Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1207.

## Thursday, April 14, 2016

### Geometry Problem 1207: Triangle, Circle, Incenter, Circumcenter, Excenter, Circumradius, Perpendicular

Labels:
circle,
circumcenter,
circumradius,
excenter,
incenter,
perpendicular,
triangle

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http://s22.postimg.org/nia3yt2fl/pro_1207.png

ReplyDeleteLet L, M, N , P are points as shown on the sketch

Observe that AL= NC= half perimeter of triangle ABC- BC

P is the midpoint of arc AC => M is the midpoint of AC and LN

In trapezoid ILND , MO is the mid-base => O is the midpoint of ID

Triangle IOP similar to triangle IDE ..( case AA)

Since O is the midpoint of ID so DE= 2 x OP= 2.R

Lets assume touch point of incircle and excircle are F and G, and Midpoint of AC is M. It is easy to see that FM=GM=(a-c)/2 Also IF,OM and DG are parallel to each other ( all are perpendicular to AC), hence O is midpoint of ID.

ReplyDeleteB, I and E are collinear, join BE and let it intersects circumcircle at point H, and IH=HE, thus H is midpoint of IE.

Consider triangle IED, O is midpoint of ID and H is midpoint of IE,

Hence DE=2*OH, since OH=R, DE=2R.

Join BIE. Let it cut circle(O)at M. Angle ECI is a right angle.

ReplyDeleteM is the midpoint of arc AMC. So OM bisects AC at right angles.

We are done if we can show O is the midpoint of ID.

Let X, Y, Z be the projections of I. O. D on AC respectively.

It is easy to see that XY = b/2 - (s -a) = (c -a)/2 = YZ.

IX, OY, DZ being //, OYM is a midline // to DE in Triangle IDE.

Hence DE = 2 OM = 2R.

N Vijaya Prasad

Rajahmundry - INDIA.

Bring IK⊥AC,MN⊥AC, EL⊥AC(M is midpoint arc AC).Is <MIC=<ABC/2+<ACB/2 , <MCI=<MCA+<ACI=<ABC/2+<ACB/2.Therefore <MIC=<MCI.So MI=MC.But IC⊥CE.

ReplyDeleteThen <MIC+<MEC=90 and <MCI+<MCE =90 therefore MI=MC=ME.So MO//ED

Is IO=OD. Therefore ED=//2OM=2R.