{"id":352,"date":"2020-02-21T14:07:46","date_gmt":"2020-02-21T20:07:46","guid":{"rendered":"http:\/\/blogs.acu.edu\/mu_sigma\/?p=352"},"modified":"2020-02-28T11:11:17","modified_gmt":"2020-02-28T17:11:17","slug":"problem-10-spring-2020","status":"publish","type":"post","link":"https:\/\/blogs.acu.edu\/mu_sigma\/2020\/02\/21\/problem-10-spring-2020\/","title":{"rendered":"Problem 10 – Spring 2020 – Where Does it Happen?"},"content":{"rendered":"

Where Does It Happen?<\/b><\/span><\/p>\n

This problem is from Jason Holland.<\/i><\/span><\/p>\n

Let\u00a0x<\/i>1<\/sub>,\u00a0x<\/i>2<\/sub>,\u00a0x<\/i>3<\/sub>,\u00a0x<\/i>4<\/sub>\u00a0be real numbers such that\u00a0x<\/i>2<\/sub>\u00a0–\u00a0x<\/i>1<\/sub>\u00a0=\u00a0x<\/i>3<\/sub>\u00a0–\u00a0x<\/i>2<\/sub>\u00a0=\u00a0x<\/i>4<\/sub>\u00a0–\u00a0x<\/i>3<\/sub>\u00a0= 1. Prove that the product\u00a0x<\/i>1<\/sub>x<\/i>2<\/sub>x<\/i>3<\/sub>x<\/i>4<\/sub>\u00a0is never less than -1, but can equal -1. Find all lists (x<\/i>1<\/sub>,\u00a0x<\/i>2<\/sub>,\u00a0x<\/i>3<\/sub>,\u00a0x<\/i>4<\/sub>) for which the product is equal to -1.<\/p>\n

Please submit all solutions to mathpotw@acu.edu by Thursday, February 27th by 5:00 PM.<\/p>\n

 <\/p>\n

Solution to\u00a0Where Does It Happen?<\/b><\/span><\/p>\n

Correct solutions were submitted by Wyatt Witemeyer.<\/em><\/p>\n

Consider a function<\/p>\n

f<\/i>(x<\/i>) = (x<\/i>-1)(x<\/i>)(x<\/i>+1)(x<\/i>+2).Then the derivative of\u00a0f<\/i>(x<\/i>) is<\/p>\n

f<\/i>\u00a0‘<\/sup>(x<\/i>) = 4x<\/i>3<\/sup>\u00a0+ 6x<\/i>2<\/sup>\u00a0– 2x<\/i>\u00a0– 2.If\u00a0f<\/i>\u00a0‘<\/sup>(x<\/i>) = 0, then we obtain the equation<\/p>\n

2x<\/i>3<\/sup>\u00a0+ 3x<\/i>2<\/sup>\u00a0–\u00a0x<\/i>\u00a0– 1 = 0.All local maxima and minima must occur at roots of this equation. The only potential rational roots are 1, -1, 1\/2, and -1\/2. A manual check shows that\u00a0x<\/i>\u00a0= -1\/2 works as a root, and it is the only one of the four that does. Using the fact that\u00a0x<\/i>\u00a0= -1\/2 is a root we can factor:<\/p>\n

2x<\/i>3<\/sup>\u00a0+ 3x<\/i>2<\/sup>\u00a0–\u00a0x<\/i>\u00a0– 1 = (x<\/i>+1\/2)(2x<\/i>2<\/sup>\u00a0+ 2x<\/i>\u00a0– 2).We check to see that\u00a0f<\/i>\u00a0”<\/sup>(-1\/2) is negative, so the function\u00a0f<\/i>\u00a0has a local maximum at -1\/2. Therefore the two places where\u00a0f<\/i>\u00a0has local minimums are the roots of the quadratic equation<\/p>\n

2x<\/i>2<\/sup>\u00a0+ 2x<\/i>\u00a0– 2.These roots are\u00a0r<\/i>1<\/sub>\u00a0= (-1 + SQRT(5))\/2 and\u00a0r<\/i>2<\/sub>\u00a0= (-1 – SQRT(5))\/2. A manual check verifies that\u00a0f<\/i>(r<\/i>1<\/sub>) =\u00a0f<\/i>(r<\/i>2<\/sub>) = -1. Therefore the two lists of four numbers are<\/p>\n

((-3 + SQRT(5))\/2, (-1 + SQRT(5))\/2, (1 + SQRT(5))\/2, (3 + SQRT(5))\/2) and
\n((-3 – SQRT(5))\/2, (-1 – SQRT(5))\/2, (1 – SQRT(5))\/2, (3 – SQRT(5))\/2).Comments:<\/b><\/p>\n