# Die Hard - Algebra Geometry Problem

One Of the toughest Question Of Maths and Geometry and algebra. Can you solve?

Now, to solve this question, you have to solve some parts individually. PART(A): If we have a circle of radius R and its inscribed triangle XYZ that is acute as well as scalene. XY is the longest side.XA,YB, ZC are the altitudes of the triangle XYZ. Let,L be the symmetric point of A w.r.t. BC and M be the symmetric point of B in AC.P is the intersection of XL and YM. H is the orthocenter of triangle XYZ. Then OP.OH = f(R). You need the function f from this question to solve 'THE QUESTION'. Note: Symmetric Point of Q w.r.t. to line EW is the point that is reflection of Q in line EW.

Part(B): {bn} is a real number sequence that depends on {an} in the following way: b=a and for k ≥ 1,b[k+1] = a[k+1]-√c, where c = Σ ( a[i]a[i] ) ; i={1, 2, 3...k}. Let G be the smallest number ≥ 0 such that for all real number sequences {a_n} and all positive integers x: (1/x) Σ ( a[i]a[i] ) ≤ Σ ( pow(G,x-j)b[j]*b[j] ); i,j={1, 2, 3...x} You need G from this question to solve 'THE QUESTION'.

Part(C): Let Q be the least possible value of XY if X, Y, Z are three non-collinear points that are planar, and all of the three points have integral coordinates and all the three lengths XY, YZ, ZX have integral lengths.You need Q from this question to solve 'THE QUESTION'.

'THE QUESTION' : Let A={1,2...m} and let x belong to the set A. Then let P( that depends on m and x) be the number of functions f from A on itself, such that for every a belonging to the set A,there is a k≥0 such that fk(a) ≤ x.Note that fk is the function that is obtained by composition of the function f, k number of times i,e. f_2(x) is the function f(f(x)). Now, you need to print P(m,x)/f(m) where f is the function obtained in Part A. Put x as the value pow(G, Q) and m as the (pow(Q, pow(Q-1, G-1)) + pow(G, 2)), where G is obtained from Part B and Q is obtained from Part C.Note that you need to print the greatest integer less than or equal to the actual exact answer. Note by Parv Parkhiya
6 years, 9 months ago

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