How to Draw a Circle Around a Triangle

(A new question of the week)

Final week we looked at a question about a triangle inscribed in a semicircle. Not long later that question, the same student, Kurisada, asked a question well-nigh triangle inscribed in a circumvolve, which had some connections to the other. As we enjoy doing, we led the student through several possible approaches to a solution. Information technology too illustrates a situation where different methods can atomic number 82 to what announced to be entirely different answers, yet they may be identical.

Finding the sides of a triangle in a circle

Here is the new problem, from the very end of last Dec:

A circle O is circumscribed effectually a triangle ABC, and its radius is r. The angles of the triangle are CAB = a, ABC = b, BCA = c. When a = 75°, b = 60°, c = 45° and r = 1, the length of sides AB, BC, and CA are calculated as ____, ____, ____ without using trigonometric functions.

Here is a moving picture showing all the data we accept:

Using trigonometry, we could find the sides if we knew one of them; but the only length we accept is the circumradius (the radius of the circumscribed circle). With no formula for this radius, and no trigonometry, how are nosotros to do this?

Since all we were given was the problem, Doctor Rick responded with just a hint, and the usual request to see work:

Hello, Kurisada. This is some other interesting trouble!

Equally a start, I propose constructing the radii OA, OB, and OC, and determining the interior angles of the triangles AOB, BOC, and COA. I tin think of several ways to practise this.

Many of the angles you will now find in these three triangles will be familiar angles that you know how to work with. The most challenging may bring to mind 1 of the problems we have discussed with y'all before. See what you can exercise at present.

Cartoon in the radii, as I already did above, is a standard first step, as they must be involved in the solution. Physician Rick'south work, equally suggested, involved a triangle similar to ane from last week'due south problem, but that is not the simply way.

Kurisada replied:

I found that AOB is ninety° and thus, AB is √two.

I've also found another angle just I wasn't able to notice AC and BC without using trigonometry ratio.

Triangle AOC has the angles 120°, 30°, and 30°.

And triangle BOC has the angles 150°, 15°, and 15°.

Did I miss something?

Here is what nosotros have now:

Equally Physician Rick said, at that place are several ways to take found these angles; one is to employ the fact that a primal bending is twice the inscribed angle, so that for instance ∠AOB = ii∠ACB = 90°. Since the triangle is isosceles, the other angles are both 45°.

Physician Rick replied, having merely started piece of work on actually solving the problem himself, simply adding more hints on the harder two triangles:

You've done well so far. You've got the easiest side, AB.

For side Ac, consider that triangle AOC is isosceles, and construct the distance to Air conditioning.

Side BC is the nigh challenging part that I mentioned. Notice that when you lot construct the altitude to BC, y'all'll have the same correct triangle that turned out to exist the answer in the triangle-in-a-semicircle problem: 15-75-90. Applying things we learned in that location can help united states of america find thearea of triangle BOC pretty easily, but I'grand not sure how much that helps. Let'due south both work on this!

Here is the figure with those two altitudes added; the showtime yields 30-threescore-xc triangles, which are easily solved, and the second gives the triangles we saw in the other problem:

I had some other thought, and jumped in briefly:

Hither is an culling: Having constitute AB, construct the distance from A to BC. Several things work out nicely.

Doctor Rick past at present had finished his piece of work, and added:

I found a fairly elementary way to complete the work I started … it involves extending BO to the other side of the circle and constructing the perpendicular from C to this line.

Then, retrieve our piece of work on the triangle in a semicircle, and construct the radius OC as well, which makes another 30-60-xc triangle.

Even so, my solution has nested square roots, whereas Doc Peterson's solution has a sum of square roots. It tin be shown that the 2 solutions are equal, but his is "nicer" — we don't really like nested roots.

From here on, the bodily interaction mingled work on the two approaches in a manner that is very hard to follow, so I am going to suspension with tradition and untangle these into two separate threads. (It was non like shooting fish in a barrel, peculiarly considering there were also several typos and consequent confusion to edit out.) First, we'll follow the discussion of Doctor Rick's idea.

Method ane

Kurisada now showed his work:

Focusing on the doctor'southward statement about 30-60-xc, then I idea that there is a fixed ratio of the sides of 30-lx-90 triangle.

