answer:**Analysis**: Since the highest digit cannot be 0, there are three possibilities for the thousandth place. Since the digits can repeat in different positions, each of the other positions (hundreds, tens, and ones) has four possibilities. By applying the principle of multiplication, we can find the solution. Therefore, the total number of four-digit numbers that can be formed is 3 times 4 times 4 times 4 = 192. So, the final answer is boxed{192}.

answer:Applying the product-to-sum identities, First, simplify the numerator: [sin{30^circ} + sin{60^circ} = 2sin{left(frac{30^circ+60^circ}{2}right)} cos{left(frac{30^circ-60^circ}{2}right)} = 2sin{45^circ}cos{-15^circ}.] Simplify the denominator: [cos{30^circ} + cos{60^circ} = 2cos{left(frac{30^circ+60^circ}{2}right)} cos{left(frac{30^circ-60^circ}{2}right)} = 2cos{45^circ}cos{-15^circ}.] Thus, the expression simplifies to: [frac{sin{30^circ}+sin{60^circ}}{cos{30^circ}+cos{60^circ}} = frac{2sin{45^circ}cos{-15^circ}}{2cos{45^circ}cos{-15^circ}} = frac{sin{45^circ}}{cos{45^circ}} = boxed{tan{45^circ}}.] Conclusion: Therefore, we get boxed{tan 45^circ} as the simplified expression for the given trigonometric function.

answer:1. **Define Tangency Points:** Let ( A_1 ) be the tangency point of the incircle ( gamma ) with side ( AB ). Similarly, define ( B_1, C_1, D_1 ) as the tangency points of ( gamma ) with sides ( BC, CD, DA ) respectively. 2. **Orthogonality Condition:** We need to show that the circle ( (EXF) ) and the incircle ( gamma ) are orthogonal with respect to the circle ( (KL) ). This means that the power of point ( X ) with respect to ( (EXF) ) and ( gamma ) should be equal. 3. **Harmonic Conjugates:** It is a well-known configuration that the pairs ( (E, F) ) and ( (K, L) ) are harmonic conjugates. This implies that inversion with respect to the circle ( (KL) ) swaps ( E ) and ( F ). 4. **Pascal's Theorem Application:** By applying Pascal's theorem to the hexagon ( A_1A_1B_1D_1D_1C_1 ), we know that the lines ( A_1B_1 ), ( C_1D_1 ), and ( AC ) are concurrent. This concurrency point is ( K ). 5. **Concurrency on ( EF ):** Similarly, by applying Pascal's theorem to the hexagons ( A_1A_1B_1C_1C_1D_1 ) and ( A_1B_1B_1C_1D_1D_1 ), we see that the concurrency point lies on ( EF ), confirming that it is ( K ). 6. **Polars and Inversion:** The lines ( A_1B_1 ) and ( C_1D_1 ) are the polars of points ( B ) and ( D ) with respect to the incircle ( gamma ). Therefore, ( BD ) is the polar of ( K ). Similarly, ( AC ) is the polar of ( L ). 7. **Self-Mapping Under Inversion:** Since ( K ) lies on ( AC ) and ( L ) lies on ( BD ), the line ( KL ) maps to itself under inversion with respect to ( gamma ). This implies that the circle ( (KL) ) is orthogonal to ( gamma ). 8. **Conclusion:** Since ( (EXF) ) and ( gamma ) are both orthogonal with respect to ( (KL) ), it follows that ( (EXF) ) is tangent to ( gamma ). (blacksquare)

answer:To solve this problem, we'll break it down into detailed steps, closely following the solution provided: 1. First, we need to determine the integer part of sqrt{6}. We know that: [ 4 < 6 < 9 ] This implies: [ 2 < sqrt{6} < 3 ] Therefore, the integer part of sqrt{6} is 2, so we have: [ a = 2 ] 2. Next, we need to find the values of b and c from the equation 2 + sqrt{6} = b + c, where b is an integer and 0 < c < 1. Given that 2 < sqrt{6} < 3, we can add 2 to each part of the inequality: [ 4 < 2 + sqrt{6} < 5 ] This means 2 + sqrt{6} is a number between 4 and 5, so b = 4 because b is the integer part of 2 + sqrt{6}, and c must be the fractional part, which is: [ c = 2 + sqrt{6} - 4 = sqrt{6} - 2 ] 3. To find the length of the third side of the right-angled triangle with sides a and b, we consider two cases: - **Case 1:** When b=4 is one of the legs, the length of the third side is calculated using the Pythagorean theorem: [ sqrt{a^2 + b^2} = sqrt{2^2 + 4^2} = sqrt{4 + 16} = sqrt{20} = 2sqrt{5} ] - **Case 2:** When b=4 is the hypotenuse, the length of the third side is: [ sqrt{b^2 - a^2} = sqrt{4^2 - 2^2} = sqrt{16 - 4} = sqrt{12} = 2sqrt{3} ] 4. In conclusion, the length of the third side of the right-angled triangle can be either 2sqrt{5} or 2sqrt{3}. Therefore, the final answer is encapsulated as: [ boxed{2sqrt{5} text{ or } 2sqrt{3}} ]