Separating line

*Solution. * Suppose $P$ is any point in the interior of the triangle $ABC$ . The criteria of point $Q$ so that $A$ and $C$ are on the same side of $PQ$ is when the line $PQ$ does not intersect the segment $AC$ . Connect and extend the line $AP$ such that it intersects the side $BC$ at point $E$ . Also, connect and extend the line $CP$ such that it intersects the side $AC$ at point $F$ . Then the set of point $Q$ that satisfies the criteria is just the union of the two triangles $CPE$ and $APF$ . We denote by $[XYZ]$ the area of a triangle $XYZ$ .

We use a coordinate system so that $A=(0,1), B=(0,0), C=(1,0), D=(1,1)$ , and let $P=(x,y) \in \triangle ABC$ .

Then, notice that

$\frac{[CPE]}{[APC]}=\frac{PE}{AP}=\frac{x}{1-x},$ $\frac{[APF]}{[APC]}=\frac{PF}{CP}=\frac{y}{1-y},$ $[APC] = \frac{1-x-y}{2}$

we have $[CPE]+[APF] = \frac{1}{2}(1-x-y)(\frac{x}{1-x}+\frac{y}{1-y}) =:f(x,y).$

Therefore the probability that $P$ is in triangle $ABC$ and $A$ , $C$ lie on the same side of $PQ$ is $I:=\int\int_{(x,y)\in \triangle ABC} f(x,y)dxdy.$

By symmetry, the probability that $P$ lies in the triangle $ADC$ and that $A$ , $C$ lie on the same side of $PQ$ is also $I$ . So the final answer is $1-2I$ . Now we calculate $2I$ . Note that

$2f(x,y)=x+y-\frac{xy}{1-x} -\frac{xy}{1-y}$ $= 2x + y(1-\frac{x}{1-x}) - \frac{x}{1-y}$

So

$2I = \int_{0}^{1} dx \int_{0}^{1-x} [2x + y(1-\frac{x}{1-x}) - \frac{x}{1-y}]dy$ $= \int_{0}^{1} [2x(1-x) + (1-\frac{x}{1-x})\cdot \frac{(1-x)^2}{2} + x \ln (1-(1-x))]dx$ $= \int_{0}^{1} [ \frac{3x(1-x)}{2} + \frac{(1-x)^2}{2} + x\ln x] dx$ $= \frac{1}{2}\int_{0}^{1} [ -2x^2 +1 + (2x\ln x + x)]dx$ $= \frac{1}{2}\int_{x=0}^{1} d[-\frac{2}{3}x^3 + x + x^2\ln x]$ $= \frac{1}{2}(-\frac{2}{3} + 1 + 0) = \frac{1}{6}.$

Therefore $1-2I = \frac{5}{6}$ .

Consider the square in a Cartesian coordinate system. Without loss of generality, let points A, B, C, D be at (0,0), (1,0), (1,1) and (0,1) respectively.

Now consider the case when P lies in ABC. Join line AP to meet BC at D, and join line CP to meet AB at E. Then A and C will be on the same side as PQ if and only if Q lies within triangle BPD or triangle APE. If point P is at (X,Y), then we can find the equations of lines AP and CP to be $y = \frac{Y}{X}x$ and $y = \frac{1-Y}{1-X}x + 1 - \frac{1-Y}{1-X}$ respectively. This would allow us to find the coordinates of D and E to be $(1, \frac{Y}{X})$ and $(1 - \frac{1-X}{1-Y}, 0)$ respectively. The area of triangle APE will thus be $\frac{1}{2}(1 - \frac{1-X}{1-Y})(Y)$ while the area of triangle BPD is $\frac{1}{2}(1 - \frac{Y}{X})(1-X)$ . Since the total area of ABCD is 1, the probability that A and C are on the same side as Q is thus the sum of these two areas, which is $\frac{1}{2}(1-X-\frac{Y}{X}+2Y-\frac{Y(1-X)}{1-Y})$ . Since X varies from 0 to 1 and Y varies from 0 to X (given that P is in ABC), we have the probability that A and C are on the same side of PQ given that P is in ABC being $\dfrac{\int_0^1 \int_0^X \frac{1}{2}(1-X-\frac{Y}{X}+2Y-\frac{Y(1-X)}{1-Y}) \, dY \, dX}{\frac{1}{2}}$ . The denominator is $\frac{1}{2}$ because point P is only chosen from ABC, which has area $\frac{1}{2}$ . This fraction is equal to $\int_0^1 \int_0^X (1-X-\frac{Y}{X}+2Y-\frac{Y(1-X)}{1-Y}) \, dY \, dX$ .

