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question:For a group of n people, where k1 are females, k2 females are in management roles, and k3 is the average years of experience in their field for the females in the group. The rest are males with l1 of them in management roles and l2 is their average years of experience in their field. The index for a certain phenomenon in group dynamics for members of that sex is measured by: Index_females = ((n k1 + k2) / n) * (1 + k3/10) Index_males = ((n (n k1) + l1) / n) * (1 + l2/10) In a group that consists of 35 people, 15 of whom are females with 5 of them in management roles and an average of 8 years of experience in their field, 6 males are in management roles with an average of 10 years of experience in their field. By how much does the index for the females exceed the index for the males in the group?
answer:First, let's calculate the index for females using the given formula: Index_females = ((n - k1 + k2) / n) * (1 + k3/10) Given: n = 35 (total number of people) k1 = 15 (number of females) k2 = 5 (females in management roles) k3 = 8 (average years of experience for females) Index_females = ((35 - 15 + 5) / 35) * (1 + 8/10) Index_females = ((25) / 35) * (1 + 0.8) Index_females = (25/35) * 1.8 Index_females = (5/7) * 1.8 Index_females = 9/7 Now, let's calculate the index for males: Index_males = ((n - (n - k1) + l1) / n) * (1 + l2/10) Given: n = 35 (total number of people) n - k1 = 35 - 15 = 20 (number of males) l1 = 6 (males in management roles) l2 = 10 (average years of experience for males) Index_males = ((35 - 20 + 6) / 35) * (1 + 10/10) Index_males = ((21) / 35) * (1 + 1) Index_males = (21/35) * 2 Index_males = (3/5) * 2 Index_males = 6/5 Now, let's find the difference between the indices: Difference = Index_females - Index_males Difference = (9/7) - (6/5) To subtract these fractions, we need a common denominator. The least common multiple of 7 and 5 is 35, so we convert both fractions to have the denominator of 35: Difference = (9/7) * (5/5) - (6/5) * (7/7) Difference = (45/35) - (42/35) Difference = (45 - 42) / 35 Difference = 3/35 Therefore, the index for the females exceeds the index for the males by boxed{3/35} .
question:Given a triangle ABC. The incircle (I) of triangle ABC touch the sides CA,AB at E,F. A point P moving on the segment EF . The line PB intersects CA at M ; the line MI intersects the line passing through C and perpendicular to AC at N. Prove that: if P is moving, the line passing through N and perpendicular to PC always passes a fixed point.
answer:1. **Setup and Definitions**: - Let the incircle (I) of triangle ABC touch CA at E and AB at F. - Let P be a point moving on the segment EF. - The line PB intersects CA at M. - The line MI intersects the line passing through C and perpendicular to AC at N. - We need to prove that the line passing through N and perpendicular to PC always passes through a fixed point as P moves. 2. **Key Points and Lines**: - Let the line passing through C and perpendicular to BC intersect the line perpendicular to PC at N at point T. - Let BM and DE intersect at S. - Let the perpendicular to BC at E and IS intersect at L. - Let CP intersect AB at K. 3. **Tangency and Polarity**: - MK is tangent to the incircle (I) at X (a well-known result). - S in FX. - The polar of K and the polar of C intersect at S, so by La Hire's theorem, the polar of S is CK. - Thus, SI perp CK or LI parallel NT. 4. **Homothety and Collinearity**: - triangle IEL and triangle NCT are homothetic, so M, L, T are collinear. - We need to show that B, I, T are collinear. 5. **Menelaus Theorem**: - Apply Menelaus' theorem on triangle CDE with transversal overline{BMS}: [ frac{BD}{CB} times frac{SE}{DS} times frac{MC}{EM} = 1 ] - Since LE parallel CT and LE parallel ID, we have: [ frac{BD}{CB} times frac{LE}{r} times frac{CT}{LE} = 1 ] - Simplifying, we get: [ frac{r}{BD} = frac{CT}{BC} ] - Therefore, triangle BDI sim triangle BCT by SAS similarity, implying B, I, T are collinear. 6. **Conclusion**: - The perpendicular to CP at N passes through the intersection point of the line passing through C and perpendicular to BC and BI, which is a fixed point. blacksquare
question:A pet store had 4 snakes. They had the snakes in cages with a certain number of snakes in each cage. The pet store had 2 cages. How many snakes were in each cage?
answer:If the pet store had 4 snakes and 2 cages, and the snakes were distributed evenly, then there would be 4 snakes ÷ 2 cages = boxed{2} snakes in each cage.
question:Given a triangle ( ABC ), a circle ( k ) is tangent to the lines ( AB ) and ( AC ) at ( B ) and ( P ). Let ( H ) be the foot of the perpendicular from the center ( O ) of ( k ) to ( BC ), and let ( T ) be the intersection point of ( OH ) and ( BP ). Prove that ( AT ) bisects the segment ( BC ).
answer:Given triangle (ABC), a circle (k) is tangent to lines (AB) and (AC) at points (B) and (P), respectively. Let (O) be the center of the circle (k), (H) be the foot of the perpendicular from (O) to (BC), and (T) be the intersection point of (OH) and (BP). We need to prove that (AT) bisects the segment (BC). Let's denote: - (X) as the intersection of (AT) and (BC), - (S) as the intersection of (BC) and (AO), - (Y) as the intersection of (BP) and (AO). 1. **Orthocenter Considerations**: We know that (BY perp SO) (since (O) is the center of the circle tangent to (AB) and (AC), and thus (A, B, Y,) and (P) lie on a circle with diameter (AO), making (BY) perpendicular to (SO)) and (OH perp BS) (by definition, (H) is the foot of the perpendicular from (O) to (BC)). Therefore, (T) is the orthocenter of (triangle BOS). 2. **Parallel Lines**: Because (T) is the orthocenter, (ST perp BO). Given that (BO) (the radius) is perpendicular to (AB), we conclude that (ST parallel AB). 3. **Triangle Ratio**: By properties of similar triangles: [ frac{ST}{AB} = frac{TX}{XA} ] However, we also know (AP = AB) because the circle is tangent at (B) and (P) with (O) as the center. Therefore, by properties of similar triangles again: [ frac{ST}{AP} = frac{TY}{YP} ] 4. **Parallelism and Midpoints**: Since (frac{ST}{AB} = frac{TY}{YP}) and (frac{TX}{XA}), it follows from the parallel line considerations (with similar triangles (SYT) and (AYP)) that (XY parallel AP). Furthermore, because (Y) is the midpoint of (BP) (since (O) is the center and tangents from a point to a circle are equal in length), (XY) is the midsegment of (triangle CBP), implying that (BX = XC), proving (X) is the midpoint of (BC). Conclusion: Since (X) is the midpoint of (overline{BC}), (AT) bisects (overline{BC}). [ boxed{}