# Thread: Another geometry puzzle

1. ## Another 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.

2. Before I start posting my incredibly long and unpractical (probably far from optimal) solution, is the answer 5pi over 6?

3. Originally Posted by RaisinBread
Before I start posting my incredibly long and unpractical (probably far from optimal) solution, is the answer 5pi over 6?
No; answer is an integer; hint: over 100 degrees.
If you're involved in something "long and complicated", then STOP

4. 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 , $\displaystyle \angle APB = 60^o + 90^o = 150^o$

5. 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.

6. 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.

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.

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.

(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; . 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!

Very long and impractical solution, but I can't say I didn't have fun doing it!

7. Nice...you made me tired just reading it!!

8. 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.