pizer@ecsvax.UUCP (07/15/84)
References: <ecsvax.2915> Here's the solution to the third problem: #3 >>First start with right triangle ABC, where A is rt. and where AC is the >>longer leg. Then draw point D on side AC, such that DB=DC. Finally, draw >>point F on BC such that AF is perpindicular, and such that CF=CD. Given: >>BD=DC=FC=1 and that AB is prpndclr to AC and that AF is prpndclr to BC, FIND: >>AC. Good luck, leave answer in radical form. My solution to this problem was simple, but used a lot of algebra, it goes as follows: Let X=AC AF = SQRT(AC^2 - FC^2) AF = SQRT(X^2 - 1) BA = SQRT(BD^2 - AD^2) BA = SQRT(1 - (X-1)^2) BF = SQRT(BA^2 - AF^2) BF = SQRT(1 - (X-1)^2 - (X^2 - 1)) BC = BF + FC BC = SQRT(1 - (X-1)^2 - (X^2 - 1)) + 1 BC^2 = BA^2 + AC^2 --BC^2-- --BA^2-- --AC^2-- (SQRT(1 - (X-1)^2 - (X^2 - 1)) + 1)^2 = 1 - (X-1)^2 + X^2 (SQRT(1 - (X^2 - 2X + 1) - X^2 + 1) + 1)^2 = 1 - (X^2 - 2X + 1) + X^2 (SQRT(-2X^2 + 2X + 1) + 1)^2 = 2X -2X^2 + 2X + 1 + 1 + 2SQRT(-2X^2 + 2X + 1) = 2X 2SQRT(-2X^2 + 2X + 1) = 2X^2 - 2 SQRT(-2X^2 + 2X + 1) = X^2 - 1 -2X^2 + 2X + 1 = X^4 - 2X^2 + 1 X^4 - 2X = 0 X^3 - 2 = 0 (X cannot be 0) X^3 = 2 X = 2^1/3 . AC equals 2^(1/3) . . Q.E.D. Another good answer was sent to me by Kamal Abdali (allegra!tekchips!abdali) and goes as follows: >Let the length AD be denoted by x, and the angle ACB by theta. From the >right triangle ACF, we get cos(theta) = CF/AC, i.e. > cos(theta) = 1/(1+x). (1) >The angle ADB equals the sum of angles DBC and DCB. So angle ADB is >2*theta. From right triangle ADB, we get cos(2*theta) = AD/DB, i.e, > cos(2*theta) = x. (2) >Write c for cos(theta), hence 2*c^2-1 for cos(2*theta). Then (1) and >(2) become: > c = 1/(1+x), that is, x = 1/c - 1, > 2*c^2 - 1 = x. >Combining, we get 2*c^2 = 1/c, i.e, c = 1 / 2^(1/3). Now AC=AD+DC=1+x=1/c >by (1). Hence, the answer is AC = cuberoot(2). Several other people sent me different methods of solving the problem, however, much to my dismay, I don't have room to list their solutions. Thank you to those people for taking the time to solve the problems and send me their ideas. Billy Pizer ({decvax,akgua,ihnp4,burl}!mcnc!ecsvax!pizer) -- "What holds things together?" "Neutrons and protons are held together by velcro." --