Triangle ConstructionDate: 03/11/2002 at 03:29:26 From: Louise Hjarnaa Subject: Construction of a triangle Let ABC be a triangle with sides a, b, c. Let r be the radius of the incircle and R the radius of the circumcircle. How can I construct the triangle, in a classical way using only ruler and compass, when I know a, R, r ? Date: 03/11/2002 at 05:07:31 From: Doctor Floor Subject: Re: Construction of a triangle Hi, Louise, Thanks for your question. Start with segment BC of length a. We know that the circumcenter O must lie on the perpendicular bisector p of BC: construct p. We also know that the distance from B to O must be equal to the given length R, so draw the circle with radius R and center B, intersect it with p and find O. Now that we have O, we can draw the circumcircle. Now we choose A to be on either side of AB (both sides may yield possible triangles, so you must repeat the construction for the other side). On that side construct line d parallel to BC at a distance of the given r. The incenter I must be on this line d. Let M and M' be the points where p meets the circumcircle: M at the same side of BC as A, M' at the opposite side. The angle bisector from A must bisect the arc BC of the circumcircle opposite to A, so the angle bisector passes through M'. Since all possible A on this side of BC yield the same angle <BAC, we see that the angle <BIC is fixed as well, since <BIC=180-(<B+<C)/2= 180-(180-<C)/2. But that means that the possible I pass through an arc. This arc contains B and C, since these are limit points of I when A approaches B or C. Of course the arc also contains the incenter I' of triangle BMC, which we can construct easily. So I must lie on r as well as on the circle through B, C and I'. The two intersections give the possible I. Intersecting M'I with the circumcircle gives the possible A, as desired. That gives A on one side of BC. Of course you will have to repeat the construction for the other side, which may yield more solutions. If you have more questions, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ Date: 03/12/2002 at 03:30:34 From: Louise Hjarnaa Subject: Construction of a triangle Dear Doctor Floor, Thank you for your nice and very fast answer to my problem. It seems that the center of the circumcircle for BI'C is always point M'. Why? Best regards and thanks in advance, Louise Date: 03/12/2002 at 10:15:14 From: Doctor Floor Subject: Re: Construction of a triangle Hi, Louise, Let's see. Angle BIC = 180 - (<B+<C)/2 = 180 - (180-<A)/2= 90 + <A/2. Since M' is on the circumcircle of ABC opposite to A, <BM'C= 180 - <A. Now let's produce segment IM' to meet again the circumcircle of BIC in Q. Then of course <BQC = 180 - <BIC because they are opposite angles in a cyclic quadrilateral. But then <BQC = 180 - (90 + A/2) = 90 - A/2. The central angle on BC in the circle (BICQ) must now be 2*<BQC = 180 - <A. From this we see that the circumcenter of ABC must be on the arc BM'C. Also it must be on the perpendicular bisector of BC (triangle BM'C is isosceles). The point of intersection of these is M', and thus M' is indeed the center of BIC. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/