Using simple integration for the first 4 terms in the integrand and integration by parts for the final term, we can evaluate the integral to be $\int_0^1 \frac{3}{2}X - X^2 + (1-X)ln(1-X) \, dX$ . Again, using simple integration for the first two terms and integration by parts for the final term in the integrand, we obtain $[-\frac{1}{3}X^3 + X^2 - \frac{1}{2}X - \frac{1}{2}(X^2+1)ln(1-X)]_0^1$ . Keeping in mind that $\displaystyle \lim_{X \rightarrow 1}[(X^2+1)ln(1-X)] = 0$ , we can evaluate this integral to be $\frac{1}{6}$ . This is the probability that A and C are on the same side of PQ given that P is in ABC.

If P is in ADC, it is easy to realise that we are faced with a similar situation as when P is in ABC, so the probability that A and C are on the same side of PQ given that P is in ADC is also $\frac{1}{6}$ . Thus, the probability that A and C are on the same side of PQ is $\frac{1}{6}$ , and the probability that A and C are on different sides of PQ is $\frac{5}{6}$ . Our desired answer is thus $5+6=11$ .

WLOG set ABCD has area 1.

We'll calculate the probability that A and C are on the same side of $L_1$ , and then subtract from 1 to find out the answer. We'll do this assuming that P lies in ABC, because the situation is symmetric

Suppose that P lies within triangle ABC. Say AP meets BC at M, and CP meets AB at N. So the event that A and C are on the same side of $L_1$ corresponds to the statement that Q lies within triangle PNA or PMC. Thus, it has probability equal to the area of those triangles.

Now, say that the perpendicular distances from P to BC and AB are x and y. So the area of PMC is $\frac{x}{2(1-x)} (1-x-y)$ . So, to work out the probability that Q lies within PMC, we must integrate $\frac{x}{(1-x)} (1-x-y)$ over the triangle, where we've multiplied by 2 because P is distributed uniformly on ABC with area $\frac{1}{2}$ .

We do the y integral from 0 to 1-x first to get $\frac{x(1-x)}{2}$ and then we do the x integral from 0 to 1 and it comes out to $\frac{1}{12}$ . Finally, we double this because Q could have been within PNA, to get final probability $\frac{1}{6}$ .

Finally, we subtract from 1 to get answer $\frac{5}{6}$ .

One randomly chooses one point $P(x,y)$ . Besides, let the length of one side be $1$ . Further $f$ is the line connecting $A$ and $P$ , $E$ is the intesection of $f$ with side $BC$ . $g$ is the line connecting $C$ and $P$ and $F$ the intersection of $g$ with side $CD$ .

Obviously, the second point $Q$ has to be in the area $APCD$ or $FCEP$ to fulfill the condition that $A$ and $C$ lie on different sides on the line $PQ$ .

One can easily calculate the sum of these areas: $A=1-\frac{(1-x)(x-y)}{2 x}-\frac{1}{2} y \left(1-\frac{1-x}{1-y}\right)$

Hence the probability is: $p=\int_0^1 \int_0^1 1-\frac{(1-x)(x-y)}{2 x}-\frac{1}{2} y \left(1-\frac{1-x}{1-y}\right) \mathrm{d} x\mathrm{d} y=\frac{5}{6}$

So, the answer is $5+6=11$ .

