Find the segment $BQ$ in the figure below

Question

For reference: Let a circle $\omega$ (not labelled in the graph) centered at $P$ tangent to $AB$, and $T$ is point of tangency. $\angle APB=90^\circ$. Let $K$ (not labelled in the graph) be some point on the circle $\omega$, the semicircle with diameter $BK$ intersects $PB$ at $Q$. Let $R$ be the radius of that semi-circle. If $4R^2-AT\cdot TB=10$ and $PQ=\sqrt2$, calculate $BQ$.(Answer:$2\sqrt3$)

My progress: $PT \perp AB\\ \triangle PTB:BT^2+PT^2 = PB^2\\ PT^2+AT^2=PA^2\\ PB=PQ+QB$ ???....

Answer 1

Since $Q$ be the intersection of semi-circle and $BP$, we have $KQ\perp BP$. So, we have $KP^2-BK^2=(KP^2-QK^2)-(BK^2-QK^2)=PQ^2-BQ^2$. Notice that $\angle APB=90^\circ$ and $TP\perp AB$, we have $KP^2=KT^2=AT\times TB$, and $BK^2=4R^2$. So we have $BQ^2=PQ^2+(BK^2-KP^2)=PQ^2+(4R^2-AT\times TB)=2+10=12$, thus $BQ=2\sqrt 3$.

Answer 2

I found JetfiRex’s answer a little dense and hard to follow, so here’s how I approached it.

Add segments $\mathrm{PK}$, $\mathrm{PT}$, and $\mathrm{KQ}$ to the initial diagram, like so:

($\mathrm{PT} \perp \mathrm{AB}$ since they are a radius of and tangent to $\omega$, respectively, and $\mathrm{KQ} \perp \mathrm{PB}$ by Thales’ theorem.)

Also assign letters to the segment lengths as shown, for brevity. We wish to find length $x$, and we can express the given relationship as $4R^2 - pq = 10$, or better yet: $$ \begin{align} (2R)^2 &= pq + 10 \\ \therefore pq &= (2R)^2 - 10 \tag 1 \end{align} $$

From $\triangle \mathrm{BKQ}$, we have $(2R)^2 = x^2 + y^2$, which we can substitute into (1) to obtain: $$ pq = x^2 + y^2 - 10 \tag 2 $$

From $\triangle \mathrm{KPQ}$ and $\triangle \mathrm{APT}$, we have: $$ \begin{align} r^2 &= y^2 + 2 \tag 3 \\ &= n^2 - p^2 \\ \therefore p^2 &= n^2 - y^2 - 2 \tag 4 \end{align} $$

And from $\triangle \mathrm{BPT}$ and $\triangle \mathrm{ABP}$, we have: $$ \require{cancel} \begin{align} (x + \sqrt{2})^2 &= q^2 + r^2 \\ &= (p + q)^2 - n^2 \\ \therefore n^2 + \cancel{q^2} + r^2 &= (p + q)^2 \\ &= p^2 + 2pq + \cancel{q^2} \\ \therefore n^2 + r^2 &= p^2 + 2pq \end{align} $$

Substituting all of (2), (3) and (4) into this gives: $$ \require{cancel} \begin{align} \cancel{n^2} + \cancel{y^2} + 2 &= n^2 - y^2 - 2 + 2(x^2 + y^2 - 10) \\ &= n^2 - y^2 - 2 + 2x^2 + 2y^2 - 20 \\ &= \cancel{n^2} + 2x^2 + \cancel{y^2} - 22 \\ \therefore 2 &= 2x^2 - 22 \\ \therefore 2x^2 &= 24 \\ \therefore x \phantom{^2} &= \sqrt{12} = 2\sqrt{3} \end{align} $$

Answer 3

Another solution:

$4R^2=10+AT.TB(I)\\ \triangle APB: PT^2=AT.TB(II)\\ \triangle PDQ: PD^2 = PQ^2+QD^2=(\sqrt2)^2+QD^2=2+QD^2 (III)\\ \triangle QBD: BD^2 = BQ^2+QD^2 \implies 4R^2 - BQ^2=QD^2(IV)\\ (II)em(I): 4R^2 = 10+\underbrace{PT}_{=PD}^2=10+PD^2=10+2+QD^2(V)\\ (III)em(V): 4R^2 = 12+4R^2-BQ^2\implies \boxed{BQ= \sqrt12 = 2\sqrt3}$