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 In triangle ABC, point P is on side BC such that PA = 13, PB = 14, PC = 4, and the circumcircles of triangles APB and APC have the same radius. Find the area of triangle ABC.

 

i've tried everything i know of. can someone give me a hint of where to start??

 Apr 7, 2020
 #1
avatar+932 
0

Did you try sketching a diagram to get a visual of what it looks like?

 Apr 7, 2020
 #2
avatar+109520 
+2

I graphed it using GeoGebra, and this is what I got.

(GeoGebra is free and fun to use)

 

I do not see any remotely easy way to get the exact answer.   indecision

 

 

 Apr 7, 2020
 #3
avatar+109520 
+1

Just looking at what i have drawn ABC appears to be an isosceles triangle.

If you can prove that (assuming it is true) then finding the area would probably become relatively simple.

I am just thinking aloud here, I am sharing ideas with you.

 Apr 7, 2020
 #4
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+2

Diagram:                                                                                                                                                                                    

 

Note that \(\overline{AP}\) is a chord of both circles. Since both circles have the same radius, chord \(\overline{AP}\) must subtend the same angle, i.e. \(\angle ABP = \angle ACP\). Thus, triangle \(ABC\) is isosceles with \(AB=AC\).

Let \(M\) be the midpoint of \(\overline{BC}\). We see that\(BC = BP + PC = 14 + 4 = 18\) , so \(BM = CM = BC/2 = 9\). Then \(PM = CM - CP = 9 - 4 = 5\). Since \(\overline{AM}\) is also an altitude, we can apply Pythagoras to right triangle \(AMP\) to get \(AM=12\).

Hence, the area of triangle \(ABC\) is \(1/2 \cdot AM \cdot BC = 1/2 \cdot 12 \cdot 18 = \boxed{108}\).

 Apr 7, 2020
 #5
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0

@HELPMEEEEEEEEEEEEE here you go :)

 Apr 7, 2020
 #6
avatar+932 
0

Thanks!

HELPMEEEEEEEEEEEEE  Apr 7, 2020
 #7
avatar+20981 
+2

I got the same answer that guest got, 108, but not in so nearly elegant way.

 

I placed the diagram on a coordinate axis, with P(0,0), C(4,0), and B(-14,0).

Since PA = 13, it is on a circle with center P and radius 13:  x2 + y2  =  13

-- I called the x-value of this point 'a', so the y-value became  sqrt(169 - a2).

 

I used the theorem that the center of a circle is on the perpendicular bisectors of the chords.

Since one chord is BP, its midpoint has x-value -7 and its y-value is on the line x = -7.

Since another chord is PC, its midpoint has x-value 2 and its y-value is on the line x = 2.

Another chord is AP: using its endpoints, I found its midpoint and its slope.

From these values, I could find the equation of its perpendicular bisector (not a lot of fun, for it had a lot of a-terms and square roots).

Then, I found the intersection of this perpendicular biscector with the line x = -7 and the intersection of this perpendicular bisector with the line x = 2 (there were still a lot of a-terms and square roots).

With these two points (the centers of the circumcircles), I could find the distance from one center to point P and the distance from the other center to point P (By this time, I was on a firt-name basis with the a-terms and square roots).

Since these are equal, by solving this equation, I could find the value of a, which was -5.

This is the x-value of point A, allowing me to find the y-value of point A, 12, which is the height of triangle ABC.

With the height 8 and the base 18, the area is 108.

 Apr 7, 2020
 #8
avatar+932 
0

Any ways counts as long as it's accurate and easy to understand.

HELPMEEEEEEEEEEEEE  Apr 7, 2020
 #9
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+1

It's funny how the guest just copied and pasted AOPS's solution. Just a thought.

 Apr 11, 2020

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