Was I on the right track with this circle-tangent geometry problem solution? How are you supposed to solve it?

Question

Given concentric circles with center O and a point P outside the circles, draw segment PO and also draw lines through P tangent to the circles at S and T, as shown. If OS=10, PT bisects angle ∠OPS, and area(△TOP)=35, then determine distance OP.

I have managed to create the system of equations: $$5(OP)^2 - OT\cdot(OP)^2 + 5(OT)^2 = 0$$ $$(OT)^2((OP)^2 - (OT)^2) = 4900$$

I tried plugging these into Wolfram alpha to see if they yield the correct solution and they do, but then I tried to solve it. And I managed to figure out that $$(OP)^2 = \frac{490 + \sqrt{(OT)^6 + 240100}}{OT}$$.

I am completely and utterly stuck. Was I on the right track or did I get sucked down a wormhole? How are you supposed to solve this problem

Answer 1

Trigonometric solution. Let $OT/OP=\sin(\alpha)$ with $\alpha=\angle OPT$, then $$\begin{cases} \frac{1}{2}OT\cdot OP\cos(\alpha)=35\\ OP\sin(2\alpha)=10 \end{cases}$$ Since $\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$ it follows that $$\begin{cases} OT\cdot OP\cos(\alpha)=70\\ OT\cos(\alpha)=5 \end{cases}$$ and after dividing the first equation by the second one we get $$OP=\frac{70}{5}=14.$$

Answer 2

Here is a way to arrive at the answer without using trigonometry:

Note that $\angle OTP$ and $\angle OSP$ are both right angles on the same segment $OP$,
so if you drew a circle $\Gamma$ with diameter $OP$, then the points $S,T$ would lie on that circle (since angle subtended by a semicircle is $90^\circ$).

Construction: Take $M$ to be the midpoint of $OP$, join $MT$ and let $MT\cap OS=X$. Draw the circle $\Gamma$ with diameter $OP$. (this is not necessary though)

So, $M$ is the center of $\Gamma$ and $MT$ is a radius of circle $\Gamma$.

$$\text{angle at the center } \angle TMO = 2\angle TPO, \text{ by Inscribed Angle Theorem} \\ \implies \angle TMO = \angle SPO \ (\because PT \text{ bisects } \angle SPO) \\ \implies MT \parallel SP \\ \implies MT \perp OS \ (\because SP \perp OS) \qquad \qquad (1) $$

$M$ being the center of $\Gamma$, a perpendicular from $M$ to a chord $OS$ of $\Gamma$ bisects $OS$,
so $OX=\dfrac12\cdot OS = 5 \qquad \qquad \qquad \qquad (2)$

Again, $TM$ is a median in $\triangle TOP$, so it divides the area of the triangle into half, so we must have $$\text{area}[\triangle MTO]=\dfrac{\text{area}[\triangle TOP]}2 \\ \implies \dfrac12\cdot MT \cdot OX = \dfrac{35}2 \ (\text{from }(1)) \\ \implies MT=7\ (\text{using }(2))$$

so the radius of the circle $\Gamma$ is $7$, and it's diameter $OP=14$.