Point P is completely inside equilateral triangle ABC.

AP = 3, BP = 4, CP = 5.

Without calculating the area and the side length of the triangle,

find size of angle APB.

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- Mar 24th 2011, 05:21 PMWilmerAnother geometry puzzle
Point P is completely inside equilateral triangle ABC.

AP = 3, BP = 4, CP = 5.

Without calculating the area and the side length of the triangle,

find size of angle APB. - Mar 25th 2011, 06:54 PMRaisinBread
Before I start posting my incredibly long and unpractical (probably far from optimal) solution, is the answer 5pi over 6?

- Mar 25th 2011, 07:43 PMWilmer
- Mar 25th 2011, 09:08 PMsimplependulum
It is an old problem , my idea is to reflect $\displaystyle P $ across the three sides $\displaystyle BC , CA , AB $, let the reflection be $\displaystyle A' , B' , C' $ respectively .

I find that $\displaystyle \Delta AB'C' ~,~ \Delta BC'A' ~,~ \Delta CA'B' $ are $\displaystyle 30^o - 120^o - 30^o $ triangles . Then $\displaystyle \Delta A'B'C' $ has the sides ratio $\displaystyle 3:4:5 $ so it is right-angled .

$\displaystyle \angle APB = \angle AC'B = 30^o + 90^o + 30^o = 150^o $ which it is also the answer given by RaisinBread .

The second solution is more elegant , but not by me .

The idea is to rotate $\displaystyle \Delta APC $ by $\displaystyle 60^o $ , by sending $\displaystyle C $ to $\displaystyle B $ , $\displaystyle P $ to $\displaystyle P' $ . Then , $\displaystyle \Delta AP'P$ is equilateral and $\displaystyle PP' = 3 $ so $\displaystyle \angle BPP' = 90^o $ . Therefore , $\displaystyle \angle APB = 60^o + 90^o = 150^o $ - Mar 26th 2011, 02:07 AMWilmer
Yes.

QUOTE=simplependulum;633536]I$\displaystyle \angle APB = \angle AC'B = 30^o + 90^o + 30^o = 150^o $ which it is also the answer given by RaisinBread .

[/QUOTE]

Sorry RaisinBread; didn't realise you were giving me radians. - Mar 26th 2011, 05:12 AMRaisinBread
It's all good, I'm happy to have found the answer! I see the solution is really simple, but believe it or not, I solved the whole thing algebraically with functions and intersections.

In short, I did this:

1. I represented the triangle in the x-y plane, such that one side is resting on the x axis, and that the summit opposed to that side is on the y axis. I can thus make any equilateral triangle with the following equations:

$\displaystyle y={\sqrt{3}} x+a$,$\displaystyle y=-{\sqrt{3}} x+a$ and $\displaystyle y=0$, where a is any real number.

Attachment 21268

2. Let the left summit be A, the top be B and the right be C. I centered 3 circles of respective radii 3, 4 and 5 on these summits.

Attachment 21269

I know I have to find the point a (upon which all my equations depend) at which the three circles intersect at a single point, that point being P.

3. I knew that three non concentric circles could only intersect at the intersection of their radical axis, so i isolated the radical axis of the circles with center A and C, using trigonometry, which gave me the x coordinate of where the circles would intersect.

I then proceeded to solve quadratic equations to find at which y did the circles around A and B intersected, and then the same thing with the circles around B and C. I got two solutions each time, but I could see that in both case, the lowest intersection would be the one where the three circles intersected.

Attachment 21270

(These are the x and y coordinates of the point P, provided the two y are equal)

I then found the value of a by conditioning that the two y coordinates be equal, and isolated $\displaystyle a=\frac {\ {\sqrt{36{\sqrt{3}}+75}}}{2}$

4. Putting these results in the graph I had a confirmation that I was on the right track; Attachment 21271. Now all I needed to do was find the equations of the line from A to P, and the line from P to B, which was easy to do now since I had the value of a. I then changed these slopes into polar coordinates to find what angle they make with the y axis, subtracted the angle values and eureka, got 5pi over 6 radians!

Attachment 21272

Very long and impractical solution, but I can't say I didn't have fun doing it! - Mar 26th 2011, 11:24 AMWilmer
Nice...you made me tired just reading it!!

- Mar 28th 2011, 09:09 AMInigo
That's amazing that you solved for an exact value using functions and intersections...you make my head feel dizzy...

Well done, I went the philosophical route simplependulum took, knowing that it is an equilateral triangle.