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The problem is as follows:

In the figure find $\angle{ABC}$. It is known $AB=BC=AD$.

Sketch of the problem

The choices given in my book are:

$\begin{array}{cc} 1.110^{\circ}-3\alpha\\ 2.115^{\circ}-2\alpha\\ 3.100^{\circ}-\alpha\\ 4.150^{\circ}-3\alpha\\ 5.120^{\circ}-2\alpha\\ \end{array}$

According to the official answers sheet the answer for this problem is choice 5. But how do you get there?

What kind of construction is needed here?. Since this problem belongs to a chapter that has not yet introduced circumpherence I think it can be solved without that. So how to solve this without using trigonometry.

The only thing I can remember is that in these cases you can say:

$\angle{ADC}=\angle{BAD}+\angle{ABC}+\angle{BCD}$

But other than that I am out of ideas. I don't know what to do here. Can someone help me please?.

Chris Steinbeck Bell
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2 Answers2

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A synthetic approach that does not require trigonometry.

First, consider a more basic scenario. You have a circle centered at $O$, with a chord given by the points $A, B$ describing a minor arc. Let the arc measure be $2X$. Let's say you have a point $C$ somewhere on the minor arc described by these points. Then, the claim is that $\angle AOC = 2\angle CBA$. There are a lot of isoceles triangles here. $\triangle AOB$, $\triangle AOC$ and $\triangle COB$. Then it follows that:

$$ \angle CBA = \angle CBO - \angle ABO = (90 - X + \frac{1}{2} \angle AOC)- (90 - X) = \frac{1}{2}\angle AOC$$ from which the result follows.

enter image description here

With this in mind, let $\omega$ be a circle centered at $E$ passing through $B, C$ and $D$. Then, the angle $\angle DEB = 2\alpha$. Note that this angle is the angle $\angle BAD$. What's more, $AB = AD$ and $EB = ED$ together with the opposite angles equal (why?) implies that the shape $ABED$ is a rhombus! Thus, $AB = AD = EB = ED = BC = EC$. So $\triangle EBC$ is equilateral. That means that $\angle BEC = 60$. This therefore implies that $\angle DBC = 30 - \alpha$. (due to the result about the point on a minor arc.) Now, you already know that $AB = AD$ gives $\triangle ABD$ isoceles and therefore $\angle ABD = 90 - \alpha$.

enter image description here

$\angle ABC = \angle ABD + \angle DBC = (90 - \alpha) + (30 - \alpha) = 120 - 2 \alpha$, and you're done.

egglog
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    +1: Really great answer, very nicely presented. My Geometry knowledge is basic high school only, with $50$ years of cobwebs on it, and even I could follow your answer. Just out of curiosity: [1] What software did you use to generate the diagrams? [2] What Geometry book(s) [or online PDF's] would you recommend for me, as a next step? – user2661923 Jun 22 '22 at 04:50
  • @user2661923 The software I used to generate the diagrams was Desmos (I just took screenshots). As the for books, I am no expert in geometry. However, the ones I have looked at are Coxeter's Introduction to Geometry (for theory and an understanding) and Evan Chen's Euclidean Geometry in Mathematical Olympiads (for problem-solving). Have fun exploring the wonderful subject of geometry! – egglog Jun 22 '22 at 08:12
  • @egglog The wording in the first part is confusing. The chord $AB$ belongs to the arc $\widehat{AOB}$ but that's not $2x$. From reading it makes to sound that way. Instead you define the arc $\widehat{AOC}=2x$. Assuming this is set, then how do you get to $\angle{CBO}=\left(90-x+\frac{1}{2}\angle{AOC}\right)$ and $\angle{ABO}=90-x$?. Yes I've read about the isosceles formed but I am still stuck. This requires explanation please. – Chris Steinbeck Bell Jun 23 '22 at 08:50
  • @egglog The second portion is easy, assuming that the intended theorem used for solving is the degree measure theorem in the circumpherence. But, Is there a way to prove that $EB=AB$ without requiring using the property associated with a rhombus?. I say this because this problem belongs to a section in my book where rhombus and quadrilaterals have not yet been introduced. Thus this could be solved I believe maybe using congruency. Is there some way to prove it using this route?. Can you attend this doubt as well please?. – Chris Steinbeck Bell Jun 23 '22 at 08:53
  • @ChrisSteinbeckBell The problem can also be solved by reflecting C across the line BD. You will see that the reflected point C' is the same as the point D reflected across AC. – egglog Jun 23 '22 at 15:29
  • @egglog I am interested in your new solution. What should I do after noticing that point C is the same as D after doing the reflexion? Can you please guide me through?. How do you find the requested angle?. – Chris Steinbeck Bell Jun 25 '22 at 00:38
  • @egglog Please, my question from above has not yet been attended. How did you found $\angle{CBO}=\left(90-x+\frac{1}{2}\angle{AOC}\right)$ and $\angle{ABO}=90-x$?. This step is skipped in your answer. – Chris Steinbeck Bell Jun 25 '22 at 00:40
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I found this to be a very challenging problem, which is why I am providing an answer, even though the OP (i.e. original poster) hasn't really shown any work.

