[Problem submitted by Kee Lam, LACC Professor of Mathematics. Source: Kee Lam]

Transcription

1 Prolem ) Find the first integer of four consecutive positive integers such tht the difference of the sum of the squres of the lst three integers nd the sum of the squres of first three integers is 7. [Prolem sumitted y Kee Lm, LACC Professor of Mthemtics. Source: Kee Lm] Solution: Let n e the first of the four consecutive positive integers, then n is the fourth integer. ( n ) is the second integer, ( n ) is the third integer, nd ( ) [( n ) ( n ) ( n ) ] n ( n ) ( n ) ( n ) ( n ) ( n ) n ( n ) ( n ) ( n ) n [ ] n 9 7 n 7 Hence, the first integer of the four consecutive positive integers is 7.

2 Prolem ) Evlute [Prolem sumitted y Kee Lm, LACC Professor of Mthemtics. Source: Kee Lm] Solution: The numertor nd denomintor re ech n rithmetic series, series in which the nth term is n ( n ) d with d eing clled the common difference. The sum n( ) of such series is n n. In the numertor d nd ( n ). So, n 507. The numertor is 507(00 04) In the denomintor d nd ( n ). So, n 04. The denomintor is 04(00 04) In its simplest form the frction is. 05

3 Prolem ) Solve [Prolem sumitted y Vin Lee, LACC Professor of Mthemtics. Source: Sint Mry s College Mthemtics Contest Prolems for Junior nd Senior High School y Brother Alfred Brousseu, 97] Solution: Simplify the eqution to get ( 5 )

4 define f ( ) Prolem 4) For { 0,, / } ( f ) ( ) f ( f ( ) ) simplify f ( 04 ) ( ), ( ) ( ) f ( f ( f ( ) )) f (. Define f ) ( ) f ( ), n n, nd in generl f f f. Find nd ( ) ( ) ( ) ( ). (Do not other to rtionl denomintors, ut simplify frctions.) [Prolem sumitted y Kent Merryfield, CSULB Professor. Source: Kent Merryfield] ( Solution: ) f ( ) ( ) ( ) f ( ) ( ) ( 4 ) f ( ) ( 4 We see tht f ) ( ) is the identity function. Hence, f ( 6 ) ( ( ) f ) ( ), nd in generl, 04 (mod 4), we hve tht f ( n ) ( m ( ) f ) ( ) f ( 5 ) ( ( ) f ) ( ), when n m (mod 4). Since ( 04 ) ( ( ) ) f f ( ) or. Comment: Liner frctionl trnsformtions compose like mtrices multiply. I ws looking for periodicity, so I ws looking for mtri whose eigenvlues were roots of ± ± i unity. I ws specificlly looking for 8th roots of unity,. So this function ws reverse-engineered from mtri with chrcteristic polynomil λ λ, nmely 0. More sophisticted solutions invoking these mtrices nd their eigenvlues re possile.

5 Prolem 5) Show tht tringle with sides whose lengths re 04, 46, nd 06 hs n re given y rtionl numer. Find the re. [Prolem sumitted y Vin Lee, LACC Professor of Mthemtics. Source: Sint Mry s College Mthemtics Contest Prolems for Junior nd Senior High School y Brother Alfred Brousseu, 97] Solution: We will use Heron s formul, which epresses the re of tringle in terms of the lengths of its sides: A s( s )( s )( s c). Let 04, 46, c c 06, nd s which is clled the semiperimeter. So, s s s s c s ( s ) ( s )( s c) s ( s )( s )( s c) A 50

6 Prolem 6) Let m nd s e the lengths of the longest nd shortest ltitudes, respectively, of tringle ABC, nd let P e point in the interior of the tringle. Suppose perpendiculrs PX, PY, nd PZ re dropped from P to the sides of the tringle. (If the ngles of the tringle re not cute, it my e necessry to etend the sides s shown.) Prove tht s PX PY PZ m nd tht oth inequlities re strict unless the tringle is equilterl. [Prolem sumitted y Iris Mgee, LACC Professor of Mthemtics. Source: Iris Mgee] Solution: Let,, nd c e the lengths of the ltitudes of tringle ABC from vertices A, B, nd C respectively, nd let K denote the re of the tringle. Then K BC CA c AB. Now drw lines PA, PB, PC, s indicted, so tht ΔABC is split into the three tringles PAB, PBC, nd PCA. Since the res of these three tringles sum to K nd since PX, PY nd PZ re their respective ltitudes from P, we get K ( PX )( AB) ( PY )( BC) ( PZ )( CA). Sustituting BC K, CA K, nd AB K PX PY PZ, we otin, upon dividing y K, the eqution. Now, c c,, c s, so this yields ( PX PY PZ ) ( PX PY PZ ). Thus resulting in m s the desired inequlities. Furthermore, if either of these is n equlity, then hence we would hve BC CA AB c nd

