answer:The inradius r of triangle ABC is given by r = frac{ssqrt{3}}{6}, where s = 800. Thus, r = frac{800 sqrt{3}}{6} approx 133.33sqrt{3}. The circumradius R of triangle ABC is given by R = frac{s}{sqrt{3}} = frac{800}{sqrt{3}} approx 266.67sqrt{3}. P, Q, and O must lie on the line perpendicular to plane ABC through the circumcenter K of triangle ABC since they are equidistant from each vertex of the triangle. Let's denote KP = x, KQ = y, and KO = d. Since O is equidistant from P and Q, O is the midpoint of line segment PQ. Therefore, d = frac{x+y}{2}. Considering the dihedral angle to be 90^circ, the relation between x, y, and d can be obtained by setting the angle angle PKQ = 90^circ. Thus, x^2 + y^2 = 4d^2 (using Pythagorean theorem in triangle PKQ). Solve for x and y: Assuming symmetry and the fact that O is the midpoint, x = y. So, 2x^2 = 4d^2, thus x^2 = 2d^2 and x = y = sqrt{2}d. Therefore, d = frac{x}{sqrt{2}}. Given the symmetry and the geometric setup, we can calculate d = R sqrt{2} = 266.67sqrt{3} cdot sqrt{2} approx 377.96. Thus, boxed{377.96}.

answer:First, let's calculate the number of turnips Keith grows in a week. Since he grows 6 turnips every day, we multiply the number of turnips he grows per day by the number of days in a week: Keith's turnips = 6 turnips/day * 7 days/week = 42 turnips Next, let's calculate the number of turnips Alyssa grows in two weeks. Since she grows 9 turnips every two days, we need to find out how many two-day periods there are in two weeks. There are 7 days in a week, so there are 14 days in two weeks. We divide the number of days by 2 to find the number of two-day periods: Two-day periods in two weeks = 14 days / 2 days/period = 7 periods Now we multiply the number of turnips Alyssa grows per two-day period by the number of periods: Alyssa's turnips = 9 turnips/period * 7 periods = 63 turnips Finally, we add the number of turnips Keith and Alyssa grew to find the total: Total turnips = Keith's turnips + Alyssa's turnips Total turnips = 42 turnips + 63 turnips = 105 turnips Keith and Alyssa grew a total of boxed{105} turnips.

answer:- The area, A, of an equilateral triangle with side length s = 6 units is given by: [ A = frac{6^2 sqrt{3}}{4} = frac{36 sqrt{3}}{4} = 9 sqrt{3} text{ square units} ] - The perimeter, P, of this triangle is: [ P = 3 times 6 = 18 text{ units} ] - The ratio of the perimeter to the area is: [ text{Ratio} = frac{P}{A} = frac{18}{9 sqrt{3}} = frac{18}{9 sqrt{3}} times frac{sqrt{3}}{sqrt{3}} = frac{18 sqrt{3}}{27} = frac{2 sqrt{3}}{3} ] Conclusion with boxed answer: [ boxed{frac{2 sqrt{3}}{3}} ]

answer:1. **The Initial Setup**: Let's consider a triangle ( ABC ). - Let ( A_1 ), ( B_1 ), and ( C_1 ) be the feet of the altitudes from vertices ( A ), ( B ), and ( C ) respectively. - Let ( M ) be the orthocenter of the triangle ( ABC ), which is the point where all three altitudes intersect. 2. **Defining the Cyclic Quadrilaterals** and their properties: - The quadrilaterals ( BCBA_1 ), ( ACAB_1 ), and ( ABCA_1 ) are concyclic, meaning that each can be inscribed in a circle. The circles in which these quadrilaterals are inscribed are the ones with diameters ( AA_1 ), ( BB_1 ), and ( CC_1 ) respectively. 3. **Intersections of the Cyclic Quadrilaterals**: - Within each of these cyclic quadrilaterals, the point ( M ) (the orthocenter) satisfies certain power of a point relations. From the power of a point theorem applied to ( M ) with respect to each of the circles, we have: [ MA cdot MA_1 = MB cdot MB_1 = MC cdot MC_1 ] Here, these equations state that the power of the orthocenter ( M ) with respect to each circle is the same. 4. **Common Chords and Power of a Point**: - The common chords of these circles are critical. By the property of the power of a point, since the point ( M ) has equal power with respect to all three circles, it must lie on the radical axes (common chords) of the pairs of circles. - Specifically, the radical axis (line consisting of points that have equal power with respect to two circles) of any two circles is the line joining their common chords. - For instance, the common chord of the circles with diameters ( AA_1 ) and ( BB_1 ) intersects the circle with diameter ( CC_1 ) at point ( M ). 5. **Conclusion**: - Therefore, the common chords of these circles, corresponding to the sides of the cyclic quadrilaterals, intersect each other at the orthocenter ( M ). Hence, it is proven that the common chords of the circles drawn with the altitudes of a triangle as diameters intersect at the orthocenter of the triangle: [ boxed{} ]