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The problem is as follows:

On the interior of an isosceles triangle $\triangle{ABC}$ where $\angle{B}=110^{\circ}$ it is situated a point $M$ such as $AB=MC$ and $\angle{BAM}=5^{\circ}$. Using this information find $\angle{MCA}$.

The choices in my book are as follows:

$\begin{array}{cc} 1.&10^{\circ}\\ 2.&15^{\circ}\\ 3.&20^{\circ}\\ 4.&25^{\circ}\\ 5.&30^{\circ}\\ \end{array}$

According to the official answers sheet the answer is choice 4. But how would you get there?.

I've been looking at this figure and I am out of ideas. Can someone help me with a sketch for this problem and how to solve it?.

This problem should be solved relying only in euclidean geometry constructions, is there a way to do that to solve this?. Since I am not good with that I will really appreciate someone could help me here.

Chris Steinbeck Bell
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3 Answers3

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draw Idea: I feel that the problem gives $m(\widehat{MAB})=5^{\circ}$, not $m(\widehat{MAC})=30^{\circ}$ to get us out of the way. Then, to use this angle, I have to get equilateral or right triangles somewhere. In order to use it effectively, I can prefer a circle around it.

Solution: Let $O$ be the center of the circle of circles of the triangle $MAC .$ $\\$ Since $|CO|=|MO|=|AO|$ and $2m(\widehat{MAC})=m(\widehat{MOC})=60^{\circ}$(relationship between central angle and inscribed angle), $CMO$ is an equilateral triangle. So $m(\widehat{OCA})=m(\widehat{CAO})=60^{\circ}-m(\widehat{MCA}).$ Now, let's draw perpendicular to $[CA]$ from $O$ and call the point $F$ where the drawn line intersects $[CA].$ Since $|CB|=|BA|$ and $CO|=|AO|$, $|CF|=|FA|$ passes through $[OF$] or rather $B$ (Originally this $ABCO$ is a property of a rhombus, the diagonals intersect perpendicularly). Where $m(\widehat{CBO})=m(\widehat{OBA})=55^{\circ}$ and $|BA|=|AO|$ we get $m(\widehat{BAO})=70^{\circ}$ using $m(\widehat{OBA})=55^{\circ}$ and $|BA|=|AO|$ we get $m(\widehat{BAO})=70^{\circ}$. Now, the problem is solved.

$$m(\widehat{CAO})=60^{\circ}-m(\widehat{MCA})=70^{\circ}-35^{\circ}$$ $$m(\widehat{CAO})=60^{\circ}-m(\widehat{MCA})=35^{\circ}$$ $$m(\widehat{MCA})=25^{\circ} \therefore$$

adzetto
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  • I did not understood your comment about $\angle{BAM}=5^{\circ}$ not $\angle{MAC}=30^{\circ}$ to get us out of the way. Did you meant that is $\angle{BAM}$ key to solve this problem and not $\angle{MAC}$? I just want to say that I was confused why $\angle{MOC}=2\angle{MAC}$, it turns out you were using the degree measure theorem in $\triangle{MAC}$ and $\triangle{MOC}$ as $O$ is center. – Chris Steinbeck Bell Jun 22 '22 at 02:16
  • Since $BF$ is perpendicular to $AC$ and belongs to $\triangle{ABC}$ which is also isosceled then it becomes into a perpendicular bisector. Thus $AF=FC$. The rest is simple but I would say that it would only suffice to say that $\angle{OCA}=35^{\circ}$ solves the problem. I recommend incorporating in your solution that you used that degree measure theorem otherwise someone can get lost as me in the beginning. – Chris Steinbeck Bell Jun 22 '22 at 02:19
  • By the way how did you made this drawing?. Its pretty and very helpful. I appreciate you attend these doubts. – Chris Steinbeck Bell Jun 22 '22 at 02:21
  • Thanks for the advice I refactored the solution. The program I use is GeoGebra. It is very easy to use I recommend it. – adzetto Jun 22 '22 at 09:44
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    I also recommend adding that because $\triangle{ABC} \cong \triangle{AOC}$ as case $\textrm{SSS}$ this will make the perpendicular in $BF$ when extended will meet point $O$ thus ensuring $BFO$ are collinear. Also point $M$ is in that position because $\triangle{BCM}$ is isosceles. If we assume $C$ is the center of a circle where $CB$ is radius $CM$ is also radius and this ensures point $M$ to be inside the region made by $\triangle{BFA}$. This will help to understand better the answer. – Chris Steinbeck Bell Jun 23 '22 at 09:49
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Let $M'$ be a point inside $\triangle ABC$, such that $BC = M'C$, and $\angle BCM' = 10^\circ$. We shall show that $M' = M$. Let $A'$ be the reflection of $A$ over $BM'$. Since $\triangle BCM'$ is isosceles, we get $\angle M'BC = 85^\circ$. So, $\angle M'BA = 25^\circ = \angle M'BA'$. Therefore, $\angle A'BC = 85^\circ - 25^\circ = 60^\circ$. Moreover, $BC = BA = BA'$, so $\triangle A'BC$ is equilateral. So, $CA' = CB = CM'$. Therefore, $C$ is the circumcenter of $\triangle A'M'B$. Now, degree measure theorem yields $\angle BAM' = \angle BA'M' = \frac12 \angle BCM' = 5^\circ$. Reconstruction of <span class=$M$" /> So, $\angle BAM' = 5^\circ$, and $BC = M'C = AB$, and this characterizes the point $M'$. So, $M' = M$, and thus, $\angle MCA = \angle M'CA = 35^\circ - 10^\circ = 25^\circ$.

