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The problem is as follows:

In figure 1. there is a circle as shown. The radius is equal to 10 inches and its center is labeled with the letter O. If $\measuredangle PC=30^{\circ}$. $\textrm{Find AB+BC}$.

Diagram of the problem

The existing alternatives in my book are:

  • $3\left( \sqrt{2}+\sqrt{6}\right)$
  • $4\left( \sqrt{6}-\sqrt{2}\right)$
  • $5\left( \sqrt{3}-\sqrt{2}\right)$
  • $5\left( \sqrt{3}+\sqrt{2}\right)$
  • $5\left( \sqrt{2}+\sqrt{6}\right)$

After analyzing the drawing the figure from below shows all all the relationships which I could found and it is summarized as follows:

Diagram of the solution

The triangle $\textrm{COP}$ is isosceles since it shares the same side from the radius of the circle and since $\measuredangle PC=30^{\circ}$, then all is left to do is to apply the identity which it says that the sum of inner angles in a triangle must equate to $180^{\circ}$.

$$2x+30^{\circ}=180^{\circ}$$ $$x=\frac{150^{\circ}}{2}=75^{\circ}$$

Since $\measuredangle OCP = \measuredangle OPC$, its supplementary angle would become:

$$180^{\circ}-75^{\circ}=105^{\circ}$$

Since it is given from the problem:

$$\measuredangle COA = 90^{\circ}$$

therefore its complementary angle with $\measuredangle COP = 30^{\circ}$ would become into:

$$\measuredangle POA = 60^{\circ}$$

Since $PO = OA$ this would also make another isosceles triangle and by recurring to the previous identity:

$$2x+60^{\circ}=180^{\circ}$$ $$x=\frac{180^{\circ}-60^{\circ}}{2}=60^{\circ}$$

Therefore the triangle POA is an equilateral one so,

$$\textrm{PA=10 inches}$$

As $\measuredangle OPB = 105 ^{\circ}$ and $\measuredangle OPA = 60^{\circ}$ then its difference is: $\measuredangle APB = 45^{\circ}$.

From this its easy to note that $\measuredangle PAB = 45^{\circ}$.

Since the vertex $\textrm{B}$ of the triangle $\textrm{ABP}$ is $\measuredangle = 90 ^{\circ}$. I did identified a special right triangle with the form $45^{\circ}-45^{\circ}-90^{\circ}$ or $\textrm{k, k,}\,k\sqrt{2}$.

By equating the newly found side $\textrm{PA = 10 inches}$ to $k\sqrt{2}$ this is transformed into:

$$k\sqrt{2} = 10$$

$$k = \frac{10}{\sqrt{2}}$$

From this is established that:

$$AB = \frac{10}{\sqrt{2}}$$

Since we have $\textrm{AB}$ we also know $\textrm{PB}$ as $AB = PB = \frac{10}{\sqrt{2}}$

Therefore all that is left to do is to find $\textrm{CP}$ as $CP+PB = BC$

To find $CP$ I used cosines law as follows:

$$a^{2}=b^{2}+c^{2}-2bc\,\cos A$$

Being a, b and c the sides of a triangle ABC and A the opposing angle from the side taken as a reference in the left side of the equation.

In this case

$$(CP)^{2}= 10^{2}+10^{2}-2(10)(10)\cos30^{\circ}$$ $$(CP)^{2}= 10^{2} \left(1+1-2\left(\frac{\sqrt{3}}{2}\right)\right)$$ $$CP = 10 \sqrt{ 2-\sqrt{3}}$$

Therefore $CP = 10 \sqrt{ 2-\sqrt{3}}$ and we have all the parts so the rest is just adding them up.

$$CP+PB= BC = 10 \sqrt{ 2-\sqrt{3}} + \frac{10}{\sqrt{2}}$$

$$AB= \frac{10}{\sqrt{2}}$$

$$AB + BC = \frac{10}{\sqrt{2}} + 10 \sqrt{ 2-\sqrt{3}} + \frac{10}{\sqrt{2}}$$

And that's how far I went, but from then on I don't know if what I did was correct or did I missed something? as my answer doesn't appear within the alternatives.

The best I could come up with by simplifying was:

$$\frac{10\sqrt{2}}{2}+10\sqrt{2-\sqrt{3}}+\frac{10\sqrt{2}}{2}$$

$$10\sqrt{2}+10\sqrt{2-\sqrt{3}}$$

$$10\left(\sqrt{2}+\sqrt{2-\sqrt{3}}\right)$$

and, that's it. But it doesn't seem to be in the choices given. Can somebody help me to find if did I do something wrong?. If a drawing is necessary please include one as I'm not savvy enough to notice these things easily.

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    $\begingroup$ If $∡PC=30º$ then I think $\angle AOP=60º$ since the angle to the center is twice as the angle to the side. $\endgroup$ – abc... Sep 13 '18 at 4:41
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Notice that $$\angle ACB=\angle OCB-\angle OCA=75^o-45^o=30^o,$$and $$AC=10\sqrt{2}.$$Thus $$AB+BC=AC\cdot(\sin \angle ACB+\cos \angle ACB)=5(\sqrt{2}+\sqrt{6}).$$

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Assuming your work is correct so far, I think we may be able to simplify further.

$$\sqrt{2-\sqrt3}=\sqrt{\frac{4-2\sqrt3}{2}}=\frac{\sqrt{\sqrt3^2-2(1)\sqrt3+1^2}}{\sqrt2}=\frac{\sqrt{(\sqrt3-1)^2}}{\sqrt2}=\frac{\sqrt3-1}{\sqrt2}$$.

Now let's see if that helps.

$$10\left(\sqrt2+\sqrt{2-\sqrt3}\right)=5\sqrt2\left(2+\sqrt3-1\right)=5(\sqrt2+\sqrt6)$$

Again, assuming everything you've done is correct, the last answer is the solution.

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  • $\begingroup$ I'm still stuck at why $\sqrt{\frac{{4-2\sqrt{3}}}{2}}=\frac{\sqrt{3}-1}{\sqrt{2}}$ ? Perhaps can you expand with steps how this reverse rationalization works?. Because I tried all the methods I know and I am still unable to reach what you have found. Does it exist an identity am I unaware of?. $\endgroup$ – Chris Steinbeck Bell Sep 14 '18 at 3:48
  • $\begingroup$ @ChrisSteinbeckBell The numerator is simply $\sqrt3^2-2(1)\sqrt3+1^2=(\sqrt3-1)^2$. $\endgroup$ – Mike Sep 14 '18 at 4:37
  • $\begingroup$ Thanks! but I voice my opinion that what have you just added in the comments should be added as part of the answer since, well maybe is not too obvious. Other than that is great to know that I was on the right track. $\endgroup$ – Chris Steinbeck Bell Sep 14 '18 at 4:51
  • $\begingroup$ @ChrisSteinbeckBell All right, I've added the requested intermediate steps. $\endgroup$ – Mike Sep 14 '18 at 5:20
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I have checked your work and it is correct.

Your answer is correct and $$10\left(\sqrt2+\sqrt{2-\sqrt3} \right)=5\left( \sqrt{2}+\sqrt{6}\right)=19.31851653$$

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