Problem: Let triangle $ABC$ be a right triangle in the xy-plane with a right angle at $C_{}$ . Given that the length of the hypotenuse $AB$ is $60$ , and that the medians through $A$ and $B$ lie along the lines $y=x+3$ and $y=2x+4$ respectively, find the area of triangle $ABC$

Solution: We first seek to find the angle between the lines $y = x + 3$ and $y = 2x + 4$ [asy] import graph;  size(150);  Label f;  f.p=fontsize(6);  xaxis(-8,8,Ticks(f, 2.0));  yaxis(-8,8,Ticks(f, 2.0));  real f(real x)  {  return (x + 3); }  real g( real x){ return (2x + 4); } draw(graph(f,-8,5),red+linewidth(1)); draw(graph(g,-6,2),blue+linewidth(1)); [/asy] Let the acute angle the red line makes with the $x-$ axis be $\alpha$ and the acute angle the blue line makes with the $x-$ axis be $\beta$ . Then, we know that $\tan \alpha = 1$ and $\tan \beta = 2$ . Note that the acute angle between the red and blue lines is clearly $\beta - \alpha$ . Therefore, we have that: \[\tan (\beta - \alpha) = \frac{2 - 1}{1 + 2} = \frac{1}{3}\] It follows that $\cos (\beta - \alpha) = \frac{1}{\sqrt{10}}$ and $\sin (\beta - \alpha) = \frac{3}{\sqrt{10}}$ . From now on, refer to $\theta = \beta - \alpha$ [asy] import graph; size(4cm);  real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */  pen dotstyle = black; /* point style */  real xmin = -5.96, xmax = 24.22, ymin = -10.06, ymax = 14.62;  /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451);   draw(arc((2,2.6666666666666665),0.6,290.55604521958344,326.3099324740202)--(2,2.6666666666666665)--cycle, linewidth(1));   draw((0,8)--(0,0), linewidth(1) + wrwrwr);  draw((0,0)--(6,0), linewidth(1) + wrwrwr);  draw((6,0)--(0,8), linewidth(1) + wrwrwr);  draw((0,8)--(3,0), linewidth(1) + wrwrwr);  draw((0,4)--(6,0), linewidth(1) + wrwrwr);   dot((0,0),dotstyle);  label("$A$", (0.08,0.2), NE * labelscalefactor);  dot((6,0),dotstyle);  label("$B$", (6.08,0.2), NE * labelscalefactor);  dot((0,8),dotstyle);  label("$C$", (0.08,8.2), NE * labelscalefactor);  dot((3,0),linewidth(4pt) + dotstyle);  label("$D$", (3.08,0.16), NE * labelscalefactor);  dot((0,4),linewidth(4pt) + dotstyle);  label("$E$", (0.08,4.16), NE * labelscalefactor);  label("theta ", (2.14,2.3), NE * labelscalefactor);  clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);   [/asy] Suppose $ABC$ is our desired triangle. Let $E$ be the midpoint of $CA$ and $D$ be the midpoint of $AB$ such that $CE = EA = m$ and $AD = DB = n$ . Let $G$ be the centroid of the triangle (in other words, the intersection of $EB$ and $CD$ ). It follows that if $CG = 2x$ $GD = x$ and if $GB = 2y$ $GE = y$ . By the Law of Cosines on $\triangle GEB$ , we get that: \[GE^2 + GC^2 - 2GE \cdot GC \cdot \cos \theta = CE^2 \Longleftrightarrow\] \[(2x)^2 + y^2 - (2x)(y)(2)\cos \theta = m^2\] By the Law of Cosines on $\triangle GDB$ , we get that: \[DG^2 + GB^2 -2 DG \cdot GB \cdot \cos \theta = DB^2 \Longleftrightarrow\] \[(2y)^2+x^2-(2y)(x)(2)\cos \theta = n^2\] Adding yields that: \[5x^2+5y^2 - 8xy\cos \theta = m^2 + n^2\] However, note that $(2m)^2 + (2n)^2 = 60^2 \Longleftrightarrow m^2+n^2 = 30^2$ . Therefore, \[5x^2+5y^2 - 8xy\cos \theta = 30^2\] We also know that by the Law of Cosines on $\triangle CGB$ \[CG^2 + GB^2 - 2CG \cdot GB \cdot \cos (180 - \theta) = CB^2 \Longleftrightarrow\] \[(2x)^2 + (2y)^2 + 2(2x)(2y) \cos \theta = 60^2 \Longleftrightarrow\] \[x^2+y^2+2xy \cos \theta = 30^2 \Longleftrightarrow\] \[5x^2+5y^2 + 10xy \cos \theta = 30^2 \cdot 5\] Subtracting this from the $5x^2+5y^2 - 8xy\cos \theta = 30^2$ we got earlier yields that: \[xy \cos \theta = 200\] but recall that $\cos \theta = \frac{3}{\sqrt{10}}$ to get that: \[xy = \frac{200 \sqrt{10}}{3}\] Plugging this into the $x^2 + y^2 + 2xy \cos \theta = 30^2$ , we get that: \[x^2 + y^2 = 500\] Aha! How convenient. Recall that $(2x)^2 + y^2 - (2x)(y)(2)\cos \theta = m^2$ and $(2y)^2+x^2-(2y)(x)(2)\cos \theta = n^2$ . Then, we clearly have that: \[m^2n^2 = ((2x)^2 + y^2 - (2x)(y)(2)\cos \theta)((2y)^2+x^2-(2y)(x)(2)\cos \theta = n^2) \Longleftrightarrow\] \[m^2n^2 = 17x^2y^2 - 4xy \cos \theta (5x^2 + 5y^2) + 16x^2y^2 \cos ^2 \theta + 4x^4 + 4y^4 \Longleftrightarrow\] \[m^2n^2 = 17 \cdot \frac{200^2 \cdot 10}{9} - 800(5 \cdot 500) + 16 \cdot 200^2 + 4 \left( (x^2+y^2)^2-2x^2y^2\right) \Longleftrightarrow\] \[\frac{m^2n^2}{100^2} = \frac{680}{9} - 8 \cdot 5 \cdot 5 + 16 \cdot 2^2 + 4 \left( 5^2 - 2 \cdot \frac{2^2 \cdot 10}{9}\right) \Longleftrightarrow\] \[\frac{m^2n^2}{100^2} = 4\] But note that the area of our triangle is $2m \cdot 2n \cdot \frac {1}{2} = 2mn$ . As $mn = 200$ , we get a final answer of $\boxed{400}$

Answer:  400