Question on an isosceles triangle.

Question

In triangle ABC, AB = AC = 25 and BC =14. The perpendicular distances from point P to each 3 sides in the interior of the triangle is equal. Find the distance.

Edit: Ah, this is not homework: its a competition question. Regarding to my efforts: I had found the vertical/perpendicular height (sqrt (25r^2-(14/2)^2) = 24). Then, I found the angle of each part (Angle of ABC and ACB must be 73.74 degrees while angle BAC is around 32.52 degrees). The angles are found via right triangle (25 is hypotenuse, 24 is height, and 7 is length). I tried the law of sine and thats where I am in so far.

I assume if I add 16.52 degrees clockwise to side AB and counterclockwise to side AC, it would create the perpendicular line to point P.

Edit: this is my drawing/depiction of the triangle. Goal is to solve x or y or z.

Answer 1

Hint: You found altitude $h_a=AG=24$, very well; now find altitude $BH=h_b=h_c$. Then use this formula:

$$\frac {L_1}{h_a}+\frac{L_2}{h_b}+\frac {L_3}{h_c}=1$$

where $L $is the distance of point from the sides, here $L_1=L_2=L_3=L$, and $h_b=h_c$.

Answer 2

I found the solution!

Using the law of sine:

As $\angle BAP$ and $\angle CAP$ are both $(180-(2(73.72)))/2 = 16.2602$, we can use the law of sine between $y$ and line $\overline{AP}$ as

$\sin (16.2602)/y = \sin (90)/(24-y)$, solve for $y$ and it becomes $6.72/1.28 = x = 21/4$!

I've got to admit I felt dumb upon the realization that the solution seemed this simple.

Answer 3

If the point P lies in the interior of a triangle and the perpendicular distance to each of the sides is equal, it would make that distance a radius of an incircle. Meaning we can find it using two area formulae.

Let p = 0.5 Perimeter

area = $ p * r $, on the other side, area = $ \sqrt{p(p-a)(p-b)(p-c)} $,

where a,b,c are side lengths. So $ 32 * r = \sqrt{32*7*7*18}$ meaning $ r = 21/4 = 5,25 $.