answer:1. **Reflecting Point P in Angle Bisectors**: Let P be any point in the plane of triangle ABC, and we reflect the lines PA, PB, and PC each in the angle bisectors at A, B, and C, respectively. Denote the reflections of P as P_1, P_2`, and P_3. 2. **Double Reflections and Angle Calculations**: Assume (angle BAC = alpha) and (angle BAP = vartheta). Reflecting line (AP) in the angle bisector of (angle BAC) involves rotating by (frac{alpha}{2} - vartheta), followed by a rotation of (alpha-2vartheta) to reach the position (AP_1). To achieve this, we first reflect (AP) to the position (AP_c), where (P_c) is the reflection of (P) in line (AB), then rotate by (alpha). 3. **Reflection and Angle Bisectors**: A similar process is used to reflect (AP) in line (AC) resulting in (AP_b). Given that (AP_c = AP = AP_b), the line (AP_1) becomes the perpendicular bisector of the segment (P_cP_b). 4. **Properties of Perpendicular Bisectors**: Following the same procedure for (B) and (C), designate P_a as the reflection of (P) in (BC). The lines (BP_2) and (CP_3) will be the perpendicular bisectors of the segments involving (P_a, P_b,) and (P_c). 5. **Intersection Point (K)**: The lines (AP_1, BP_2, CP_3) will therefore be the perpendicular bisectors of the triangle (P_aP_bP_c), meeting at the center (K) of this triangle unless (P_aP_bP_c) is degenerate. 6. **Cases with Specific Points**: - Choosing (P) as the orthocenter (M) of triangle (ABC), the resulting point (P' ) is the circumcenter (O). - For (P) as the circumcenter (O), (P') is the orthocenter (M). 7. **Properties with Centroid**: - Choosing (P) as the centroid (S), line (AS_1) ensures (S') such that distances from sides are proportional. 8. **Projected Reflections**: Use the Sine Rule in the solution to verify the distance relationships between reflected points. 9. **Important Note**: If certain lines (PA, PB,) or (PC) are parallel to sides (BC, CA,) or (AB), special cases apply, such as if line (PA parallel BC), (P, A,) and (C) form a trapezoid, applying the Sine Law for verification. # Conclusion: This confirms the reflections and their properties, ensuring (P') exists for given specific points P. (P') becomes the unique intersection point of the constructed perpendicular bisectors in non-degenerate cases. All steps confirm (P') always lies on a particular point corresponding to the reflection properties. If conditions meet certain triangles, vertical bisectors will intersect following Ceva’s and Sine Rule application. Therefore, ( boxed{P'} ) will emerge for valid (P) chosen inside the triangle.

answer:Since the negation of a universal proposition is a particular proposition, the negation of the proposition p: ∀x∈(0, frac {π}{2}), f(x)<0 is ¬p: ∃x₀∈R, x_{ 0 }^{ 2 }=x₀. Hence, the answer is: boxed{D}. By directly using the fact that the negation of a universal proposition is a particular proposition, we can obtain the result. This question examines the understanding of the relationship between universal and particular propositions, and is a basic question.

answer:To determine the greatest common divisor (gcd or operatorname{pgcd} in French) of the numbers represented by ( underbrace{1 ldots 1}_{m text{ chiffres}} ) and ( underbrace{1 ldots 1}_{n text{ chiffres}} ) where m and n are strictly positive integers, follow these steps: 1. **Representation of the Numbers**: First, note that ( underbrace{1 ldots 1}_{k text{ chiffres}} ) represents a number formed by k digits, all of which are 1. This number can be written as: underbrace{1 ldots 1}_{k text{ chiffres}} = frac{10^k - 1}{9} 2. **Setting Up the Euclidean Algorithm**: Suppose ( m < n ), and write the Euclidean division of n by m: n = mq + r where q is the quotient and r is the remainder (0 leq r < m). 3. **Applying the Euclidean Algorithm**: The division of ( underbrace{1 ldots 1}_{n text{ chiffres}} ) by ( underbrace{1 ldots 1}_{m text{ chiffres}} ) uses the fact that: underbrace{1 ldots 1}_{n text{ chiffres}} = underbrace{1 ldots 1}_{m text{ chiffres}} left(10^{m(q-1) + r} + cdots + 10^{m+r} + 10^r right) + underbrace{1 ldots 1}_{r text{ chiffres}} 4. **Simplifying**: The remainder when ( underbrace{1 ldots 1}_{n text{ chiffres}} ) is divided by ( underbrace{1 ldots 1}_{m text{ chiffres}} ) is ( underbrace{1 ldots 1}_{r text{ chiffres}} ). According to the Euclidean algorithm, the gcd is the same as the gcd of ( underbrace{1 ldots 1}_{m text{ chiffres}} ) and ( underbrace{1 ldots 1}_{r text{ chiffres}} ). 5. **Iterating the Process**: Continue applying the Euclidean algorithm: m = r_0 q_1 + r_1 r_0 = r_1 q_2 + r_2 vdots r_{k-1} = r_k q_{k+1} + 0 where the process ends when the remainder is zero. The gcd of the sequence will be the last non-zero remainder. 6. **Conclusion**: The gcd of ( underbrace{1 ldots 1}_{m text{ chiffres}} ) and ( underbrace{1 ldots 1}_{n text{ chiffres}} ) is given by the initial result of the Euclidean algorithm: operatorname{pgcd}left( underbrace{1 ldots 1}_{m text{ chiffres}}, underbrace{1 ldots 1}_{n text{ chiffres}} right) = underbrace{1 ldots 1}_{operatorname{pgcd}(m,n) text{ chiffres}} Oxed: boxed{ underbrace{1 ldots 1}_{operatorname{pgcd}(m,n) text{ chiffres}} }

answer:Mildred has boxed{79} oranges.