Y by Ziyadel, Yrock, mygoodfriendusesaops, idksomething
Points 
and 
are chosen uniformly and independently at random on sides 
and 
respectively, of equilateral triangle 
Which of the following intervals contains the probability that the area of 
is less than half the area of 
![$\textbf{(A) } \left[\frac 38, \frac 12\right] \qquad \textbf{(B) } \left(\frac 12, \frac 23\right] \qquad \textbf{(C) } \left(\frac 23, \frac 34\right] \qquad \textbf{(D) } \left(\frac 34, \frac 78\right] \qquad \textbf{(E) } \left(\frac 78, 1\right]$](//latex.artofproblemsolving.com/5/8/7/5874157b23f9d385f707b696eee3d0780fd8a45b.png)
and 
are chosen uniformly and independently at random on sides 
and 
respectively, of equilateral triangle 
Which of the following intervals contains the probability that the area of 
is less than half the area of 
![$\textbf{(A) } \left[\frac 38, \frac 12\right] \qquad \textbf{(B) } \left(\frac 12, \frac 23\right] \qquad \textbf{(C) } \left(\frac 23, \frac 34\right] \qquad \textbf{(D) } \left(\frac 34, \frac 78\right] \qquad \textbf{(E) } \left(\frac 78, 1\right]$](http://latex.artofproblemsolving.com/5/8/7/5874157b23f9d385f707b696eee3d0780fd8a45b.png)
;
over some interval, but it's not necessary.)
on your scratch paper, you can see that it is very below the line 
and very clearly takes up more than 
of the square
with 
and 
in ![$[0, 1]$](http://latex.artofproblemsolving.com/a/b/1/ab178d831a786b92cb4c9ddc2d33578223036f98.png)
and estimate by cutting of triangle with vertices 
,
and 
.
and 
.
.
square with a hyperbola connecting 
and 
cut out. This can be bounded between a straight line, which would give 
,
.
.![\[\int_0^1 \min(1,\frac{1}{2x}) \text dx = \int_0^{1/2} 1 \text dx + \int_{1/2}^1 \frac{1}{2x} \text dx = \frac{\log 2 + 1}{2}.\]](http://latex.artofproblemsolving.com/9/2/2/922ab27e9e04d8a10f4b05ed1077046f6c1501ba.png)
We can approximate ![\[\log 2 = \frac{\log_{10} 2}{\log_{10} e} \approx \frac{0.3}{0.434},\]](http://latex.artofproblemsolving.com/4/4/9/44977cf30a952cf0b6ae4f088eaa02a820ec5f31.png)
which is comfortably between 
and 
.
![$\frac{[APQ]}{[ABC]} = \frac{AP}{AB} \cdot \frac{AQ}{AC} $](http://latex.artofproblemsolving.com/e/2/a/e2ac06a62af06ef2c9cb4336ffc99c9504d67cb8.png)
,
.![$\left(\frac{3}{4}, \frac{7}{8}\right].$](http://latex.artofproblemsolving.com/5/4/2/5424c456fae91fafbce63d458fdcdd4297ac2506.png)
and 
are both points on the graph. We can see that the curve will connect those points. Check 
,
and 
.
!
from ap chem.
where we used 
.
.
on the origin. Line 
has slope 
,
,
.![$[APQ] = pq\frac{\sqrt{3}}{8} \leq \frac{1}{2} [ABC]$](http://latex.artofproblemsolving.com/c/4/d/c4de97f97f6d7dd4bb5d019b3cdaaac37e44c6c5.png)
.
.
which is less than 
,
.
and let 
.
given that 
.
that is within a square box with vertices 
and 
.![$[0, \frac {1}{2}]$](http://latex.artofproblemsolving.com/b/3/0/b30ba49bbbe7ee623eb592d0e98dae9d7419cd5c.png)
,![$[\frac {1}{2}, 1]$](http://latex.artofproblemsolving.com/f/4/f/f4ff2abb48424fa210dd7456057482d121a334ff.png)
,
