**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 *x* ∈ **R**.

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**”.

**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 2*d*, whereas the RHS is of degree *d*^{2}. Then 2*d* = *d*^{2}, 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*) = *ax*^{2} + *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*(*ax*^{2} + *bx* + *c*)^{2} + *b*(*ax*^{2} + *bx* + *c*) + *c*

⇒ {*a*(*x*^{2} – 2*x* + 1) + *b*(*x* − 1) + *c*}{*a*(*x*^{2} + 2*x* + 1) + *b*(*x* + 1) + *c*} = *a*(*a*^{2}*x*^{4} + *b*^{2}*x*^{2} + *c*^{2} + 2*abx*^{3} + 2*bcx* + 2*cax*^{2}) + *b*(*ax*^{2} + *bx* + *c*) + *c*

⇒ {*ax*^{2} + (*b* – 2*a*)*x* + (*a* – *b* + *c*)} {*ax*^{2} + (*b* + 2*a*)*x* + (*a* + *b* + *c*)} = *a*{*a*^{2}*x*^{4} + 2*abx*^{3 }+ (*b*^{2} + 2*ca*)* x*^{2} + 2*bcx *+* c*^{2}} + *bax*^{2} + *b*^{2}*x* + *bc *+ *c*

⇒ *a*^{2}*x*^{4} + {(*b* + 2*a*)*a *+ (*b* – 2*a*)*a*}*x*^{3} + {*a*(*a* + *b* + *c*) + *a*(*a* – *b* + *c*) + (*b*^{2} – 4*a*^{2})}*x*^{2} + {(*b* − 2*a*) (*a* + *b* + *c*) + (*b* +2*a*) (*a* − *b* + *c*)}*x* + (*a* − *b* + *c*) (*a* + *b* + *c*) = *a*^{3}*x*^{4} + 2*a*^{2}*bx*^{3 }+ {*a*(*b*^{2} + 2*ca*) + *ab*}*x*^{2} + {2*abc* + *b*^{2}}*x *+(*ac*^{2 }+ *bc *+ *c*)

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

*a*^{2}=*a*^{3 }or,*a*(1 −*a*) = 0 or,since*a*= 1*a*= 0 is not possible;- (1 +
*b*+*c*) + (1 –*b*+*c*) + (*b*^{2}– 4) = (*b*^{2}+ 2*c*) +*b*or,;*b*= – 2 - (
*a*−*b*+*c*) (*a*+*b*+*c*) =*ac*^{2 }+*bc*+*c*or,.*c*= 1

Hence *f*(*x*) = *x*^{2} − 2*x* + 1 = (*x* − 1)^{2}.

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

It’s nearly impossible to find knowledgeable people for thios

subject, but you sound like you know what you’re talking

about! Thanks

php patterns

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