 1) + b(x − 1) + c}{a(x2 + 2x + 1) + b(x + 1) + c}  = a(a2x4 + b2x2 + c2 + 2abx3 + 2bcx + 2cax2) + b(ax2 + bx + c) + c

⇒ {ax2 + (b – 2a)x + (ab + c)} {ax2 + (b + 2a)x + (a + b + c)} = a{a2x4 + 2abx3 + (b2 + 2ca) x2 + 2bcx  + c2} + bax2 + b2xbc + c

a2x4 + {(b + 2a)a + (b – 2a)a}x3 + {a(a + b + c) + a(ab + c) + (b2 – 4a2)}x2 + {(b − 2a) (a + b + c) + (b +2a) (ab + c)}x + (ab + c) (a + b + c) = a3x4 + 2a2bx3 + {a(b2 + 2ca) + ab}x2 + {2abc + b2}x  +(ac2 + bc + c)

Equating the coefficients of like powers of  on both sides, we get

  • a2 = a3   or, a(1 − a) = 0   or, a = 1 since a = 0 is not possible;
  • (1 + b + c) + (1 – b + c) + (b2 – 4) = (b2 + 2c) + b   or, b = – 2;
  • (ab + c) (a + b + c) = ac2 + bc + c      or, c = 1.

Hence f(x) = x2 − 2x + 1 = (x − 1)2.

It is easy to verify that, this is indeed the solution.

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