# Another geometry puzzle

• Mar 24th 2011, 06:21 PM
Wilmer
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.
• Mar 25th 2011, 07:54 PM
Before I start posting my incredibly long and unpractical (probably far from optimal) solution, is the answer 5pi over 6?
• Mar 25th 2011, 08:43 PM
Wilmer
Quote:

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 (Wink)
• Mar 25th 2011, 10:08 PM
simplependulum
It is an old problem , my idea is to reflect $P$ across the three sides $BC , CA , AB$, let the reflection be $A' , B' , C'$ respectively .

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

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

The second solution is more elegant , but not by me .
The idea is to rotate $\Delta APC$ by $60^o$ , by sending $C$ to $B$ , $P$ to $P'$ . Then , $\Delta AP'P$ is equilateral and $PP' = 3$ so $\angle BPP' = 90^o$ . Therefore , $\angle APB = 60^o + 90^o = 150^o$
• Mar 26th 2011, 03:07 AM
Wilmer
Yes.
QUOTE=simplependulum;633536]I $\angle APB = \angle AC'B = 30^o + 90^o + 30^o = 150^o$ which it is also the answer given by RaisinBread .
[/QUOTE]
• Mar 26th 2011, 06:12 AM
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:
$y={\sqrt{3}} x+a$, $y=-{\sqrt{3}} x+a$ and $y=0$, 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 $a=\frac {\ {\sqrt{36{\sqrt{3}}+75}}}{2}$

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, 12:24 PM
Wilmer