4
$\begingroup$

I encountered the following problem in a math problem book, in a section, entitled Geometry.

In an $\Delta ABC$ with $AB=BC$ and $\angle ABC = 80^\circ$, the point $O$ is chosen so that $\angle OAC = 30^\circ$ and $\angle OCA = 10^\circ$. Find $\angle BOC$.

I struggled repeatedly trying to solve this using some reasonable selection of methods, and failed utterly. I can construct a solution using analytic geometry, but that is not a fair choice of methods for this section.

From simple geometry, I keep writing down systems of equations using various triangles in the picture, and always keep missing one constraint, as it seems there are infinite amount of solutions to the system, which I know is wrong since the analytic geometry approach yields a unique solution.

I am happy to add my infinite solution approach using the angle constraints, if you find it helpful. Please let me know if that's the case.

I would like to solve this using methods from geometry, without using more advanced techniques. Thank you very much, I really appreciate your help.

$\endgroup$
4
  • $\begingroup$ Are you assuming O in the triangle? $\endgroup$
    – Moti
    Mar 25, 2021 at 4:37
  • 1
    $\begingroup$ May be it is easier to prove that BC=OC $\endgroup$
    – Moti
    Mar 25, 2021 at 5:19
  • $\begingroup$ yes you're right @Moti $\endgroup$ Mar 25, 2021 at 13:18
  • $\begingroup$ @Moti yes, thanks - see the answer, looking through it $\endgroup$
    – gt6989b
    Mar 25, 2021 at 20:11

3 Answers 3

5
$\begingroup$

enter image description here

Extend $AO$ to meet $BC$ at $M$ . Note that $\widehat{AMB}=80^o$ (why?) and therefore $\widehat{MOC} = 40^o$ (why?).

Draw $BP$ , bisector of $\widehat B$ . Note that $\triangle ABC$ is symmetric with respect to $BP$ . Let $CN$ be the mirror image of $AM$ with respect to $BP$. Then $$\widehat{NCB} = \widehat{MAB} = 20^o$$ $$\widehat{NCO} = \widehat{NCA} - \widehat{OCA} = 30^o - 10^o = 20^o $$ Therefore we have: $$\triangle IOC = \triangle IBC$$ It follows that $BC = OC$, which means $\triangle COB$ is isosceles, hence: $$\widehat{BOC} = \widehat{OBC} = \frac12 (180 - \widehat{OCB}) = 70^o$$ Et voila!

Fun fact: the segments shown in cyan ($AM,CN,CO,AB,CB$) are of equal length.

$\endgroup$
7
  • 3
    $\begingroup$ This answer could be improved more with better contrasting colors to replace the blues and greens and a white background $\endgroup$
    – Some Guy
    Mar 25, 2021 at 5:26
  • 1
    $\begingroup$ Not the answer - the presentation. Very nice Saeed - confirms the GeoGebra drawing. $\endgroup$
    – Moti
    Mar 25, 2021 at 6:50
  • 2
    $\begingroup$ Good answer (+1). The key constructed point is just $I$, and the diagram would be neater without the line segments $IM,IN,IP$. Thus: Let the median of triangle $ABC$ through $B$ meet $AO$ projected at $I$. By symmetry, $20^\circ=\angle IAB=\angle ICB$, and $\angle ICO=40^\circ-20^\circ=20^\circ$ too. It is similarly easy to see that $40^\circ=\angle IBA=\angle IBC=\angle IOC$. Therefore triangles $IAB$, $ICB$, and $ICO$ are congruent (ASA), and so $|CO|=|CB|$. Hence the equal base angles of the isosceles triangle $BCO$ are $70^\circ$. $\endgroup$ Mar 25, 2021 at 9:02
  • 1
    $\begingroup$ $\angle IOC=\angle OAC+\angle OCA=30^\circ+10^\circ=40^\circ$. $\endgroup$ Mar 25, 2021 at 14:00
  • 1
    $\begingroup$ +1, thank you - makes a lot of sense $\endgroup$
    – gt6989b
    Mar 25, 2021 at 20:12
1
$\begingroup$

Let $P$ be the circumcentre of $\triangle AOC$. Extend $AO$ to meet $BC$ at $D$.

Observe that, $\angle APC=80^{\circ}$ and hence $\triangle BAC\cong \triangle PAC$ and thus $BC=PC$.

Also, notice that $\triangle POC$ is equilateral and thus $OC=PC$.

Coupled with the result obtained earlier, we get $BC=PC=OC$ and hence $\triangle BOC$ is isosceles.

Therefore, $\angle BOC=\frac {180-40}{2}=\boxed {70^{\circ}}$

$\endgroup$
2
  • $\begingroup$ +1, thank you very much $\endgroup$
    – gt6989b
    Mar 25, 2021 at 20:15
  • $\begingroup$ @gt6989b Thanks. $\endgroup$
    – Limestone
    Mar 26, 2021 at 6:27
1
$\begingroup$

enter image description here

Here is an "out of triangle thought" for this question. Law of sine can be applied twice to reveal a hidden equilateral triangle $\triangle BOC$ if O is allowed to be outside $\triangle ABC$. (There might be a simple geometric solution out there someone can find.)

Set BC=1

For $\triangle ABC$ , AC=$\frac{\sin 80^\circ}{\sin 50^\circ}$

For $\triangle AOC$, CO=$\frac{AC\sin 30^\circ}{\sin 140^\circ}=\frac{\sin 80^\circ\sin 30^\circ}{\sin40^\circ\sin 50^\circ}=\frac{\sin 80^\circ}{2\sin40^\circ\cos40^\circ}$=1

Therefore, $\triangle BOC$ is an isosceles triangle, with $\angle BCO=60^\circ$$\to$ $\angle BOC=60^\circ$

$\endgroup$
6
  • $\begingroup$ OCA = 10 not AOC. $\endgroup$
    – Moti
    Mar 25, 2021 at 5:17
  • $\begingroup$ @Moti, thanks for the reminder, I reworked my answer and the result is a little surprise :-). $\endgroup$ Mar 25, 2021 at 6:32
  • $\begingroup$ GeoGebra says $70^0$ $\endgroup$
    – Moti
    Mar 25, 2021 at 6:49
  • $\begingroup$ The first word of the question is “In”. $\endgroup$ Mar 25, 2021 at 6:53
  • $\begingroup$ The answer is to explore the case when O is outside △ ABC. Since the result is$ 60^∘$, an interesting value, that’s why I shared. $\endgroup$ Mar 25, 2021 at 16:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .