Sunday, July 26, 2009

Problem 327. Right triangle Area, Incircle, Circumcircle, Square

Proposed Problem
Click the figure below to see the complete problem 327 about Right triangle Area, Incircle, Circumcircle, Square.

See more:
Complete Problem 327
Level: High School, SAT Prep, College geometry

3 comments:

1. Let ID = r and AC = b = 2R, AB = c & BC = a following the usual triangle notation.
Now, DE^2 = AD * DC since ADC is also a rt. triangle where AD = R - OD & DC = R + OD.
Hence, DE^2 = (R - OD)(R + OD) = R^2 - OD^2
But OD^2 = OI^2 - ID^2 = (R^2 - 2rR) - r^2 using Euler distance formula for OI. Hence we've: DE^2 = R^2 - (R^2- 2rR)+r^2= r^2+2rR=r^2+br=r(r+b).
In rt. triangle ABC, the in-radius r=(a+c-b)/2 & hence, DE^2 = [(a + c - b)/2]*[(a + c - b)/2 + b]
= [(a + c - b)/2]*[(a + c + b)/2]
= (1/4)*[(a + c)^2 - b^2]
= (1/4)*[a^2 + c^2 + 2ac - b^2]
= (1/4)*(2ac) since a^2 + c^2 = b^2
or DE^2 = ac/2 = Area of Tr. ABC. QED.
Ajit: ajitathle@gmail.com

2. DE^2=ADxDC;
AD=AB-r;
DC=BC-r;
(AB-r)x(BC-r) = ABxBC - rx(AB+BC) -r^2 = ABxBC - [(1/2 rxAB +1/2 rx AD + 1/2 r^2) + (1/2 rxBC + 1/2 x rxDC + 1/2 r^2) - r^2 =Area ABC;

3. AC is a diameter so AD. DC = DE^2 = (c-r).(a-r) = ( b+c-a)(a+b-c)/4 since r = (a+c-b)/2

This is = to { b^2 - (a-c)^2}/4 = ac/2 = Area of ABC (I used b^2 = a^2 + c^2 to simplify the algebra)

Sumith Peiris
Moratuwa
Sri Lanka