Joe Holbrook Memorial Math Competitio 8th Grade Solutios October 5, 07. Sice additio ad subtractio come before divisio ad mutiplicatio, 5 5 ( 5 ( 5. Now, sice operatios are performed right to left, ( 5 ( 5 7.. There are 000 meters i a kilometer, so it takes Malik. 5000. 50.5 secods to fiish 00 the race.. Let be the umber of shots he must make to reach his goal. We have the iequality 7 0.7, or 5 5. Sice must be a positive iteger, the miimum valid value for is.. Let, y be the umber of heads ad tails, respectively. The coditio fails if ad oly if < 500 ad y < 500. Summig the iequalities yields y < 000, which is a cotradictio. Thus, the probability that this occurs is 00%, so 00. 5. The factors of are:,,,, 7,,, ad. They pair up ad multiply to. There are pairs. Therefore, their product is, so.. By the idetity lcm (a, b gcd (a, b ab, it follows that ab 0 0 00. 7. Sice the sequece starts off with odd terms, every third term will be eve. Thus, all the terms that are multiples of three are the oes we are lookig for. Amog the first 00, thee are multiples of. 7 8. Betwee ay two circle, there ca at most eist itersectios. Therefore, we have. Betwee ay 7 two lie, there ca at most eist itersectio. Therefore, we have aother itersectios. Betwee a lie ad a circle, there ca eist itersectio, for a additioal we have 7 itersectios. This yields a total of 98 itersectios. 9. I those 5 umbers, there must be at least 5 s i order to make it the uique mode. I order to miimize the average of the smallest umbers, you wat ad to be part of the set (you ca t have s sice that would mea that 5 is ot the uique mode. Now, give these umbers, you eed i order to make the average 5. 0. There are ways to choose socks of the same color, ad there are differet colors. Meawhile, there are ( 8 ways to choose socks. Therefore, the probability is ( ( 8 5/8.. Lookig at the hudredth place, the two digits have to be 9 ad. Now, if the teth place of the subtrahed is, the smallest it ca be is 0, ad the miued would have to be more tha 89 0 00. Thus the teth place of the subtrahed is 0, ad so is the product of the digits.. Let the foot of the altitude from A oto side MN, be O. We get OM BM ad ONND. These two equality coditios lead to two sets of cogruet triagles, thus MAN, which has half of both sets, has a degree measure of 5 degrees.. The ceter of the square ca be located iside the square of ich iside each tile. Therefore, the probability is /.. Chord BP is a diameter of Ω as poit O lies o BP. As the radius of Ω is, the legth of BP is, ad the agle BCP 90, applyig the Pythagorea Theorem o BCP yields BC P C BP. Pluggig i the give values yields BC 7 BC 7.
8th Grade Solutios Joe Holbrook Memorial Math Competitio October 5, 07 5. As AB 8 ad BC AD, by the Pythagorea Theorem, it follows that BD 0. As M is the midpoit of BD, DM must be 5. As AN is the height to the base BD i triagle ABD, AN BD AB AD AN 5. By the Pythagorea Theorem i triagle AND, DN AD AN DN 8 5 MN 7 5.. Set side BC to be of legth. The side CD is of legth 7.5, sice the sum of the two sides is half the perimeter. The, (7.5, solvig gives us 0. Thus the area is 0 80. 7. Note that 8 7, 0 5, ad 7. The, takig the greatest commo divisor of each pair of the three values yields, 7, ad. Sice if a iteger divides a correspodig pair of itegers, the it must divide the greatest commo divisor of the itegers, we the sum the total umber of divisors of each of the gcds:, 8, ad, so 8 8 total factors. However, we are overcoutig the umbers that divide all three values. Sice the gcd of all three is, we overcouted divisors twice, for a total of 8 overcouted divisors. Thus, our aswer is 8 8 0. 8. Clearly there are 00 total outcomes. Let m be a positive iteger. We will cout the umber of poits Bessie ca choose i order to have a lie with slope m, ad sum over all positive itegers m. By defiitio of slope, the poits Bessie chooses must be of the form (k, km for some positive iteger k. Sice we must have km 0, we get k, or valid slopes. Notig that Bessie 0 0 m m caot choose ay poit such that the lie has a slope greater tha 0, we sum over the first 0 positive itegers to get 0 5 7 valid lattice poits. Dividig over the total 7 umber of cases gives the desired aswer of 00. 9. The coditio is a b b ab b, which simplifies to (a b(a b 0. Sice a ad b are differet, a b 0. f( is a b, so the desired value is. 0 0. There are 90 highways total. Cosider how may highways that are eeded i order to coect every tow. Without loss of geerality, let tow A ad B be coected. I order for tow C to be coected to both of them, tow C oly eeds to be coected to A or B, but ot both. Repeat for all the remaiig tows, ad you will get 9 required highways. Therefore, 7 of them ca be closed.. The diagoal of the kite that bisects the kite must be a diameter, so it has legth 0. This forms a right triagle with hypoteuse 0 ad a leg with legth, so the other leg must have legth 8. y y z z. We claim that is the umeric value for the largest positive value of the differece betwee terms of, y, z. To see this, cosider WLOG that y z. The y y z z y z y z z (z, ad the result follows. Hece, the maimum possible value is 0/ 0. This ca be attaied by lettig 0, y 0, z 0.. There are 0 ways to choose edges from a cube, which has edges. The edges ca be grouped based o their orietatios, with each group cotaiig parallel edges. Sice the edges have to be pairwise skew, it is imperative to choose oe edge from each group. Choose oe edge at radom from ay of the groups. There are possibilities. Two edges from each of the other two groups itersect the selected edge, so choose ay of the other two groups ad a edge at radom. There are possibilities. There is oly oe edge left from the third group. Thus there are 8 possibilities i total to make the desired -edge selectio, ad the probability is 55.. By power of a poit, we kow P B P C P A.Sice AC is a diameter, ad P A is a taget, we kow that AC P A. Thus we ca use the Pythagorea theorem to solve for P C. We have P A AC P C. Substitutig P B P C P C for P A, ad rearragig, we get P C P C 5 0. Solvig for P C ad takig the positive root we get 5.
