Kathleen's triangles

I present a solution to this problem using the Cartesian plane. Let the rectangle $ABCD$ be situated in the following manner: $\begin{aligned} A &= (0,-6) \\ B&= (0,0) \\ C&= (8,0) \\ D&=(8,-6)\end{aligned}$ First we locate the coordinates of the reflection of the point $C$ across the diagonal $BD$ . By construction, the diagonal lies on the line $y = -\frac{3}{4}x$ . Given $(x,y)$ and a line $y = ax + c$ , the point $(x', y')$ reflected across said line is given by $\begin{aligned} x' &= 2d - x \\ y' &= 2da - y + 2c \end{aligned}$ where $d= \frac{x + a(y - c)}{1 + a^2}$ . For a complete explanation of this formula, see this stackoverflow post . Thus the coordinates of the reflection $C'$ are $\frac{1}{25}(56,192)$ . Next we locate the point $P$ by solving $-6 = -\frac{24}{7}x$ which gives us $P = (\frac{7}{4}, -6)$ (recall that the line $y=-6$ is the bottom of the rectangle).

The area of triangle $BAP = \frac{1}{2}\times\frac{7}{4}\times6 = \frac{21}{4}$ . The area of the triangle $BAD = 24$ , hence the area of the triangle $BPD = 24 - \frac{21}{4} = \frac{75}{4}$ . Therefore, the ratio of the area of triangle $BAP$ to the area of triangle $BPD$ is

$\frac{21}{75} = \frac{7}{25}$

and their sum is $\boxed{32}$ .

Use of trigonometry kills the problem. Let AB=a, BC=b. Let b>a. angle PAB= angle ABD - angle DBC take tan of the angles on both sides, PA/a=(b/a-a/b)/(1+(a/b*b/a))

PA/a=(b^2-a^2)/2ab

PA=(b^2-a^2)/2b

PD=AD-PA=b-(b^2-a^2)/2b=(a^2+b^2)/2b

Required ratio of areas is PA/PD=(b^2-a^2)/(b^2+a^2)

Plot points $A(0,6) , B(0,0), C(8,0)$ and $D(8,6)$ on the Cartesian plane so that we have rectangle $ABCD$ where $AB=6$ and $BC=8$ . The line that passes through $B$ and $D$ has the equation $y=(3/4)x$ and so if $C'$ is a point obtained by reflecting $C$ across $BD$ , the line that passes through $C$ and $C'$ is perpendicular to $y=(3/4)x$ . Thus, this has the equation, $y=(-4/3)x + 32/3$ because it has slope that is negative reciprocal of the slope of the first line we have. Also, we can get the distance between point $C$ and the line $y=3x/4$ using the formula, $D=|(Ax+By+C)/(A^2+B^2)|$ which gives $D=3.84$ . This is also the distance between $C'$ and $y=3x/4$ . Therefore, the line that passes through $C'$ that is parallel to $y=3x/4$ has the equation $y=3x/4 + 6$ . This is because we have that $3.84=|(b-b')/(A^2+B^2)|$ where $b$ and $b'$ are the $y-intercepts$ of the two parallel lines. This gives $b'=6$ . So we have two lines which contains $C'$ . $y=-4x/3+32/3$ and $y=3x/4+6$ . Solving the system, we have that point $C'$ is at $(2.24,7.68)$ . The line through $B$ and $C'$ has the equation $y=7.68x/2.24$ . When $y=6$ , $x=1.75$ . Thus, we have $AP=1.75$ and $PD=6.25$ . The ratio $AP/PD$ gives $7/25$ . Thus, the answer is $32$ .

Draw a line parallel to $AD$ through $C'$ . Extend $BA$ and $CD$ to intersect this line at $F$ and $E$ respectively . Therefore , $AF = DE$ . Let $CC'$ intersect $BD$ at $O$ . By Pythagoras theorem , $BC^2 + CD^2$ = $BD^2$

$\Rightarrow$ $BD^2$ = $\sqrt{BC^2 + CD^2}$

$\Rightarrow$ $BD$ = $\sqrt{8^2 + 6^2} = 10$ units . Let

$BO$ = $x , OD$ = $10 - x$ and $OC$ = $y$ . Applying Pythagoras theorem for $\bigtriangleup{OBC}$ ,

$OB^2$ + $OC^2$ = $BC^2$ $\Rightarrow$ $x^2 + y^2 = 64$ $\rightarrow eq.(1)$ .

Similarly , for $\bigtriangleup{OCD}$ , $OC^2$ + $OD^2$ = $CD^2$

$\Rightarrow$ $(10 - x)^2 + y^2 = 36$ $\Rightarrow$ $x^2 - 20x + 100 + y^2 = 36$ $\rightarrow eq.(2)$ . Substituting $eq.(1)$ in $eq.(2)$ and solving we get , $x = \frac{32}{5} = 6.4$ units . Substituting this in $eq.(1)$ ,we get $y = \frac{24}{5} = 4.8$ units . Hence , $OC$ = $4.8$ units and as $OC'$ is the reflection of $OC$ in $BD$ , $CC' = 2OC = 9.6$ units . Let $\theta$ be $\langle{OCD}\rangle$ . For $\bigtriangleup{OCD}$ , $\sin\theta$ = $\frac{3.6}{6}$ . Also , for $\bigtriangleup{CC'E}$ , $\sin\theta$ = $\frac{C'E}{CC'}$ . Thus , $\frac{C'E}{CC'} = \frac{3.6}{6} \Rightarrow C'E = 5.76$ units . Therefore , $FC' = 8 - 5.76 = 2.24$ units . As $OC = OC' , \langle{BOC}\rangle = \langle{BOC'}\rangle$ and $OB$ is common , hence $\bigtriangleup{BOC} \cong \bigtriangleup{BOC'}$ . Therefore , $BC = BC' = 8$ units . Let $AF$ = $z$ . Applying Pythagoras theorem for $\bigtriangleup{BFC'}$ , $BF^2 + FC'^2 = BC^2 \Rightarrow (6 + z)^2 + (2.24)^2 = 64$

$\Rightarrow 6 + z = 7.68 \Rightarrow z = 1.68$ units . As $\bigtriangleup{BAP} \sim \bigtriangleup{BFC'}$ , thus , $\frac{AP}{FC'} = \frac{BA}{BF} \Rightarrow \frac{AP}{2.24} = \frac{6}{7.68} \Rightarrow AP = 1.75$ units . Thus , $PD = 8 - 1.75 = 6.25$ units . Thus , $\frac{\bigtriangleup{BAP}}{\bigtriangleup{BPD}} = \frac{\frac{1}{2} \times 1.75 \times 6}{\frac{1}{2} \times 6.25 \times 6} = \frac{7}{25} = \frac{a}{b}$ . Hence , $a + b = 32$