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Step-by-Step Solution
Step 1: Finding the Orthocenter (a, b)
The triangle has vertices:
(1, 2), (2, 3), and (3, 1).
The orthocenter is the point where the altitudes of the triangle meet.
Find the slope of the side joining (1, 2) and (2, 3):
$ \text{slope}_1 = \frac{3 - 2}{2 - 1} = 1. $
Equation of altitude from the vertex (3, 1):
The altitude is perpendicular to the side that has slope 1, so its slope is $ -1 $.
The line with slope $ -1 $ passing through (3, 1) is:
$ (y - 1) = -1 (x - 3). $
Simplifying:
$ y - 1 = -x + 3 \implies y = -x + 4. $
Find the slope of the side joining (2, 3) and (3, 1):
$ \text{slope}_2 = \frac{1 - 3}{3 - 2} = -2. $
Equation of altitude from the vertex (1, 2):
The altitude is perpendicular to the side that has slope $ -2 $, so its slope is $ \frac{1}{2} $.
The line with slope $ \frac{1}{2} $ passing through (1, 2) is:
$ (y - 2) = \frac{1}{2} (x - 1). $
Simplifying:
$ y - 2 = \tfrac{1}{2}x - \tfrac{1}{2}
\implies y = \tfrac{1}{2}x + \tfrac{3}{2}. $
Find the intersection of these two altitudes to get the orthocenter:
Altitude 1: $ y = -x + 4. $
Altitude 2: $ y = \tfrac{1}{2}x + \tfrac{3}{2}. $
Setting them equal:
$ -x + 4 = \tfrac{1}{2} x + \tfrac{3}{2}. $
Combine like terms:
$ -x - \tfrac{1}{2} x = \tfrac{3}{2} - 4
\implies -\tfrac{3}{2} x = -\tfrac{5}{2}
\implies x = \frac{-\tfrac{5}{2}}{-\tfrac{3}{2}} = \frac{5}{3}. $
Substitute $ x = \tfrac{5}{3} $ into $ y = -x + 4 $:
$ y = -\tfrac{5}{3} + 4 = -\tfrac{5}{3} + \tfrac{12}{3} = \tfrac{7}{3}. $
Hence, the orthocenter is
$ (a, b) = \left(\tfrac{5}{3},\, \tfrac{7}{3}\right). $
Step 2: Writing Down the Integrals
We have
$ a = \frac{5}{3}, \quad b = \frac{7}{3}. $
The integrals given in the question are:
$ I_1 = \int_{a}^{b} x \sin\bigl(4x - x^2\bigr)\,dx $
$ I_2 = \int_{a}^{b} \sin\bigl(4x - x^2\bigr)\,dx $
Step 3: Using Symmetry to Relate $I_1$ and $I_2$
Notice that the interval
$ \left[\frac{5}{3}, \frac{7}{3}\right] $
is symmetric about $ x = 2 $.
Also, observe:
$ 4x - x^2 = 4 - (x^2 - 4x) = 4 - (x - 2)^2. $
When a function of the form $ \sin\bigl((x-2)^2\!\bigl) $ is integrated over an interval symmetric around $ x=2 $, the average value of $ x $ over that interval is 2.
Hence, in many such integrals, the factor $ x $ in $ I_1 $ can be treated like a constant 2 for a perfectly symmetric integrand regarding $ x=2 $. More rigorously, one can show that
$ I_1 = \int_{a}^{b} x \sin(4x - x^2)\,dx
= 2 \int_{a}^{b} \sin(4x - x^2)\,dx = 2 I_2. $
Thus, $ \displaystyle \frac{I_1}{I_2} = 2. $
Step 4: Final Computation
Substitute $ \frac{I_1}{I_2} = 2 $ into
$ 36 \times \frac{I_1}{I_2}: $
$ 36 \times \frac{I_1}{I_2} = 36 \times 2 = 72. $
Therefore, the required value is
72.