Further, the OP specifically indicated that Trigonometry is not to be used. The only way that I could solve the problem was to use Trigonometry.

It could well be that there is some elegant purely geometric construction that proves the problem. Anyway...

Tools

In the diagram below, which is not drawn to scale, the challenge is to show that $(y + z) = 120^\circ - 2a.$

$(y + z) = 180^\circ - 4a - 2b = (120^\circ - 2a) + (60^\circ - 2a - 2b).$

So, the problem has been reduced to showing that

$$(60^\circ - 2a - 2b) = 0 \iff a + b = 30^\circ \iff \sin(a + b) = \dfrac{1}{2}. \tag0 $$

Since $\triangle ABC$ is isosceles,

$$S = 2R\cos(2a+b). \tag1 $$

Let $T$ denote the length of line segment $\overline{CD}.$
Then,

$$S = R\cos(b) + T\cos(a + b). \tag2 $$

Using the Law of Sines, you have that

$$\frac{T}{\sin(b)} = \frac{R}{\sin(a + b)} \implies T = \frac{R\sin(b)}{\sin(a + b)}. \tag3 $$

Combining (2) and (3) gives

$$S = R\cos(b) + \frac{R\cos(a+b)\sin(b)}{\sin(a + b)} \implies $$

$$ S = \frac{R\cos(b)\sin(a+b) + R\cos(a+b)\sin(b)}{\sin(a + b)}. \tag4$$

Using sum and difference formulas, (4) above simplifies to

$$ S = \frac{R\sin(a+2b)}{\sin(a + b)}. \tag5$$

Combining (1) and (5) gives:

$$\cos(2a + b) \times [2 \times \sin(a + b)] = \sin(a + 2b). \tag6 $$

So, the entire problem has been reduced to showing that (6) above implies the RHS of (0) above.

Let $\theta = (a + b) \implies $

  • $(a + 2b) = (2\theta - a)$
  • $(2a + b) = (\theta + a)$.

Also, using sum and difference formulas,

  • $\sin(2\theta) = 2\sin(\theta)\cos(\theta).$

  • $\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = 1 - 2\sin^2(\theta).$

So, (6) above may be re-expressed as

$$\cos(\theta + a) \times 2 \times \sin(\theta) = \sin(2\theta - a). \tag7 $$

Using sum and difference formulas, the LHS of (7) above becomes

$$[\cos(\theta)\cos(a) - \sin(\theta)\sin(a)] \times 2\sin(\theta) $$

$$ = 2\sin(\theta)\cos(\theta)\cos(a) - 2\sin^2(\theta)\sin(a). \tag8 $$

Similarly, the RHS of (7) above becomes

$$\sin(2\theta)\cos(a) - \cos(2\theta)\sin(a) $$

$$ = 2\sin(\theta)\cos(\theta)\cos(a) - [1 - 2\sin^2(\theta)]\sin(a). \tag9 $$

Comparing (8) and (9) above, notice that the first term in each expression is identical. Therefore, as a direct implication of (7) above, you have that

$$ - 2\sin^2(\theta)\sin(a) = - [1 - 2\sin^2(\theta)]\sin(a) \implies $$

$$ 2\sin^2(\theta) = [1 - 2\sin^2(\theta)] \implies $$

$$ 4\sin^2(\theta) = 1 \implies \sin^2(\theta) = \dfrac{1}{4} \implies $$

$$\sin(\theta) = \dfrac{1}{2} \implies \sin(a + b) = \dfrac{1}{2}.$$

Therefore, the equation on the RHS of (0) above has been established.

enter image description here

user2661923
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