answer:1. **Given Setup and Definitions:** - In triangle ABC, the A-excircle is tangent to BC, AB, and AC at E, D, and F respectively. - EZ is the diameter of the A-excircle. - Points B_1 and C_1 lie on DF such that BB_1 perp BC and CC_1 perp BC. - Lines ZB_1 and ZC_1 intersect BC at X and Y respectively. - EZ and DF intersect at H. - ZK is perpendicular to DF at K. - H is the orthocenter of triangle XYZ. 2. **Objective:** - Prove that H, K, X, and Y are concyclic. 3. **Using Menelaus' Theorem:** - Let the incircle of triangle ABC intersect AB, BC, and CA at D_2, E_2, and F_2 respectively. - Let DF cap BC = T and D_2F_2 cap BC = T_2. - By Menelaus' theorem applied to triangle ABC with transversal D_2F_2, we have: [ frac{BT_2}{T_2C} times frac{CF_2}{F_2A} times frac{AD_2}{D_2B} = 1 ] Simplifying, we get: [ frac{BT_2}{T_2C} times frac{CF_2}{BD_2} = frac{BT_2}{T_2C} times frac{DB}{CF} ] 4. **Applying Menelaus' Theorem Again:** - By Menelaus' theorem applied to triangle ABC with transversal TDF, we have: [ frac{CT}{TB} times frac{BD}{DA} times frac{AF}{FC} = 1 ] Simplifying, we get: [ frac{CT}{TB} times frac{DB}{CF} ] - Therefore, frac{BT_2}{T_2C} = frac{CT}{TB}, implying that the midpoint of TT_2 is M, the midpoint of BC. 5. **Harmonic Division:** - It is well-known that (B, C; E_2, T_2) = -1 since AE_2, CD_2, and BF_2 are concurrent. - Reflecting through M gives us (B, C; E, T) = -1. - Considering the pencil at the point at infinity in the direction perpendicular to BC, we get (B_1, C_1; H, T) = -1. - At point Z, we get (X, Y; E, T) = -1. 6. **Orthocenter and Cyclic Quadrilateral:** - Let XH perp YZ at P and YH perp XZ at Q. - It is well-known that if PQ cap XY = T_3, then (X, Y; E, T_3) = -1, so T = T_3. - Consider the cyclic quadrilateral XQPY. Note that angle XQY = angle XPY, so the quadrilateral is cyclic with center at N, the midpoint of XY. 7. **Brocard's Theorem and Collinearity:** - By Brocard's Theorem, we get that TH perp ZN. - Since ZK perp TH, we get Z, K, N collinear. - Let Z_1 be the point defined by reflecting Z through N. We have Z_1X parallel ZY and Z_1Y parallel ZX. - Therefore, angle HXZ_1 = 90^circ = angle HYZ_1, and we also have angle HKZ_1 = 90^circ. 8. **Conclusion:** - Hence, X, H, K, Y, Z_1 lie on the same circle, in other words, H, K, X, Y are concyclic. blacksquare

answer:# Problem: Fill the circles with the numbers from 1 to 6 such that the sum of the numbers on each straight line is equal. The diagram has three straight lines with three circles each, and two straight lines with two circles each. Then find the two-digit number overline{AB} created by placing the numbers A and B together. 1. **Set up Equations Based on the Given Conditions:** According to the problem, we have several conditions involving the sums of the numbers assigned to each circle. Let us denote the circles by A, B, C, D, E, and F. The conditions are as follows: [ begin{align*} 1) & quad A + C + D = text{constant}, 2) & quad A + B = text{constant}, 3) & quad B + D + F = text{constant}, 4) & quad E + F = text{constant}, 5) & quad E + B + C = text{constant}. end{align*} ] 2. **Using the Pair Sums:** From the problem statement, we also get the pairwise sums: [ begin{align*} 6) & quad B = C + D, 7) & quad B + D = E, 8) & quad E + C = A. end{align*} ] 3. **Solving for Specific Values:** We need to solve the above system of equations to determine the values of A, B, C, D, E, and F. Let's check two possible solutions to these equations: **Solution Set 1:** begin{align*} D & = 1, C & = 2, B & = 3, E & = 4, A & = 6, F & = 5. end{align*} Verification: [ begin{aligned} & A + C + D = 6 + 2 + 1 = 9, & A + B = 6 + 3 = 9, & B + D + F = 3 + 1 + 5 = 9, & E + F = 4 + 5 = 9, & E + B + C = 4 + 3 + 2 = 9. end{aligned} ] **Solution Set 2:** begin{align*} D & = 2, C & = 1, B & = 3, E & = 5, A & = 6, F & = 4. end{align*} Verification: [ begin{aligned} & A + C + D = 6 + 1 + 2 = 9, & A + B = 6 + 3 = 9, & B + D + F = 3 + 2 + 4 = 9, & E + F = 5 + 4 = 9, & E + B + C = 5 + 3 + 1 = 9. end{aligned} ] 4. **Extracting the Two-digit Number:** Both solution sets give us the same A and B values (A=6, B=3). Therefore, the two-digit number overline{AB}, which is formed by combining A and B, is: [ boxed{63} ]

answer:Given the parabola y = ax^2 + bx opens downwards and passes through the point P(-1, m) in the third quadrant, and point P and the origin are on opposite sides of the axis of symmetry of the parabola, we can deduce the following: 1. Since the parabola opens downwards, we have a < 0. 2. The axis of symmetry of the parabola is given by x = -frac{b}{2a}. Since point P and the origin are on opposite sides of the axis of symmetry, and P is in the third quadrant (implying x = -1 < 0), we have -frac{b}{2a} < 0. This leads to the conclusion that b < 0, because for -frac{b}{2a} to be negative while a < 0, b must also be negative. 3. At x = -1, the y-coordinate of point P is given by m = a(-1)^2 + b(-1) = a - b. Since point P is in the third quadrant, we know m < 0. Therefore, a - b < 0, which implies m = a - b < 0. Given the information above, we analyze the graph of the linear function y = (a - b)x + b: - Since a < 0 and b < 0, the slope (a - b) and the y-intercept b are both negative. - This means the line descends from left to right and starts below the x-axis because the y-intercept is negative. Considering the characteristics of the line y = (a - b)x + b, it will intersect the x-axis at some point to the right of the y-axis (since it descends from left to right) and will not re-enter the first quadrant because both the slope and y-intercept are negative. Therefore, the graph of this linear function will be present in the second, third, and fourth quadrants but not in the first quadrant. boxed{text{A}}

answer:Haley initially took a total of 50 (zoo) + 8 (museum) = 58 pictures. After deleting some pictures, she had 20 pictures left. Therefore, the number of pictures she deleted is 58 (initial) - 20 (left) = boxed{38} pictures.