answer:1. **Definition and Setup**: Let ( A_1 ) and ( B_1 ) be the feet of the perpendiculars dropped from point ( P ) to the lines ( BC ) and ( AC ) respectively. This creates right angles at points ( A_1 ) and ( B_1 ) with the segments ( BC ) and ( AC ). 2. **Inscribed Circle**: Notice that points ( A_1 ) and ( B_1 ) lie on a circle with diameter ( PC ). This follows from the fact that (angle PA_1C = angle PB_1C = 90^circ), thus both angles subtend the diameter ( PC ). 3. **Angles in the Inscribed Circle**: Let the angle (angle ACB) be constant everywhere, as given in the problem. By virtue of the circle with diameter ( PC ), and the right angles inscribed in this circle: [ sin angle A_1CB_1 = sin angle ACB. ] 4. **Chord Length**: Since (angle A_1CB_1 = angle ACB) and (angle ACB) is constant, the length of the chord ( A_1B_1 ), which subtends the angle (angle A_1CB_1), is also constant. This follows from the property of a circle where the length of a chord is determined by the measure of the angle it subtends at the circumference and the radius of the circle. 5. **Fixed Circle**: Therefore, the perpendiculars dropped from ( P ) yield chords ( A_1B_1 ) of constant lengths on the same circle with diameter ( PC ). 6. **Simson Line Property**: Given that ( A ) and ( B ) move such that the angle ( angle ACB ) remains constant, the locus of points ( A_1B_1 ) remains fixed relative to a certain circle, meaning their Simson lines also behave predictively with respect to this fixed geometric arrangement. 7. **Conclusion**: Consequently, the Simson lines of point ( P ) with respect to the triangle ( ABC ) are tangent to a fixed circle, implying that as ( A ) and ( B ) vary while maintaining the fixed angle, the tangency condition remains invariant to the system. Thus, the proof is complete. blacksquare

answer:To solve for the value of the integral int_{0}^{2} left(sqrt{4-(x-2)^2} - xright) dx, we break it down into two separate integrals and evaluate each one individually. First, we recognize that the function sqrt{4-(x-2)^2} represents the top half of a circle with a radius of 2 and center at (2,0). Specifically, it is the top semicircle of the circle (x-2)^2 + y^2 = 4. The area under this curve from x=0 to x=2 corresponds to a quarter of the circle's area. Therefore, we can compute the area directly using the formula for the area of a circle, which is pi r^2: int_{0}^{2} sqrt{4-(x-2)^2} dx = frac{1}{4} cdot pi cdot 2^2 = frac{pi}{4} cdot 4. Simplifying gives us the area of the quarter circle: int_{0}^{2} sqrt{4-(x-2)^2} dx = pi. Next, we evaluate the integral of the linear function -x from x=0 to x=2: int_{0}^{2} -x dx = -left[ frac{x^2}{2} right]_{0}^{2}. Plugging in the bounds, we get: -left( frac{2^2}{2} - frac{0^2}{2} right) = -2. Now, we combine the two integral results to find the value of the original integral: int_{0}^{2} left(sqrt{4-(x-2)^2} - xright) dx = pi + (-2) = pi - 2. Therefore, the final result is: [boxed{pi - 2}.]

