Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $M$ be the middle of the segment $[BC]$ and $P$ the middle of the segment $[BM]$. A point $A$ in the exterior of the segment $[BC]$ such that $\angle BAP=\angle PAM=\angle MAC$. Find the measure of the following angles: $\angle BAC, \angle CBA, \angle BCA$?

Can you find the measure of the angles without using Pythagora's theorem?

share|improve this question
    
Does such point $A$ with those hypothesis exist? How to prove this? –  Sigur Sep 15 '12 at 15:17
    
Don't know about whether Pythagorean Theorem is necessary. I certainly used it in solving the problem. Also used somewhat fancier tools. –  André Nicolas Sep 15 '12 at 15:29
    
Yes, can avoid using Pythagoreaan Theorem directly. Don't know about indirectly. –  André Nicolas Sep 15 '12 at 15:39
    
please, can you write a solution:) thanks –  Iuli Sep 15 '12 at 15:41
1  
@Sigur: just take half of an equilateral triangle. Picture can be found below in my answer. –  tomasz Sep 15 '12 at 15:58

3 Answers 3

up vote 3 down vote accepted

enter image description hereDenote $BAC$ by $3\alpha$, $CBA$ by $\beta$, $ACB$ by $\gamma$.

I'm not sure if it qualifies, but you can see that from the fact that the angular bisector divides the third side of a triangle in proportion equal to the proportion of the two other sides (angle bisector theorem), so the triangle $AMB$ must be isosceles, by applying it to the triangle $PAC$ you see that the cosine of $2\alpha$ is $\frac {1}{2}$.

So $2\alpha=\pi/3$, $\beta=\pi/2-\alpha=\pi/3$, $\gamma=\pi-\beta-3\alpha=\pi/6$

share|improve this answer
    
Very nice. how did you draw this ? –  Belgi Sep 15 '12 at 16:00
    
@Belgi: Geogebra. It's a good tool to cheat with to solve high school geometry problems. ;-) –  tomasz Sep 15 '12 at 16:01
    
Why $\cos(2\alpha)=1/2$? We know that it is $AP/AC$. –  Sigur Sep 15 '12 at 16:04
1  
@Sigur: by angular bisector theorem. Notice that $MC$ is twice as long as $PM$. –  tomasz Sep 15 '12 at 16:07

Let $\alpha=\angle BAP$. At least one of the triangles $BPA$, $PMA$ has an angle $\ge\frac\pi2$ at $P$ (and a nonzero angle at $B$ or $M$, respectively). From the sum of angles in a triangle, we conclude $\alpha<\frac\pi2$. Construct $X$ such that $\angle PBX=\frac\pi2-\alpha$ and $XP\perp BC$. Then $\angle BXP=\alpha$ and $A,X$ are on the same side of $BC$. Therefore by the inverse of the inscribed angle theorem, $A$ is on the circumscribed circle of $BPX$. Reflecting the figure at $BP$, we find $\angle PXM=\alpha$, hence $A$ is also on the circumscribed circle of $PMX$. These two circles intersect exactly in $P$ and $X$. From $A\ne P$ we conclude $A=X$ and from this $|AB|=|AM|$

Translate the figure by $\overrightarrow{BM}$. Then $B\mapsto M$, $M\mapsto C$, $A\mapsto A''$. Since $|A''M|=|A''C|$, let $c$ be the circle araound $A''$ through $M$ and $C$. Because $\angle MA''C=2\alpha$, the circle $c$ is the locus of all points $X$ with $\angle MXC=\alpha$. Hence $A\in c$. We conclude that $|BM|=|AA''|=|A''M|=|AB|$, i.e. $BMA$ is equilateral, hence $\alpha=\frac\pi6$. The rest is easy: $\angle BAC=3\alpha=\frac\pi2$, $\angle CBA=\frac\pi2-\alpha=\frac\pi3$, $\angle BCA=\frac\pi2-2\alpha=\frac\pi6$.

share|improve this answer

My friend with I work a lot gave me a very interesting solution. For the beginning I will introduce the draw. enter image description here

Notations: $$BP=x.$$

Because $BP=x$ it follows that $PM=x$ and $MC=2x.$ $M \in[AM)$ and $[AM)$ is bisector of the angle $\angle PAC$ $\Rightarrow$ $MQ \perp AC$ and $MQ=PM$, so $MQ=x.$ So knowing that the triangle $\triangle MQC$ is a rectangular angle and $MC=2MQ$ $\Rightarrow$ the measure of the angle $\angle C=30^{\circ}$. What it remained to prove is easy to demonstrate it. ($\triangle ABM$ is isosceles because $P$ is a middle of $BM$ and $\angle BAP=\angle PAM$ from here the measure of the angle $\angle APM =90^{\circ}$).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.