tag:blogger.com,1999:blog-6933544261975483399.post4415108333714348009..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1043: Right Triangle, Incenter, Excenter, Congruence, Metric RelationsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-43232100435169095702018-10-18T12:14:09.383-07:002018-10-18T12:14:09.383-07:00If IE=EC & ABC is a right triangle, then it ca...If IE=EC & ABC is a right triangle, then it can be proved that it is a 30-60-90 triangle (refer problem 1295)<br />So BC=3 and AB=3√3, the result follows with 3+x being the semi-perimeter of ABC<br />Sailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-26507420640452255462014-09-21T13:48:25.908-07:002014-09-21T13:48:25.908-07:00Radius of incircle=r. Drop I onto ED to obtain poi...Radius of incircle=r. Drop I onto ED to obtain point P. <br />AB=x+r=PI, so triangle PIE is congruent to triangle BAC. <br />BC=PE=x-r. DC=2x-r=6+r. Therefore<br />1) x-r=3.<br />Substitute this into (x-r)^2+(x+r)^2=36 in triangle PIE to get <br />2) x+r=3*sqrt(3).<br />Adding 1) and 2) yields 2x=3(1+sqrt(3))==>x=3(1+sqrt(3))/2Ivan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-13969177661556640032014-09-11T05:03:22.763-07:002014-09-11T05:03:22.763-07:00C, I, E are collinear
Connect I with E and A with...C, I, E are collinear<br /><br />Connect I with E and A with B<br /><br /> AE = AI = 3√2<br />in the triangle AIC we have IC = 3√3 - 3<br />triangle EBC similar to triangle AIC, EB/AI = EC/AC<br />x = 3(1 + √3)/2<br />x= 4.098<br /><br />Erina, NJAnonymousnoreply@blogger.com