I searched it and I found the ratio one : √three : ii.

I wrote the perpendicular point from C to line BO after extended as Y (sorry for my bad English in this, only I attached the moving picture below).

And I accept the triangle COY with angles 30-60-90.

Since OC = 1, and then OY = (√3)/two, and CY = ane/two.

Then I take triangle BCY.

CY = 1/2 and BY = 1 + (√iii)/2.

Then using Pythagoras Theorem, I got BC = √(two + √3).

The primal answer shows that BC = (√6 + √two)/ii.

I wonder if I did anything incorrect.

I likewise wonder if what doctor wanted to tell me is as above  or non.

I also tried to use about my previous trouble (triangle inside a semicircle), but I tin can't notice something to apply to this problem especially the non-trigonometry i. Can doctor give me a footling more clue?

Aught is wrong. Kurisada has washed well, and as mentioned earlier, the answers are equivalent.

Doc Rick replied (using a picture show I've replaced with one of my own to correct an error):

Here is my figure for this solution method:

There are several ways to bear witness that angle COY is 30°.

"Focusing on the doctor's statement most 30-threescore-90, then I idea that there is a fixed ratio of the sides of 30-60-90 triangle

I searched it and I found the ratio 1 : √3 : 2″

I had assumed y'all were already familiar with this fact, as we used it in discussing the previous trouble with you. Information technology is easily derived by starting with an equilateral triangle and amalgam an altitude (which is as well a perpendicular bisector and an angle bisector). This forms two xxx-lx-90 triangles. The side opposite the 30° angle is one-half of a side of the equilateral triangle, and hence one-half of the hypotenuse of the 30-60-90 triangle. The length of the remaining side follows via the Pythagorean Theorem.

"And I take the triangle COY with angles 30-lx-xc.

Since OC = ane, so OY = (√3)/two, and CY = 1/2.

Then I take triangle BCY.

CY = 1/2 and Past = ane + (√3)/2.

Then using Pythagoras Theorem, I got BC = √(ii + √3).

The key reply shows that BC = (√6 + √2)/2.

I wonder if I did anything wrong."

No, you haven't done anything wrong. As I said last time, this method results in an answer with a nested square root — exactly what you found, √(2 + √3) — while Dr. Peterson's method gives a sum of roots — as your answer primal does, (√6 + √2)/two. And I said that these tin can be proved to be equal, merely this is far from obvious at showtime! I suppose, therefore, that the reply in the key was obtained by something more similar Doctor Peterson's method.

"I too wonder if what doctor wanted to tell me is equally to a higher place or non."

Yous did fine using this method. If y'all finish the work by Doc Peterson'south method, yous should obtain the book's answer.

We'll become to the direct route to the answer \(\frac{\sqrt{vi}+\sqrt{2}}{2}\); but in club to run into that the two answers are equal, that is, that $$\sqrt{ii + \sqrt{iii}} = \frac{\sqrt{vi}+\sqrt{2}}{2},$$ we tin just square both sides (having observed that both sides are positive, so that squaring does not lose information): On the left, $$\left(\sqrt{ii + \sqrt{3}}\right)^2 = 2 + \sqrt{3},$$ while on the right, $$\left(\frac{\sqrt{6}+\sqrt{ii}}{2}\correct)^ii = \frac{vi + 2\sqrt{6}\sqrt{ii} + ii}{iv} = \frac{eight + 2\sqrt{12}}{4} = 2 + \sqrt{iii}.$$ And so the two sides are in fact equal.

Continuing,

"I likewise tried to apply nearly my previous problem (triangle inside a semicircle), but I can't find something to apply to this trouble especially the non-trigonometry i. Can medico give me a little more inkling?"

In my non-trig solution to that other problem, I constructed the radius equivalent to OC in this problem. If, in effigy (b), nosotros give the name F to the other intersection of BO extended with the circumvolve, and construct FC, then triangle FCB is simply the triangle inscribed in the semicircle of the other trouble. It is a fifteen-75-90 triangle; its distance OE is half the radius of the circle, as nosotros discussed in that problem (equally this makes the surface area of FCB half the maximal expanse of an inscribed triangle). Thus this new trouble is nearly the reverse of the previous problem: in that location we needed to make up one's mind the angle FBC knowing the base and distance of the triangle, whereas now we know the angles and need to determine the side lengths.