answer:Let ( x = 0.overline{abcd} ). To convert this repeating decimal into a fraction, multiply ( x ) by ( 10000 ) (since there are four digits in the repeating sequence): [ 10000x = abcd.overline{abcd} ] Therefore, subtract ( x ) from ( 10000x ): [ 9999x = 10000x - x = abcd ] [ x = frac{abcd}{9999} ] Now, consider the factors of ( 9999 ). Factoring ( 9999 ) yields: [ 9999 = 3^2 times 11 times 101 ] To determine how many different numerators are required, let's analyze the divisors: - Multiples of ( 3 ) not reduced by ( 3^2 ) - Multiples of ( 11 ) - Multiples of ( 101 ) Assessing whether ( abcd ) is divisible by ( 11 ) or ( 101 ) independently changes the aspect drastically as there are fewer multiples. Every multiple of ( 3 ), ( 11 ), and ( 101 ) among ( 1 ) to ( 9999 ) that does not contain these factors at a higher power reduces its terms uniquely when put in the denominator. Counting numbers not divisible by ( 3^2 ), ( 11 ), or ( 101 ): [ 9999 - text{count}(3^2) - text{count}(11) - text{count}(101) + text{intersections} ] Here: [ text{count}(3^2) = 1111, quad text{count}(11) = 909, quad text{count}(101) = 99 ] Calculating intersections and applying the inclusion-exclusion principle gives: [ 9999 - 1111 - 909 - 99 + text{intersections} = 7880 + text{intersections} ] Correcting this for over-count of conditions where ( abcd ) is a multiple of combinations of ( 3^2 ), ( 11 ), and ( 101 ) gives us the final count needed. In conclusion: [ boxed{6643} ] is the tentative number of different numerators required after applying adjustments for simple multiples.

answer:We start with the given recurrence relation for a_n: [ sqrt{a_{n} a_{n-2}} - sqrt{a_{n-1} a_{n-2}} = 2a_{n-1} quad (n geq 2) ] Substitute a_{n-1}, we rearrange the equation as follows: [ sqrt{a_{n} a_{n-2}} - 2a_{n-1} = sqrt{a_{n-1} a_{n-2}} ] Now, divide both sides by sqrt{a_{n-1} a_{n-2}}: [ frac{sqrt{a_{n} a_{n-2}}}{sqrt{a_{n-1} a_{n-2}}} - frac{2a_{n-1}}{sqrt{a_{n-1} a_{n-2}}} = frac{sqrt{a_{n-1} a_{n-2}}}{sqrt{a_{n-1} a_{n-2}}} ] This simplifies to: [ sqrt{frac{a_n}{a_{n-1}}} - 2sqrt{frac{a_{n-1}}{a_{n-2}}} = 1 ] Next, observe the sum of geometric series and constantly use it to simplify the recurrence. Multiply each equation of the form sqrt{frac{a_k}{a_{k-1}}}-2sqrt{frac{a_{k-1}}{a_{k-2}}}=1 from k=2 to n by 2^{k-2} and then sum all equations: [ sum_{k=2}^{n} 2^{k-2} left(sqrt{frac{a_k}{a_{k-1}}} - 2sqrt{frac{a_{k-1}}{a_{k-2}}}right) = sum_{k=2}^{n} 2^{k-2} ] This simplifies to: [ sqrt{frac{a_n}{a_{n-1}}} - 2^{n-1}sqrt{frac{a_1}{a_0}} = 1 + 2 + 2^2 + cdots + 2^{n-2} ] Using knowledge of geometric series: [ 1 + 2 + 4 + cdots + 2^{n-2} = 2^{n-1} - 1 ] Thus, the equation becomes: [ sqrt{frac{a_n}{a_{n-1}}} - 2^{n-1} cdot 1 = 2^{n-1} - 1 ] Simplifying further: [ sqrt{frac{a_n}{a_{n-1}}} = 2^{n} - 1 ] Squaring both sides to eliminate the square root: [ frac{a_n}{a_{n-1}} = (2^{n} - 1)^2 ] Therefore, we can write: [ a_n = (2^n - 1)^2 a_{n-1} ] Continuing recursively, we have: [ a_n = (2^n - 1)^2 (2^{n-1} - 1)^2 cdots (2^1 - 1)^2 cdot a_0 ] Given that a_0 = 1, we find: [ a_n = prod_{k=1}^{n} (2^k - 1)^2 ] To summarize, the general term for a_n is given by: [ left{ begin{array}{ll} 1 & text{if } n = 0, prod_{k=1}^{n} (2^k - 1)^2 & text{if } n > 0. end{array} right. ] Thus, the closed form expression is: [ boxed{ a_n = prod_{k=1}^{n} (2^k - 1)^2 } ]