Substitute these values into the first equation to get ( z + 6) + ( z + 3) + z = 27. Then solve to get

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1 Problem ) The sum of three umbers is 7. The largest mius the smallest is 6. The secod largest mius the smallest is. What are the three umbers? [Problem submitted by Vi Lee, LCC Professor of Mathematics. Source: Vi Lee] Solutio: Let be the largest umber, y the secod largest, ad z the smallest. The first three seteces i the problem may be writte as three equatios: y z 7 z 6 y z Note that z 6 z 6 y z y z Substitute these values ito the first equatio to get ( z 6) ( z ) z 7. The solve to get z 6, y 9, ad.

2 Problem ) Fid the ratio of the area a circle to the area of a iscribed square. [Problem submitted by Kee Lam, LCC Professor of Mathematics. Source: Kee Lam] Solutio: Let the measure of the side of the square be a. Sice the square is iscribed i the circle, the diagoal of the square is r. Usig the Pythagorea Theorem ad solve for r terms of a, a a r a ( ) r a r a r πa The, the area of the circle is π r. Hece, the ratio of the area a circle to the area of a πa π iscribed square is : a, which is :.

3 Problem ) sphere is iscribed withi a cube which has a surface area of square meters. secod cube is iscribed withi the sphere. What is the surface area of the ier cube? [Problem submitted by Vi Lee, LCC Professor of Mathematics. Source: 007 MC 0, problem 0.] Solutio: Let e be the legth of each edge of the outer cube. The 6e e. The the diameter of the sphere iscribed withi this cube must be the same as the legth of each edge of the cube, e. Now let d be the legth of the diagoal of the cube iscribed withi this sphere. The d. Let e be the legth of each edge of the ier cube. The d e ( e ) e. The surface area of this cube is 8 square meters.

4 Problem ) The first four terms of a arithmetic sequece are p, 9, p q, ad p q. What is the 06th term of this sequece? [Problem submitted by Ha Nguye, LCC Professor of Mathematics. Source: 00 MC, Problem 0] Solutio: Let d be the commo differece betwee cosecutive terms. The d 9 p d ( p q) 9 d p q 9 d ( p q) (p q) d q Use these equatios to fid p 5 ad q. So, the first term is 5 the commo differece is. Therefore, the th term is a " 5. The 06 th term is

5 Problem 5) Fid the sum 50 0 [Problem submitted by Kee Lam, LCC Professor of Mathematics. Source: Kee Lam] Solutio: Let s decompose the fractio of. That is, ) )( ( ) )( ( ) ( ) )( ( ) ( Hece, 0 ) ( implies 0 ad. Solvig the system of 0 ad, we get ad. The,

6 Problem 6) Fid k such that the sum of the squares of the roots of 0 k would be equal to 0. [Problem submitted by Kia Kaviai, LCC Mathematics Departmet Chairma. Source: Kia Kaviai] Solutio: ( ) ( ) ( ) 0 k k

7 Problem 7) Suppose a lie whose slope is is draw through the focus F of the parabola y 8( ). If the two poits of itersectio of the lie ad the parabola are ad, ad the perpedicular bisector of the chord itersects the -ais at poit P, what is the legth of the segmet PF? [Problem submitted by Vi Lee, LCC Professor of Mathematics. Source: Problem, page of Mathematical Olympiad i Chia by Xiog i ad Lee Peg Yee, East Chia Normal Uiversity Press, 007.] Solutio: The equatio is give i the stadard form ( y k) a( h) of a parabola opeig to the right with its verte at ( h, k) ad distace to the focus a. The verte of this parabola is (,0) ad the focus F ( 0,0). So, the equatio of the lie through F with slope is y. To fid the -coordiates of the poits of itersectio of this lie with the parabola, substitute for y i y 8( ) : ( ) 8( ) ± ± 8, The -coordiate of the midpoit of the chord is. The y-coordiate is. The distace from the focus to the midpoit is 8 Let E deote the midpoit. Note that the segmets FE, EP, ad FP form a 0, 60, 90 degree triagle whose legs are FE ad EP ad whose hypoteuse is FP. The hypoteuse is twice the 8 6 legth of the shorter leg,, which is the aswer to the questio.

