Area Problem 4

hint1: $\Delta APD \sim \Delta BPC \implies \frac{DP}{CP} = \frac{AP}{BP} = \sqrt{\frac{27}{12}} = \frac{3}{2}$

hint2: $\Delta ACQ \sim \Delta BDQ \implies \frac{AC}{BD} = \sqrt{\frac{27+37}{12+37}} = \frac{8}{7}$

hint3: $AP:BP:CP:DP = 6:4:2:3$

hint4: $\frac{S_{\Delta APD}}{S_{\Delta APB}} = \frac{DP}{BP} \quad \text{or} \quad \frac{S_{\Delta APB}}{S_{\Delta BPC}} = \frac{AP}{CP}$

1x6x6 :36 since, sin90 is 1.

$\triangle APD \sim \triangle BPC$

$\rightarrow \frac{\triangle APD}{\triangle BPC}$ = $\frac{27}{12}$ = $\frac{9}{4}$

$\rightarrow \frac{AP}{BP}$ = $\frac{DP}{CP}$ = $\frac{AD}{BC}$ = $\frac{3}{2}$

$\triangle ACQ \sim \triangle BDQ$

$\rightarrow \frac{\triangle ACQ}{\triangle BDQ }$ = $\frac{27+37}{12+37}$ = $\frac{64}{49}$

$\rightarrow \frac{AC}{BD}$ = $\frac{AQ}{BQ}$ = $\frac{CQ}{DQ}$ = $\frac{8}{7}$

Then we can find AD:DQ:CQ = 9:7:8 and AP:DP = 2:1

Also, $\triangle APB \sim \triangle DPC$

$\rightarrow \frac{\triangle APB}{\triangle DPC} = 4$

Let AQ = 9a, then DQ = 7a, CQ = 8a

$\triangle ACQ = \frac{1}{2}(7a+9a)(8a) \sin \angle Q = 27 + 37$

$\rightarrow a^2 \sin \angle Q = 1$

$\triangle CDQ = \frac{1}{2}(7a)(8a) \sin \angle Q = 28$

$\triangle DPC = 37-28 = 9$

$\triangle APB = 4 \triangle DPC = 36$

Connect $PQ$ .

$a+b=37$

We have $x:12=(27+a):b$ and $x:27=(12+b):a$

$x:108=(27+a):9b=(12+b):4a$

Using $108a+4a^2=108b+9b^2$ and $a+b=37$ ,

we get $a=21,b=16$ .

Putting back and get $x=36$

Angles subtended by the same arc at the circumference are congruent, so $\angle ADB = \angle ACB$ and $\angle DAC = \angle DBC$ . This makes $\triangle APD \sim \triangle BPC$ by AA similarity, and since their areas are $27$ and $12$ respectively, the ratio of their sides is $\sqrt{\frac{27}{12}} = \frac{3}{2}$ , so $AP = \frac{3}{2}BP$ and $DP = \frac{3}{2}CP$ .

Also, $\triangle AQC \sim \triangle BPC$ by AA similarity, and since their areas are $27 + 37 = 64$ and $12 + 37 = 49$ respectively, the ratio of their sides is $\sqrt{\frac{64}{49}} = \frac{8}{7}$ , so $AC = \frac{8}{7}BD$ .

Since $AC = AP + CP$ and $BD = DP + BP$ , $AC = \frac{8}{7}BD$ becomes $AP + CP = \frac{8}{7}(DP + BP)$ or $\frac{3}{2}BP + CP = \frac{8}{7}(\frac{3}{2}CP + BP)$ , which simplifies to $BP = 2CP$ . Since $AP = \frac{3}{2}BP$ , $AP = 3CP$ .

The area of $\triangle BPC$ is $S_{\triangle BPC} = \frac{1}{2} \cdot CP \cdot BP \cdot \sin \angle BPC = 12$ , and since $BP = 2CP$ , this means $CP^2 \sin \angle BPC = 12$ .

The area of $\triangle APB$ is $S_{\triangle APB} = \frac{1}{2} \cdot BP \cdot AP \cdot \sin (180° - \angle BPC)$ , and since $BP = 2CP$ and $AP = 3CP$ and $\sin (180° - x) = \sin x$ , $S_{\triangle APB} = 3 \cdot CP^2 \cdot \sin \angle BPC$ , and since $CP^2 \sin \angle BPC = 12$ , $S_{\triangle APB} = 3 \cdot 12 = \boxed{36}$ .

Let us denote Area APD = x, Area APB = y, Area BPC = z and Area Tr. ABQ = L.Then without using the fact that A,B,C,D are cyclic, we can use the Ladder Theorem which say that: $\frac{1}{L}$ + $\frac{1}{y}$ = $\frac{1}{x+y}$ + $\frac{1}{y+z}$ In the present case, x=27, z=12 and L= 39+37+y and thus, $\frac{1}{76+y}$ + $\frac{1}{y}$ = $\frac{1}{27+y}$ + $\frac{1}{y+12}$ which yields: y=36 as the only admissible value. Actually I used Menelaus's Theorem and arrived at: Area (CPDQ) = $\frac{xz(x+2y+z)}{(y²-zx)}$ which also yields y=36 and thus the two results should be equivalent. In addition, we can say that unless y²>zx, such a configuration is impossible.