Vol. 8 No 7 - Pi Mu Epsilon
Vol. 8 No 7 - Pi Mu Epsilon
Vol. 8 No 7 - Pi Mu Epsilon
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So&t-tton by Hmy Sedingm and ChffAZe~ R. VXmiiLe, St.<br />
Bonauen-tuAe UyU.uu~.-ity, St. BonauentuAe, New Yoik.<br />
Choose point P on BC such that 4BDP = 3BDA.<br />
Then, by ASA,<br />
triangles ABD and PBD are congruent. Therefore DF = AD = DE. We see<br />
that 4ADB = ^CDE = 60'. Then ^CDF = 60° too. Thus, by SAS,<br />
triangles CDP and CDE are congruent and we get that<br />
4BEC = WEC = 120' - WE = 120' - W F = 120' - 4ACB= 80'.<br />
SotiLtLon by Jack Gcui.6unkel, Ftiiihing, New Yo&.<br />
Since AE and AC are tangent to the circle, then OA bisects ^CAE.<br />
Similarly OC bisects 3ACF. Hence<br />
1<br />
+OAC = -^CAE<br />
1<br />
= -(180Â 2 - ~BAC) = 90' - ~ 2 B A C<br />
Ah0 tiotved by RUSSELL EULER,<br />
<strong>No</strong>Athtuut HtcAhouA^ State<br />
UiM.ue~&Xy, MmyvJUULe, JACK GARFUNKEL, F&tAh&g, NY, ROBERT C.<br />
GEBHARDT, Hopatcong, NJ, RICHARD A. GIBBS, Po& Lcitf-ci Co-fcZege, VuAango,<br />
CO, PETER GEISER, St. Ctoud State Uni.uu~.-c-ty, MN, RICHARD I. HESS,<br />
Rancho Paloh Vvidu, CA, RALPH KING, St. BonaventuAe Un-CumJULy, NY,<br />
JOHN D. MOORES, Cambtu.dge, MA, NORTHWEST MISSOURI STATE UNIVERSITY<br />
MATHEMATICS CLUB, MmyvWLe, JOHN H. SCOTT,<br />
<strong>Mu</strong>eoJLutm CoU.ege, S h t<br />
Pad, MN, WADE H. SHERARD, Fman UH-tvmJut.y, GieenuJUULe, SC, ARTHUR<br />
H. SIMONSON, E u t Texm State. UVIA.uvu.-ity a t Texo~kana, KENNETH M.<br />
WILKE, Topeka, KS, and the PROPOSER. Some of the tiotuu~. u~ed the<br />
taw of, hhu. GLEN E. MILLS, Uiange County Public Sehoot&, OnJLando,<br />
FL, iiihg decimal udue~, found the. angLe to withiin 0.05'. One 0 t h ~ ~<br />
bubm-cibion uhumed that D -LA the. midpoint of AC and a 6Lnd papa<br />
wad in appiying the. ims of thu, each obtaining a Wiong anhum.<br />
638. [Fall 19861 <strong>Pi</strong>opotied by R. S. LuthaM., UnLuWs^ty of<br />
Whcoiui.n Cintm, JanuuWLe, Whco&n.<br />
In the figure on the next page, the circle with center 0 is an<br />
excircle of triangle ABC.<br />
OA is produced to meet BK in D.<br />
quadrilateral.<br />
Then BK is drawn so that ^KBA = bloc, and<br />
Prove that OCBD is a cyclic<br />
and similarly<br />
Hence<br />
~AOC = 1800 - (WAC + ~ACO)<br />
<strong>No</strong>w we have that<br />
W C t W C = W A + W C<br />
1<br />
wo = go0 - $BCA.<br />
1 1<br />
= ~ B A C + ~ B C A = 9 0 - +AN.<br />
+ 3AOC = 4ABC + 23AOC<br />
= 3ABC + 180' - 4ABC = 180°<br />
so quadrilateral DBCO is cyclic, since a pair of opposite angles are<br />
supplementary.<br />
SAWJULLVL bo.euAc.ow4 w ~ubm-c.fcted by RICHARD A. GIBBS, FoA-t Leuici<br />
CoUege, Vmgo, CO, RALPH KING, St. BonuventuA.e Ufu.vmJULy, NY,<br />
HENRY S. LIEBERMAN, Waban, MA, JOHN D. MOORES, Camblu.dge, MA,<br />
NORTHWEST MISSOURI STATE UNIVERSITY MATHEMATICS CLUB, MoA~WLC, JOHN<br />
H. SCOTT, Macuhtm CoUege, Saint Pad, MN, MADE H. SHERARD, Fman<br />
Ufu.uuu,m, GieenvWLe, SC, ARTHUR H. SIMONSON, E u t Texa S-tote<br />
UyU.vi~^ltq at TexaAkana, and the PROPOSER.<br />
L&<br />
In Memoriam<br />
~auve taught a backbreaking load of trigonometry and algebra<br />
at Algonquin College in Ottawa, Ontario, Canada.<br />
Nevertheless, to<br />
keep from becoming mathematically stale, in 1975 he founded a small<br />
problem journal called Eureka, later renamed Crux Mathematicom, "a