8th Grade Solutios Joe Holbrook Memorial Math Competitio October 5, 07 5. Sice 0 if k <, we oly cosider the terms for which k. Thus, the desired sum is k ( From the Biomial Theorem, Hece, the desired sum is equivalet to ( ( (, 0, 0. 0 ( ( (. 7 7 7 8 7. There are possible distict rolls, so the epected value of is y.. We aim to fid may perpedicular lies. EG is ormal to BDHF. Call the itersectio poit I. Now we fid the distace from poit I to BH. That distace is clearly half of the height from poit F to side BH i the triagle BHF, which is. The aswer is thus. 7. P (gets as may heads as tails umber of ways to get strictly more heads tha tails umber of ways to get the same umber of heads ad tails total umber of possible coi flip sequeces ( 55 08. 8. Pick,,..., 99 arbitrarily, for a total of 00 99 possibilities. Let the remaider of their sum is divided by 00 be S. The 00 must be equal to 00 S, which is uiquely determied for each choice of,,, 99. 8 9. There are ways of choosig where the 0 s go. The, there are ways of choosig where the goes. Now, there are spots left. Oe of the s has to go i the rightmost ope spot i order to be pure. The, there are three ope spots left for, 7, ad 7. There are ways of choosig spots for the 7 s. ( 8 Sice these steps are doe i order, ivokig product rule yields 0. 0. a 0 is a solutio. Otherwise, we ca divide by a to get (a (a a 0. This must be a perfect square, or else there would be solutios, so (a a a, which gives a, 5, so the solutios are a 0,, 5.. As 0,. Thus, 00 ( ad 00 (, so 00 00 0.
8th Grade Solutios Joe Holbrook Memorial Math Competitio October 5, 07. It is evidet that the circle with radius X is surrouded by rest of the circles. Coectig the pairwise ceters of the circles with kow radius, the side legths of the triagle formed are, 7, 7. Droppig the altitude to the base of this isosceles triagle, the height of the triagle to the base is 7 8 5 by Pythagoras. Coect the vertices of the triagle to the ceter of the circle with radius X. By usig pythagorea theorem, we kow that the (8 X 8 X 9 5. Simplifyig this formula gives X X X. Squarig both sides yields X X X X 8X X 9 7.. Let X deote the evet that the ball selected is blue ad let Y deote the evet that the ball selected is from Bag B. By defiitio of coditioal probability, Pr[Y X] Pr[X ad Y ] Pr[X] 7 8 7 5.. Let the ceter of the circle be O. WLOG, we cosider OP Q. Let the foot of the perpedicular from O to BC be M, so BM BQ. Sice OP 5, from right triagle OBM, we have OM. Idetically, if the feet of the perpediculars from O to BC ad CA are L ad N, the OL ON. Thus, O is the iceter of ABC. Let AB. Recall that the area of ABC is rs, where r is the iradius ad s is the semiperimeter, ad is also. Thus, ( [ ABC] rs ( 57, Hece, AB 0. [ ABC], ( 57, 57, 57 5 57 5 0 57, 0, 0, 0. 5. Let AB ad CD y. We wat to fid the legth of MN y. Cosider the lie l through B parallel to AC. Let l itersect CD at E. The, BE AC ad CE AB. Sice AC BD ad AC BE, we also have BE BD. Thus, by Pythagorea theorem o right triagle EBD, we have that ( y 0 00 7, so y. Hece, MN y.. Note that 0! 08 09 00 0 0, 07! so 00 0! 0!, which implies that 0 07! 07!, so,, 80, AB, 8, 080 0 (mod 0. