I would appreciate if somebody could help me with the following problem
Q: if $\overline{OQ}\times\overline{OP}=r^2 $ then $\angle OAP=\frac{\pi}{2}$
($r$: radius of $C$, $C$: circle, $O$: center $C$)
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You would want to use the Pythagorean theorem as well as the assumption in the proposition. – user60887 Jun 26 '13 at 05:55
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The Pythagorean theorem is not necessary. Proportionality of sides of similar triangles is enough to derive the relation in virtually a single step. – David H Jun 26 '13 at 06:05
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Oops I was thinking in hyperbolic geometry. – user60887 Jun 26 '13 at 18:02
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$AP$ is tangent to circle $C, \angle OAP$ is right angle. It can be proved by similar triangles $ OAQ, OPA$ – Narasimham Feb 20 '18 at 04:06
3 Answers
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Hint: Show that triangle $OQA$ and $OOAP$ are similar by SAS.
Hence $\angle OAP = \angle OQA = \frac{\pi}{2}$.
Calvin Lin
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Hint: Show that $\Delta AOQ \sim \Delta POA$ by observing that: $$ \overline{OQ}\times\overline{OP}=r^2 \iff \dfrac{r}{\overline{OP}} = \dfrac{\overline{OQ}}{r} $$
Adriano
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If the usage of Law of Sine is allowed,
let $\angle OAQ=\beta, \angle OPA=\angle QPA=\alpha$
$\frac{OQ}{ \sin\beta}=\frac r1, \frac{OP}{\sin(\frac\pi2+\beta-\alpha)}=\frac r{\sin\alpha}$
As $OQ\cdot OP=r^2, \sin\alpha=\sin\beta\cdot \sin(\frac\pi2+\beta-\alpha)=\sin\beta\cdot \cos(\beta-\alpha)$
$$\implies 2\sin\alpha=2\sin\beta\cdot \cos(\beta-\alpha)=\sin(2\beta-\alpha)+\sin\alpha$$
$$\implies \sin\alpha=\sin(2\beta-\alpha)$$
$$\implies \alpha=2\beta-\alpha\text{ as } \frac\pi2+\beta-\alpha<\pi\iff \beta-\alpha<\frac\pi2$$
$$\implies \alpha=\beta$$
$$\implies \angle OAP=\angle OAQ+\angle QAP=\beta+\frac\pi2-\alpha=\frac\pi2$$
lab bhattacharjee
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