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Tuesday, 22 July 2014

ISI 2014 SOLUTIONS
  • Let PQR be a triangle. Take a point A on or inside the triangle. 
  • Let f(x, y) = ax + by + c. Show that \mathbf { f(A) \le max \{ f(P), f(Q) , f(R)\} }

Discussion:
Basic idea is this: First we take A on a side, say PQ. We show \mathbf { f(A) \le max \{ f(P), f(Q) \} \implies f(A) \le max \{ f(P), f(Q) , f(R)\} }
Next we take A in the interior. Lets join RA and produce it to meet PQ at T. Using the previous argument we show \mathbf { f(A) \le max \{ f(T), f(R) \} \text{but} f(T) \le max \{ f(P), f(Q) \} \implies f(A) \le max \{ f(P), f(Q) , f(R)\} }
Hence it is sufficient to show for a point A on the line PQ (and the rest will follow):
Let \mathbf { A = (x, y), P = (x_p , y_p) , Q = (x_q, y_q) } Since A is on PQ, it is possible to write \mathbf { x = \frac{m x_p + n x_q }{ m+n} , y = \frac{ m y_p + n y_q} {m+n}, m,n \in \mathbb{R}^{+} }
Suppose f(A) is larger than both f(P) and f(Q). Then f(A) – f(P) and f(A) – f(Q) are both positive.
Let \mathbf { \lambda_1 = \frac{m}{m+n} , \lambda_2 = \frac{n}{m+n} \implies \lambda_1 + \lambda_2 = 1 }
\mathbf { f(A) - f(P) = ax+by+c - ax_p - by_p -c = a(x-x_p) + b(y - y_p) = \newline a(\lambda_1 x_p + \lambda_2 x_q - x_p) + b (\lambda_1 y_p + \lambda_2 y_q - y_p) }
\mathbf {\implies f(A) - f(P) = a(\lambda_2 x_q - (1-\lambda_1) x_p) + b(\lambda_2 y_q - (1-\lambda_1) y_p) = a \lambda_2 (x_q - x_p) + b \lambda_ 2 (y_q - y_p) }
Similarly we can show \mathbf { f(A) - f(Q) = a \lambda_1 (x_p - x_q) + b \lambda_1 (y_p - y_q) }
\mathbf { f(A) - f(Q) = \frac{-\lambda_2}{\lambda_1} (f(A) - f(P) ) }
Hence f(A) – f(Q) and f(A) – f(P) are opposite signs. The conclusion follows.
  • Prove that sum of any 12 consecutive integers cannot be perfect square. Give an example where sum of 11 consecutive integers is a perfect square

Discussion: Suppose a, a+1, a+2 , … , a+ 11 are 12 consecutive integers.
Sum of these 12 integers are 6(2a + 11). This is an even integer. If it is square, it must be divisible by 4. But 6(2a+11) = 2 times odd (hence not divisible by 4). Thus it is never a perfect square.
For 11 consecutive integers the sum is \mathbf{ \frac {11}{2} (2a + 10) = 11(a+5) } . a = 6 gives a perfect square.
  • Let \mathbf { y = x^4 + ax^3 + bx^2 + cx +d , a,b,c,d,e \in \mathbb{R}}. it is given that the functions cuts the x axis at least 3 distinct points. Then show that it either cuts the x axis at 4 distinct point or 3 distinct point and at any one of these three points we have a maxima or minima.

Discussion:
Since all the coefficients are real, complex roots occur as conjugates. Hence the fourth root (it is a four degree polynomial hence has a fourth root), must be real (if it is complex then we must have at least one more complex root, but all the other three roots are given to be real).
Let l, m, n be the three roots. Then the fourth root is either distinct from l, m, n or it is equivalent to exactly one of them say ‘n’.
If it is equal to n then we may rewrite the polynomial as \mathbf { y = (x-n)^2 (x-l)(x-m) }
We take first and second derivative of the y with respect to x.
\mathbf { y' = (x-n)^2 (x-l) + (x-n)^2 (x-m) + 2(x-n) (x-l)(x-m) }. At x=n the first derivative vanishes. Hence x=n is a critical point. We want to show that this is also a point of maxima or minima. For that we must show that the second derivative at x=n is positive or negative (not zero).
\mathbf { y'' = 2(x-n)(x-l) + (x-n)^2 + (x-n)^2+ 2(x-n) (x-m) + 2(x-n) (x-m) + 2(x-l)(x-m) + 2(x-n) (x-l) }
Hence at x=n \mathbf { y'' = 2(n-l)(n-m) } . Since n is distinct from m or l, hence the second derivative is either positive or negative and not zero. Hence we have maxima or minima at that point.
Proved.
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