Find the square root of
The surface area of a circular cone is given by , where s cm is the slant height, r cm is the radius of the base and is . Find the radius of the base if a cone of surface area 93.5 square cm has a slant height of 5 cm.
Subtract and divide the difference by
Solve the following for the unknown x:
Find the square root of the following:
; also, cube the result.
Refer to question posted on blog of Mar 13 2018, reproduced here for your convenience. Compare your solution with the one given here:
Show that the following expression: remains constant in the interval . Find this constant value.
Let y equal the given expression for x in the prescribed interval. Then, taking cube of both sides, we write
The only real value of , a constant. The roots of the quadratic equation for are complex. It is easy to check that for both and 1.
Alternately, derive the square roots of the expression within the radical; you can use the method of undetermined coefficients for this.
Prove that if the polynomial can be written as the product of two monic polynomials with real non-negative coefficients, then those coefficients are all 0 or 1.
Proof by Andrei Stefanescu:
We call any polynomial symmetrical if and only if for all i. We will use the following lemma:
LEMMA: If P is a polynomial which can be written as , where Q and R are symmetrical polynomials, then P is symmetrical too.
Let and . We have that and . Since Q and R are symmetrical, it follows that and . Hence, for any i, and thus P is symmetrical.
Let and assume that , where g and h are nonconstant polynomials with non-negative coefficients. Let and . Since , all the complex roots of g and h have absolute value 1. Also, if , then . It follows that both g and h are a product of symmetric terms of the form or , hence by lemma, g and h are symmetrical.
If all the coefficients of g and h are in we are done. Otherwise, let k be the least number such that is not a subset of . Since , it follows that , thus . As g is symmetrical, . Computing the coefficient of gives us . Since and all the terms of the sum are non-negative, we obtain , thus one of and (say, )is 0. Computing the coefficient of , we have . As for and , all the terms but must be in . It follows that must be in , which leads to a contradiction with the definition of k.
Several Olympiad problems deal with functions defined on certain sets of points. These problems are interesting in that they combine both geometrical and algebraic ideas.
Let be an integer and a function defined on the set of points in the plane, with the property that for any regular n-gon
Prove that f is the zero function.
In Euclidean geometry, the only motions permissible are rigid motions — translations, rotations, and reflections.
Let A be an arbitrary point. Consider a regular n-gon . Let k be an integer, . A rotation with center A of angle sends the polygon to , where and is the image of for all
From the condition of the statement, we have
Observe that in the sum the number appears n times, therefore,
On the other hand, we have
since the polygons are all regular n-gons. From the two equalities above we deduce , hence, f is the zero function.
Ferrari’s solution of the Biquadratic.
Writing the equation …Equation A
we assume that …Equation B
Expanding and equating coefficients, we have
and and ….Equation C
Eliminating m, n,
which reduces to
The second term can be removed by the substitution …Equation E
and the equation D becomes …Equation F, which is the “reducing cubic”.
Equations C become
Call the above three equations as G.
Thus, if is a root of F and
and , the equation
can be put in the form
and its roots are the roots of the quadratics
It should be noted that the three roots of F correspond to the three ways of expressing u as the product of two quadratic factors.
Homework: Solve the equation .
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