8 Problem 8) Suppose log ( y) log ( y). Fid the miimum value of y. [Problem submitted by Vi Lee, LCC Professor of Mathematics. Source: Problem, page 8 of Mathematical Olympiad i Chia by Xiog i ad Lee Peg Yee, East Chia Normal Uiversity Press, 007.] Solutio: Sice the domai of the log fuctio is (0, ), y > 0 ad - y > 0. ddig these iequalities shows > 0. lso, the secod iequality idicates y <. The equatio may be rewritte as log ( y)( y) which implies ( y)( y) y y This is the stadard form of the equatio of a hyperbola cetered at the origi with vertices (±,0) ad asymptotes y ±. However, sice > 0, the graph of the equatio is oly the right side of the hyperbola with the sigle verte (,0). Sice the graph is symmetric about the -ais, without loss of geerality we may cosider oly the case of y 0 for which y y. Let u y ad substitute u y for i the equatio y to get ( u y) y u uy y y u uy y y uy u 0. Now use the quadratic formula to solve this equatio for y i terms of u. u ± u ( u ) y 6 u ± u y. This equatio has real solutios if ad oly if u or u. Recall from the first paragraph of this solutio that y <. Sice > > y ad > 0, y > 0. Therefore, u. So, is the miimum value of y.

9 Problem 9) How may itegers from 000 to 06 whe tripled have o eve digits? [Problem submitted by Vi Lee, LCC Professor of Mathematics. Source: February/March 05 MTYC Studet Mathematics League] Solutio: Note that whe multiplyig each of the itegers from 000 through 06 by, all of the products are digit umbers whose first digit is,, 5, or 6. Sice we are asked how may of these products have o eve digits, we eed oly cosider those whose first digit is or 5. Usig the fact that a positive iteger is divisible by if ad oly if the sum of its digits is divisible by, for digit itegers with odd digits which are divisible by ad whose first digit is, the possibilities for the last digits are: permutatio 7 permutatios 77 permutatios 5 6 permutatios 59 6 permutatios permutatio 57 6 permutatios 9 permutatios 99 permutatios 555 permutatio permutatios 777 permutatio 999 permutatio This is a total of umbers. Now cosider the last digits of positive digit itegers with odd digits which are divisible by ad whose first digit is 5: 5 permutatios permutatios 9 6 permutatios 57 6 permutatios 99 permutatios 7 permutatios 55 permutatios 79 6 permutatios 559 permutatios 577 permutatios 799 permutatios This is a total of umbers. So the aswer to the questio is 8.

10 Problem 0) Suppose a circle of radius oe is iscribed iside a equilateral triagle, at each verte there is a secod smaller circle taget to the larger circle ad two sides of the triagle, at each verte there is a third smaller circle taget to the secod circle ad two sides of the triagle, ad so o with o ed to this sequece of smaller ad smaller circles at each verte. What fractio of the triagle is occupied by the ifiitely may circles (icludig the circle of radius )? That is, epress the sum of the areas of the circles as a fractio of the area of the triagle. [Problem submitted by Vi Lee, LCC Professor of Mathematics. Source: Vi Lee] Solutio: Oriet the equilateral triagle with oe of its sides horizotal ad the opposite verte above the horizotal side. First cosider the circle of radius. Draw a lie segmet whose edpoits are the ceter of the circle ad the lower right verte of the triagle. Let be the distace from the itersectio of this lie segmet with the circle to the lower right verte of the triagle. Now draw a vertical lie segmet from the ceter of the circle to the base of the triagle. Note that this forms a 0,60, 90 triagle whose hypoteuse has a legth of ad shorter leg a legth of, Let l be the legth of the loger leg. For this triagle vertical horizotal l l l hypoteuse vertical

11 Now cosider the first (largest) of the smaller circles approachig the lower right verte. Draw a lie segmet from the ceter of this circle to the lower right verte of the triagle. Let r be the radius of this circle ad be the distace from the itersectio of the lie segmet with the circle to the lower right verte of the triagle. Now draw a vertical lie segmet from the ceter of the circle to the base of the triagle. Note that this forms a 0,60, 90 triagle the legth of whose r hypoteuse is r ad shorter leg r. For this triagle hypoteuse vertical r r r Note that r. Therefore, r. Net cosider the secod of the smaller circles approachig the lower right verte. Draw a lie segmet from the ceter of this circle to the lower right verte of the triagle. Let r be the radius of this circle ad be the distace from the itersectio of the lie segmet with the circle to the lower right verte of the triagle. Now draw a vertical lie segmet from the ceter of the circle to the base of the triagle. Note that this forms a 0,60, 90 triagle the legth of whose hypoteuse is r ad shorter leg r. hypoteuse r For this triagle r vertical r Note that r. Therefore, r. 9 So the radii of the smaller circles at this verte are,,, π π π π π The sum if their areas is 9 π π The area of all the circles withi the equilateral triagle is π π 8 8 bove we foud that the height of the equilateral triagle is ad the base is ; so, its area is. The area of the circles divided by the area of the triagle is π, which is approimately.87086, slightly less tha 6 5. r r

CALCULUS BASIC SUMMER REVIEW

CALCULUS BASIC SUMMER REVIEW NAME rise y y y Slope of a o vertical lie: m ru Poit Slope Equatio: y y m( ) The slope is m ad a poit o your lie is, ). ( y Slope-Itercept Equatio: y m b slope= m y-itercept=