Could someone help me with this geometry (triangle) problem?

Question

Let ABC be a triangle with sides AB = 5 f, BC = 8 f, AC = 4 f and let C be the circle inscribed in ABC. Let t be a tangent to C that intersects AB and AC at the points M and N respectively. Find the perimeter of triangle AMN

Answer 1

Let's label some more things:
BC is tangent to the circle at O
AB is tangent to the circle at P
AC is tangent to the circle at Q
MN is tangent to the circle at R

The two tangents from any exterior point to a circle are equal, right? Let's call the lengths of AP and AQ length a, BP and BO length b, and CO and CQ length c.

a+b=5
b+c=8
a+c=4

Solve those simultaneous equations and we find that...
a = .5 f (AP and AQ)
b = 4.5 f (BP and BO)
c = 3.5 f (CO and CQ)

NQ=NR
MP=MR
(...since the two tangents to a circle from an exterior point are equal.)

The perimeter of AMN = AM + MN + AN
= AM + (MR + NR) + AN
= AM + (MP + NQ) + AN
= (AM + MP) + (NQ + AN)
= AP + AQ
= .5 f + .5 f
= 1 f

Edit 6/13/07: Moved the "Let's label some more things" section to the top.

Answer 2

As others correctly pointed out, the answer is 1 f. Let's try to find a general solution. Let AB = c, AC = b and BC =a. Let O the point at which C intersects AB. We observe that the circle C is inscribed in ABC and exinscribed to AMN, it touches sides AM and AN outside the area of AMN. If P is the perimeter of ABC and p is the perimeter of AMN, then, according to the triangles properties, we have

p = AO (from triangle AMN)

Now, since C is inscribed in ABC, we have

AO = P - BC = P - a = (a+ b +c)/2 - a = (b + c -a)/2. Therefore, the perimeter pr of AMN is pr = 2p = b + c -a, which, actually, independs of the point where t intersects the circle C. The perimeter of AMN is invariant for any tangent t.

In our case, the perimeter is 5 + 4 - 8 =1, as others correctly found.

Answer 3

I think you've got two similar triangles here. In other words, ABC and AMN have corners of the same angle in the same order. As a result, you can use a ration to convert the lenghts of ABC's sides to the lengths of AMN's sides if you can get a ration. Draw a line from A that intersects BA at a 90 degree angle, that will give you right triangles to work with. Call that intersection on BC D. You know ADC is 90 degrees and AC is 4. Bust out your mad sin, cos & tan skills to find the length of CD which will be proportional to the portion of MN on one side of the AD line. Find the length of AN and make a ratio with that using the 4 between the triangles. Repeat for the other side and you'll have all the numbers you need for the perimeter.

Answer 4

The answer amazingly is 1. Or 1 f.

Let X and Y be the points where the circle touches AC and AB respectively. O = center of circle.

You can show that AX = AY = 0.5.
Angle XAY = 125.1 deg (cosine rule)
Using various rules, XY = 0.8874
r = OX = OY = 0.9625.

Let's draw the tangent MN parallel to XY, just to make life easy. AMN is similar to AXY.
You can determine that the ratio between them is 0.5298.

Perimeter of AXY = 0.5 + 0.5 + 0.8874 = 1.8874
So perimeter of AMN = 0.5284*1.8874 = 1.

**EDIT**
For the more general case, ryanker's solution is right on.

Answer 5

I submit to in concepts that there is a thorem that if the two sides of a traingle are equivalent then the perspective opposite to them are additionally same. Please see the thorem. So the bisectors make a isoscele traingle with the part in between the angles have been getting bisected.i visit later write you the evidence of the theorm. with the aid of fact the each and each a million/2 of the angles are equivalent, then the entire angles are equivalent. right here you discovered the two angles are equivalent. As consistent with definition: in a traingle if the two angles are equivalent, then that traingle is named a isoscele traingle.

Answer 6

2.2f