KVS02
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    Why $\angle{BCM'}=10^{\circ}$?. I mean how did you came up with this at the beginning?. Please explain. There is a typo, I believe you meant $\angle{A'BC}=85-25=60^{\circ}$. Then $\angle{BAM'}=\angle{BA'M'}$ is because $\triangle{ABM'}\cong\triangle{BA'M'}$ as it comes from the reflection you mentioned. – Chris Steinbeck Bell Jun 22 '22 at 02:27
  • Then only $\angle{BA'M'}=\frac{1}{2}\angle{BCM'}$ by the degree measure theorem in the circumcircle for $\triangle{BA'M'}$. I think this later part requires this mention. Finally you end up concluding $M'=M$ but again, from where did you arrive to $\angle{BCM'}=10^{\circ}$ in the first place?. Because without it you can't find $\angle{M'BC}$. I appreciate you please attend these doubts. – Chris Steinbeck Bell Jun 22 '22 at 02:29
  • $\angle MCB = 10^\circ$ is a guess coming from an accurate figure. In general, the angles are "nice" because the question asks you to find them. $10^\circ$ is a reasonable guess, because the angle seems (and is) slightly larger than $5^\circ$. It is easier to work with the $10^\circ$ and isosceles construction $M'$, as you are able to figure out a lot more angles. Moreover, $10^\circ$ is twice $5^\circ$, which motivates us to find some triangle with circumcenter $C$. – KVS02 Jun 22 '22 at 09:50
  • Another example of the "guessing" is the following (easier) problem: Let $ABCD$ be a square, and let $E$ be a point such that $\angle EDA = 60^\circ$, and $\angle EBC = 15^\circ$. Find $\angle EAB$. – KVS02 Jun 22 '22 at 09:56
  • Honestly I would have never "guessed" that such angle was $10^{\circ}$. I mean it could had been any value that it might be slightly larger than $5^{\circ}$. I think your approach works best using a protractor and a ruler and a compass, and lots of patience. – Chris Steinbeck Bell Jun 23 '22 at 10:00
  • About your problem, well I am stuck. I think it might be something less than $45^{\circ}$. If both $\angle{EDA}$ were the same as $\angle{EBD}$. Then the diagonal would fit. But because $\angle{EBD}$ is half of that on $\angle{EDA}$ then could it be that is $30^{\circ}$?. I say this because is off by $15^{\circ}$. As this causes the diagonal to be more tilted. But if it is the right answer. How you do it?. What is the right way to solve this problem?. Please help. – Chris Steinbeck Bell Jun 23 '22 at 10:06
  • Try to make a guess about $E$, something which can give you more to work with. Maybe construct an $E'$ using the guess, from which proceed to show it is the same as $E$. – KVS02 Jun 23 '22 at 18:07
  • I am a novice in this sort of guessing. If you don't mind. Can you share the solution?. This will help me better to understand how to do that. – Chris Steinbeck Bell Jun 25 '22 at 00:05
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Here is a further solution based on the symmetries of a regular polygon. In this case, a regular $36$-gon $0123\dots$, needed to cover the $5^\circ$ angle from the problem. The prime-notation means reflection w.r.t. the symmetry line $0\ 18$. We start with the following picture, which realizes the given $\Delta ABC$ as $\Delta 707'$ with vertices among those of a $36$-gon:

mathematics stackexchange regular polygon 4476963

Claim: The chords $1'7$, $0\;12$, $1\; 14'$, $27'$ are concurrent in a point $M$.

