H and R is me

Geometry Level 4

abc is a triangle sides=13,14,15.if h is the incentre and r is its circum radius then find ah * bh * ch divided by r


The answer is 64.

Harshi Singh by Harshi Singh , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

link text
The above link proves that .
I n r a d i u s r i n = 4 r ∗ S i n A / 2 ∗ S i n B / 2 ∗ S i n C / 2 (The link has used r in place of inradius and R in place of r that we use for our problem.) . s = 1 2 ∗ ( 13 + 14 + 15 ) = 21 , ( A r e a ) 2 = s ∗ ( s − 13 ) ( s − 14 ) ( s − 15 ) = 8 4 2 . r i n = A r e a s . a h = r i n S i n A / 2 , b h = r i n S i n B / 2 , c h = r i n S i n C / 2 , ∴ a h ∗ b h ∗ c h r = 4 r r i n r i n 3 ∗ 1 r = 4 ∗ r i n 2 = 4 ∗ A r e a 2 s 2 = 4 ∗ 16 = 64 Inradius\ \ r_{in}=4r*SinA/2*SinB/2*SinC/2\\ \text{(The link has used r in place of inradius and R in place of r that we use for our problem.)}.\\ s=\frac 1 2*(13+14+15)=21,\ \ \ \ (Area)^2=s*(s-13)(s-14)(s-15)=84^2.\ \ \ \ r_{in}=\dfrac{Area} s. \\ ah=\dfrac{r_{in}}{SinA/2},\ \ \ bh=\dfrac{r_{in}}{SinB/2},\ \ \ ch=\dfrac{r_{in}}{SinC/2},\ \ \ \\ \therefore\ \dfrac{ah*bh*ch} r = \dfrac{4r}{r_{in}}r_{in}^3*\dfrac 1 r=4*r_{in}^2=4*\dfrac{Area^2}{ s^2}=4*16=64\\\ \ \ \\

It might be better, in my opinion, if triangle is named ABC, incenter as I and circumcenter as R. That will result in asking for,
AI * BI * CI divided by R.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...