And so Doc Rick'due south method gives a correct answer, and ties into what we looked at last calendar week.

Method two

Now allow'due south look at the discussion of my method, which was interlaced with that. Kurisada said:

I drew the altitude Advert, and found that Ad = DC since ADC is 90°, 45°, 45°.

But, I as well did : BD x CD = AD^ii, resulting BD = AD which I recall is impossible as the angles are 90°, 60°, thirty°.

I also tried to practice AC ÷ AB = DC ÷ AD, but it resulted Air conditioning = AB which I remember is as well impossible due to the same reason as above.

Rick replied:

"I drew the altitude Advertising, and constitute that AD = DC since ADC is 90°, 45°, 45°."

Correct.

"But, I as well did : BD x CD = Advertisement^2, resulting BD = AD which I recall is impossible as the angles are xc°, lx°, thirty°."

The geometric hateful property we discussed earlier [in the semicircle problem] applies but to a correct triangle; ABC is non a right triangle. Or am I misunderstanding what you did here? The angles y'all cite are for triangle ADC.

But that, in fact, is exactly what Doc Peterson was getting at (in function) — you can utilize the side ratios for a 30-lx-90 triangle to decide your OC, and the side ratios of a 45-45-90 triangle to determine your OB. (This is subsequently you've determined Air-conditioning and AB as you indicated earlier.)

"I also tried to exercise AC ÷ AB = DC ÷ AD, only it resulted Air-conditioning = AB which I recollect is besides impossible due to the same reason equally to a higher place."

Presumably you are still talking virtually the theorem most a right triangle, in which in that location are three similar correct triangles. That doesn't use hither.

Kurisada answered:

I didn't realise about the fact that the geometric mean is but applicable to right angle so what I did is wrong.

Rick answered (again, I had to replace his picture with i that is labeled correctly):

Doctor Peterson gave you alink to Wikipedia which calls the theorem the "right triangle altitude theorem or geometric hateful theorem". Information technology'south important to be enlightened of the givens when you seek to utilize a theorem!

Here is the figure for this solution:

Hither, D is the foot of the perpendicular from A to BC, as Dr. Peterson had in mind.It should be obvious that triangle ABD is a 45-45-90 (correct isosceles) triangle, since bending ABD = ABC is given as 45° and ADB is a right bending; and as well obvious that triangle ACD is a 30-threescore-90 triangle since angle ACB = ACD is given as lx°. At present, early on, nosotros discussed finding the lengths of AB and AC, and so you should know those — do you? Y'all said AB = √2, which is correct; perhaps you never finished finding AC.

Here'due south what I said in my second message almost that: "For side AC, consider that triangle AOC is isosceles, and construct the altitude to AC." What exercise you notice? I hope you'll recognize two more of those 30-sixty-90 triangles that I had assumed you already understood.

And so CD = Air-conditioning/√2, and BD = AB/two, by the side ratios for the 2 "special triangles". Add these and you'll get the length of BC, which is what we're looking for.

Let's finish the piece of work. Here is a moving-picture show with that altitude to AC, OE:

From triangle CEO, we see that \(CE = \frac{\sqrt{iii}}{ii}\), so $$AC = \sqrt{3}.$$ Then, going dorsum to the previous picture, from triangles CAD and BAD we have \(CD = \frac{\sqrt{iii}}{\sqrt{2}} = \frac{\sqrt{half-dozen}}{2}\), and \(BD = \frac{Advertising}{2} = \frac{\sqrt{2}}{two}\), and so $$BC = BD + CD = \frac{\sqrt{six}+\sqrt{ii}}{two}$$ as before. Those are our final answers.

Now Kurisada was satisfied:

Give thanks You lot very much Physician Rick!

At present I understand about this problem

And sorry considering it took besides long fourth dimension

The word concluded on January ane.

We practise not heed taking time over a problem; we like going deeper to brand sure a educatee understands the concepts fully. Thanks for sticking with this, and have a happy New year's day!

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Source: https://www.themathdoctors.org/triangles-in-a-circle-two-methods/

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