First, draw $AC$ . Without a loss of generality, let $P$ be in $\triangle ABC$ . Extend $AP$ to meet $BC$ at $E$ , and extend $CP$ to meet $AB$ at $F$ . It can be easily seen that the locus of points $Q$ that satisfy the conditions are the interiors of the quadrilaterals $ADCP$ and $BEPF$ . The ratio of the area of the locus to the area of the square is then the probability.

For now, we will disregard $\triangle ADC$ (since that is in common to all of the loci), and focus on $\triangle ABC$ .

Let the area of a locus at a point $P$ be denoted as $[L]_P$ . Draw the medians $AX$ , $BY$ , and $CZ$ of $\triangle ABC$ . We note that at the median $G$ of $\triangle ABC$ , $[L]= \frac{2}{3} [ABC]$ . Further, if $P$ is located in $\triangle CMY$ , $\triangle YMA$ , $\triangle MBX$ , or $\triangle MBZ$ (let these points be called $P_1$ ), $\frac{2}{3}[ABC]\le [L_{P_1}]\le [ABC]$ . However, if $P$ is located in $\triangle CMX$ or $\triangle AMZ$ (let these points be called $P_2$ ), we know that $0\le [L]\le \frac{2}{3}[ABC]$ .

Since $[L]$ is continuous throughout the triangle, we may map any point $P_2$ to a point $P_1$ and its reflection across $BY$ (we will call this $P_1^{\prime}$ ), such that $[L_{P_2}]+[L_{P_1}]+[L_{P_1^{\prime}}]=[L_{P_2}]+2[L_{P_1}]=2[ABC]\qquad (1)$ . Likewise, we may map any point $P_1$ to its reflection across $BY$ and a point $P_2$ , with the same condition as above.

Next, we note that $[CMY]+[YMA] +[MBX] +[MBZ]=2([CMX]+[AMZ])$ . Since we are mapping two points of $P_1$ to one point in $P_2$ , we can create a "bijection" between these sets that satisfy $(1)$ because the sets are of proportional areas. Thus, the expected value of $[L]$ for all points is $\frac{2}{3}[ABC]=\frac{1}{3}[ABCD]$ (only considering $\triangle ABC$ ).

The same is true if $P$ is located in $\triangle ADC$ , so the probability that $A$ and $C$ are on opposite sides of $PQ$ is $\frac{\frac{1}{3}[ABCD]+\frac{1}{2}[ABCD]}{[ABCD]}=\frac{5}{6}$ , where the $\frac{1}{2}[ABCD]$ comes from the triangle we ignored earlier. Therefore, the answer is $5+6=\boxed{11}$ .

We will assume that the square has unit side length. By symmetry, we may assume the point $P$ is contained in triangle $ABC$ . Let $X$ be the point where the line through $A$ and $P$ intersects $BC$ , and $Y$ the point where the line through $C$ and $P$ intersects $AB$ . We see that when $Q$ is in $PXC$ or $PYA$ that the line through $P$ and $Q$ has $A$ and $C$ on the same side of it, and when $Q$ is in $BXPY$ or $PCDA$ , the line through $P$ and $Q$ has $A$ and $C$ on opposite sides.

The desired answer follows from the following two lemmas (the proofs are left as exercises to the reader):

Lemma 1. Let $S$ be a point chosen at random from triangle $TUV$ . Then the expected value $E([STU]) = \frac{[TUV]}{3}$ .

Lemma 2. Let $S$ be a point chosen at random from triangle $TUV$ and $W$ the point where the line through $S$ and $T$ intersects $UV$ .
Then the expected value $E([TWU])= \frac{[TUV]}{2}$ .

From Lemma 1, we see that $E([APC]) = \frac{[ABC]}{3} = \frac{1}{6}$ . From Lemma 2, we see that $E([AXC]) = \frac{[ABC]}{2} = \frac{1}{4}$ . Combining these, and using Linearity of expectation, we have that $E([PXC]) = \frac{1}{12}$ . By symmetry, we also get that $E([PYA]) = \frac{1}{12}$ . Thus, the probability that the line does not separate $A$ and $C$ is $\frac{1}{6}$ , so the probability that it does is $\frac{5}{6}$ , and so $a + b = 5 + 6 = 11$ .