Parallel Tangent

Tangents drawn from a point to a circle are equal in length image image

So, $20+9=29$

It's not hard to see that $\triangle ABC$ and $\triangle AMN$ are similar.

Let's call the height of $\triangle ABC$ , $h$ and its inradius $r$ . That means the height of $\triangle AMN$ is $h-2r$ .

Now we can write,

$\frac{h-2r}{h}=\frac{MN}{BC}$

Now we're going to make some substitutions.

Notice that $h=\frac{2[ABC]}{BC}=\frac{2[ABC]}{4}$ and $r=\frac{[ABC]}{\text{the semiperimeter of} \triangle ABC}=\frac{[ABC]}{9}$ . [ $[ABC]$ denotes the area of $\triangle ABC$ .]

So, $\frac{h-2r}{h}=\frac{MN}{BC}$

$\Rightarrow 1-\frac{2r}{h}=\frac{MN}{4}$

$\Rightarrow 1-\frac{\frac{2[ABC]}{9}}{\frac{2[ABC]}{4}}=\frac{MN}{4}$

[I hope this line is not confusing. I basically did all the substitution at once.]

The $[ABC]$ 's cancel out and we're left with an equation with one unknown $MN$ . After a little bit of rearranging, we get

$MN=\frac{20}{9}$ .

So, $a+b$ is $20+9=\boxed{29}$ .

Note that clearly $\triangle AMN \sim \triangle ACB$ .

By Heron's formula, the area of triangle ABC is $3\sqrt{15}$ . As the area of a triangle equals the inradius multiplied by the semiperimeter, the radius of the incircle is $\frac{\sqrt{15}}{3}$ .

Note that the altitude from A to BC is $\frac{3\sqrt{15}}{2}$ . As the diameter that is perpendicular to MN and BC is parallel to this altitude, we may arrive at the conclusion that the altitude from A to MN is $\frac{3\sqrt{15}}{2} - \frac{2\sqrt{15}}{3} = \frac{5 \sqrt{15}}{6}$ .

We can set up a proportion as follows: let the length of MN be $x$ . Then the ratio of the length of $x$ and the length of $BC$ is equal to the altitudes from A of triangle $ABC$ and $AMN$ , or, mathematically

$\frac{\frac{5 \sqrt{15}}{6}}{\frac{3 \sqrt{15}}{2}} = \frac{x}{4}$

The computation is left to the reader, which produces a final answer of $x = \frac{20}{9}$ , which means that the desired sum is $\boxed{29}$ .

MN+BC=NC+MB; AN+MN+AM+2BC=AB+AC+BC; P(AMN)=18-8=10; But the perimeter of AMN is propotional to ABC.Let the constant be x. then P(ABC)*x=10; x= $\frac{5}{9}$ ; MN=xBC= $\frac{20}{9}$ ; thus a+b=29.

r = V((s-a)(s-b)(s-c)/s) = 1/3V15. Area triangle = Vs(s-a)(s-b)(s-c) = 3V15, therefore height from A = 3/2V15 height incircle : height from A = 2/3V15 : 3/2V15 = 4 : 9 . Therefore MN: CB = 5 : 9. So MN = 20/9. Answer = 29

The area of triangle ABC = pr/2 = h x BC/2 with p is the perimeter and h is the altitude drawn from A to the side BC, and r is the radius of the circle, then pr = h x BC => 18r = sinC x AC x BC = 32sinC => r=16/9 x sinC. Since MN and BC are parallel, then MN/BC = (h- 2r)/h => MN =20/9 = a/b => a+b=29

Take C as origin, B as (4,0) and A as(5.5,135^.5/2)...(u can get it by cosine formula). Consider the parallel line be y=k and center be (a,b). Find N and M in terms of k , i.e, N(11k/sqrt135,k) and M(4+3k/sqrt135, k). Clearly 2b=k(since b is radius of circle). Also perpendicular distance from the lines AC and AB to (a,b) is b. So we get 3 equations and 3 unknowns in a,b and k. Solve them and find k. Find MN as it was already in terms of k. answer is 20/9

its application of similarity and tangents...from a point...are equal...

Let r = the radius of the incircle for triangle ABC and h = the altitude from A to side BC. The semiperimeter of triangle ABC is half of the perimeter, or 9. Therefore we can express the area in two equivalent ways: $Area = 9r = \frac{1}{2}(4h)$ $h=\frac{9}{2}r$ Since the incircle is tangent to both MN and BC, the incircle's diameter is perpendicular to both segments. Therefore, the distance between MN and BC is (2r), and thus the altitude from A to side MN is $\frac{9}{2}r - 2r = \frac{5}{2}r$ Since the diameter is perpendicular to both MN and BC, MN is parallel to BC, and therefore triangles ABC and AMN are similar. The ratio of the altitudes on triangle ABC and AMN is equal to the ratio of side BC to MN: $\frac{\frac{5}{2}r}{\frac{9}{2}r} = \frac{side MN}{4}$ $side MN = \frac{20}{9}$ So the final answer is 29 .

Let... the line segment AB tangent to the incircle at r, the line segment AC tangent to the incircle at s, the line segment BC tangent to the incircle at t,

ِAccording to the conditions, we get that: Ar = As, Br = Bt, Cs = Ct,

∵ Bt + Ct = BC = 4 ∴ Br + Cs = 4

∵ (Br + Ar) + (Cs + As) = AB + AC = 6 + 8 = 14 ∴ (Br + cs) + Ar + As = 14, I.e. (4) + Ar + As = 14,,,,,,,,,,,,,, Then, Ar + As = 10 ∴ Ar = As = 5 ------> *

∵ the line segment MN ∥ the line segment BC ∴ Δ ABC ∼ Δ AMN

∴ AB/AM = BC/MN = AC/AN

∴ 6/AM = 4/MN = 8/AN

∴ 6 MN = 4 AM, & 8MN = 4AN

∴ 3MN = 2AM, & 2MN = AN --------> **

Let... the line segment MN tangent to the incircle at h, and... hM = X, & hN = Y, So, MN = X + Y

ِAccording to the conditions, we get that: Mr = hM = X, & Ns = hN = Y,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, Therefore, AM = Ar - X = 5 - X, AN = As - Y = 5 - Y

From ** we can get: 3(X + Y) = 2(5 - X), & 2(X + Y) = 5 - Y, 3X + 3Y = 10 - 2X, & 2X + 2Y = 5 - Y

∴ 5X + 3Y = 10, 2X + 3Y = 5 By subtraction we get: 3X = 5 -----> X = 5/3

∴ 2(5/3) + 3Y = 5 ∴ 10/3 + 3Y = 5, 10 + 9Y = 15, 9Y = 5 -------> Y = 5/9

∴ MN = X + Y = 5/3 + 5/9 = 15/9 + 5/9 = 20/9 ∴ a + b = 20 + 9 = 29