Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10 5/3/83; site asgb.UUCP Path: utzoo!watmath!clyde!burl!mgnetp!ihnp4!zehntel!hplabs!sdcrdcf!sdcsvax!bmcg!asgb!vasudev From: vasudev@asgb.UUCP Newsgroups: net.puzzle Subject: Solution to triangles in a square Message-ID: <527@asgb.UUCP> Date: Mon, 13-Aug-84 14:33:12 EDT Article-I.D.: asgb.527 Posted: Mon Aug 13 14:33:12 1984 Date-Received: Wed, 15-Aug-84 06:53:00 EDT Organization: Burroughs Corporation, San Diego Lines: 37 Use the following result: A triangle inscribed in a semi-circle is a right angled triangle. The corollary to which is: If the apex of the triangle lies outside the semi-circle then it is an acute angled triangle. Solution: Let ABCD be the square. Let E be the mid-point of AB. Let F be the mid-point of BC. Let G be the mid-point of CD. Let H be the mid-point of DA. Construct the semi-circles AH, HD AB, DC CF, FB such that they all lie in the square. Let semi-circle AH intersect semi-circle AB at I. Let semi-circle HD intersect semi-circle DC at J. Join IJ. Construct I'J' such that I'J' < IJ and is completely within IJ. The triangles are: AI'B, DJ'C, AI'H, HI'D, HI'J' and BI'F, I'J'F and FJ'C. Since the apexes of all the triangles lie outside the semi-circle, they are all acute. -QED -asgb!vasudev