answer:[asy] pair X,Y,Z; X = (0,0); Y = (14,0); Z = (0,2); draw(X--Y--Z--X); draw(rightanglemark(Y,X,Z,23)); label("X",X,SW); label("Y",Y,SE); label("Z",Z,N); label("100",(Y+Z)/2,NE); label("k",(Z)/2,W); label("7k",Y/2,S); [/asy] Since triangle XYZ is a right triangle with angle X = 90^circ, we have tan Z = frac{XY}{XZ}. Since tan Z = 7, we have XY = 7k and XZ = k for some value of k, as shown in the diagram. Applying the Pythagorean Theorem gives (7k)^2 + k^2 = 100^2, so 50k^2 = 100^2, which gives k^2 = 100^2/50 = 200. Since k must be positive, we have k = sqrt{200} = 10sqrt{2}, so XY = 7k = boxed{70sqrt{2}}.

answer:Note that the area is given by Heron's formula and it is 20sqrt{221}. Let h_i denote the length of the altitude dropped from vertex i. It follows that h_b = frac{40sqrt{221}}{27}, h_c = frac{40sqrt{221}}{30}, h_a = frac{40sqrt{221}}{23}. From similar triangles we can see that frac{27h}{h_a}+frac{27h}{h_c} le 27 rightarrow h le frac{h_ah_c}{h_a+h_c}. We can see this is true for any combination of a,b,c and thus the minimum of the upper bounds for h yields h = frac{40sqrt{221}}{57} rightarrow boxed{318}.

answer:The formula for the interior angle of a regular sided polygon is frac{(n-2)180}{n}. Thus, frac{frac{(r-2)180}{r}}{frac{(s-2)180}{s}} = frac{59}{58}. Cross multiplying and simplifying, we get frac{58(r-2)}{r} = frac{59(s-2)}{s}. Cross multiply and combine like terms again to yield 58rs - 58 cdot 2s = 59rs - 59 cdot 2r Longrightarrow 118r - 116s = rs. Solving for r, we get r = frac{116s}{118 - s}. r ge 0 and s ge 0, making the numerator of the fraction positive. To make the denominator positive, s < 118; the largest possible value of s is 117. This is achievable because the denominator is 1, making r a positive number 116 cdot 117 and s = boxed{117}.

answer:We begin by drawing a diagram: [asy] pair A,B,C,D,E,X; A=(0,9); B=(0,0); C=(13,0); E=(A+C)/2; D=(A+B)/2; X = intersectionpoint(B--E,D--C); label("X",X,N); fill(A--E--X--D--cycle,rgb(135,206,250)); fill(B--X--C--cycle,rgb(107,142,35)); draw(A--B--C--cycle); draw(C--D); draw(B--E); draw(rightanglemark(A,B,C,15)); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,W); label("E",E,NE); label("13",(6.5,0),S); label("9",(-2,4.5),W); draw((-2.7,5.3)--(-2.7,9),EndArrow(TeXHead));draw((-2.7,3.7)--(-2.7,0),EndArrow(TeXHead)); [/asy] Since D and E are midpoints, overline{CD} and overline{BE} are medians. Let F be the midpoint of overline{BC}; we draw median overline{AF}. The medians of a triangle are always concurrent (pass through the same point), so overline{AF} passes through X as well. [asy] pair A,B,C,D,E,X,F; A=(0,9); B=(0,0); C=(13,0); E=(A+C)/2; D=(A+B)/2; X = intersectionpoint(B--E,D--C); label("X",X,N); F=(B+C)/2; draw(A--F,dashed); label("F",F,S); draw(A--B--C--cycle); draw(C--D); draw(B--E); draw(rightanglemark(A,B,C,15)); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,W); label("E",E,NE); [/asy] The three medians cut triangle ABC into six smaller triangles. These six smaller triangles all have the same area. (To see why, look at overline{BC} and notice that triangle BXF and triangle CXF have the same area since they share an altitude and have equal base lengths, and triangle ABF and triangle ACF have the same area for the same reason. Thus, triangle ABX and triangle ACX have the same area. We can repeat this argument with all three sizes of triangles built off the other two sides overline{AC} and overline{AB}, to see that the six small triangles must all have the same area.) Quadrilateral AEXD is made up of two of these small triangles and triangle BXC is made up of two of these small triangles as well. Hence they have the same area (and this will hold true no matter what type of triangle triangle ABC is). Thus, the ratio of the area of quadrilateral AEXD to the area of triangle BXC is 1/1=boxed{1}.