Proof: The angle bisectors in $\Delta 0\, 2\, 14'$ are concurrent, these are the last three chords in the list. Let us denote by $M$ this incenter. It is also the orthocenter of $\Delta 1\, 7'\,12$, built from the other three vertices involved in the chords.

mathematics stackexchange regular polygon 4476963 equilateral triangles

(Why does $1'7$ go through this point, too?! The simplest solution i found also constructs the point from the other solutions, but in a different setting with a different argument.)

Angles between chords in the picture are easily computed, for instance the angle $\widehat{0M7'}$ between $27'$ and $0\;12$ is the mean of the arcs measured they delimit, $\frac 12((0'-7')+(12-2))\cdot10^\circ=85^\circ$. This is the same measure as for $\widehat{M07'}$, so $\Delta 7'0M$ isosceles.

Let now $M^*$ be the reflection of $M$ w.r.t. $77'$. Since $M\in 7'2$, the reflection $M^*$ is on the reflected line $7'12$. We have $7'M^*=7'M=7'0$, and the angle $\widehat{07'M^*}=\widehat{07'12}=\frac 12\cdot (12-0)\cdot 10^\circ=60^\circ$. So $\Delta 07'M^*$ is equilateral: $$ \tag{$\dagger$} 0M^*=7'M^*=7'M=7'0=70\ , $$ and $M^*$ is also on $0\; 17$. We compare now the triangles $$ \begin{aligned} &\Delta 0MM^* &&\text{ and}\\ &\Delta 0M7 \ .\\ \end{aligned} $$ $07$ is a common side, $0M=0\;12$ is the angle bisector of $\widehat{70M^*}=\widehat{7\;0\;17}$, given two congruent angles in $0$, and $(\dagger)$ gives a third side, $0M^*=07$. We compute now $$ \widehat{M70} = \widehat{MM^*0} = \widehat{M^*\;0\; 18} = \widehat{17\;0\; 18} = 5^\circ\ , $$ so $M$ is also on the chord $71'$, the one building a $5^\circ$ degree angle with $70$ inside $\Delta 707'=\Delta ABC$. This finishes the proof.

$\square$

Note: From $\Delta 0MM^* \cong\Delta 0M7$ we obtain $M7=MM*$, so $\Delta 7MM^*$ is isosceles in $M$, which together with the knowledge of $\widehat{7MM^*}=2\cdot\widehat{7\;M\;12}=2\cdot 30^\circ=60^\circ$ makes it equilateral. We get $7,M^*,13'$ colinear. As already observed, $M^*\in 7'12$, and it seems that $M^*$ is also on a fourth chord, $14'6$.

dan_fulea
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  • I think the answer it is a little bit too much for my knowledge but I like the way that it explains the symmetry. – Chris Steinbeck Bell Jun 25 '22 at 00:11
  • The answer should be simple to digest, else it completely misses the target. The story behind is as follows. Very many exercises in synthetic geometry involving angles which are multiples of $10^\circ$ or $6^\circ$ or $5^\circ$ or ... are related to properties of the associated regular polygons, which provide inscribed angles or angles at center with the given angles. A first step is always to decide about the starting regular polygon. In our case, the angles are $110^\circ$ and twich $35^\circ$, so a regular $36$-gon can be used. In the first picture, note there are between the vertices... – dan_fulea Jun 27 '22 at 16:43
  • ... between the vertices $B=0$ and $A=7$ exactly $7$ arcs, each inscribed angle "seeing this arc" $\overset\frown{AB}=\overset\frown{07}$ has measure $(7-0)\cdot 5^\circ=35^\circ$. Same happens for $B=0$ and $C=7'$. And using the guidelines of the regular polygon it is easy to compute the angle between any two of its chords (with vertices among those of the regular polygon). Then the message of the proof is that angle computations for points related to the regular polygon are in a geometrically structural way equivalent to concurrence of lines / diagonals of the regular polygon. – dan_fulea Jun 27 '22 at 16:47
  • The framework of the regular polygon is in some sense the good place to search for helper points that may lead to the solution. – dan_fulea Jun 27 '22 at 16:49