7 Prolem 7) Suppose is rel numer, > 0, nd. Find ll pirs of rel numers (, y) such tht log ( y) log log y. Justify your nswer. [Prolem sumitted y Vin Lee, LACC Professor of Mthemtics. Source: Vin Lee] Solution: Since the domin of log is ( 0, ), > y > 0. Use the reltionship log log y log to sustitute into the eqution given in the y prolem getting log ( y) log. So, y y y y y 0 y y. Use the qudrtic formul to solve this eqution for y in terms of. ± 4 y Since y is rel numer, 4 0 ( 4) 0 Since > 0, 4 0. So, 4. Any solution must hve y greter thn ero, nd this 4 is oviously true for y. However, we must determine for wht vlues of 4 is y > 0: > 0 0 > 4 > 4 > 4 4 > 0. 4 So, y > 0 for ny > 0. Therefore, ± 4 y. log ( y) log log y for every (, y) such tht 4 nd

8 Prolem 8) Five squres re rrnged s shown elow. Demonstrte tht the re of the shded squre is equl to tht of the shded tringle. [Prolem sumitted y Roger Wolf, LACC Professor of Mthemtics. Source: Roger Wolf] Solution: Are of the shded tringle ( ) ( ) ( ) ( ) Are of the shded squre

9 Prolem 9) Solve for in terms of nd find the domin of the function () f. [Prolem sumitted y Vin Lee, LACC Professor of Mthemtics. Source: Vin Lee] Solution: ) ( log log Since the domin of log is positive numers 0 > which implies < <.

10 Prolem 0) In the digrm the length of ech side of the squre is nd the ottom of the squre is the dimeter of the semicircle. The circle is tngent to the semicircle nd to two sides of the squre s shown in the digrm. A) Compute (with proof) the rdius of the second circle. B) Then s we dd the smller circle, tngent to the semicircle, the first circle nd the side of the squre s shown, show tht the rdius of the new smller circle is ectly / the rdius of the lrger circle. [Prolem sumitted y Iris Mgee, LACC Professor of Mthemtics. Source: Iris Mgee] Solution to Prt A: Let R e the rdius of the circle. Connect the center of the circle nd the center of the semicircle s shown elow. To find the rdius of the circle, R, we cn solve y using Pythgoren Theorem. R R R 4 4R R ( ) ( ) ( ) R R R R R 8R 4 0 R 4

11 Solution to Prt B: Drw verticl line through the center of the smll circle nd form two right tringles s shown. The hypotenuse of the upper tringle joins the centers of the two circles nd hypotenuse of the lower tringle joins the center of the smll circle nd the center of the semicircle. Let R nd r e the rdii of the lrge circle nd smll circle respectively, recll tht R 4. Also, of course, the rdius of the semicircle is. The length of the hypotenuse of the upper tringle is R r, nd its horiontl side hs length R r. It follows y the Pythgoren Theorem tht the squre of the length of the verticl side is ( R r) ( R r) 4rR. Similrly, for the lower tringle, the lengths of its hypotenuse nd horiontl side re respectively r nd r, nd hence the squre of the length of the verticl side is 4r. We see from the picture tht the totl of the lengths of the verticl sides of the two tringles is R, nd this yields the eqution 4rR 4r R, where the lst equlity follows from the known vlue of ( ) R. Simplifiction yields r ( R ), nd so r ( ) /( R ). Oserve tht ( ) 4 R This yields r R / nd so r R /, s we wnted! nd since > 0, we see tht R. Agin using the two right tringles shown ove, the Solution to Prt B my e summried s R ( R r) ( R r) ( r) ( r R 4Rr 4r Now sustitute R 4 from prt A to get r ( R ) r ( R ) * Oserve tht ( ) 4 R So, R Sustituting this result into * ove gives R r( ). Therefore, R r r R / )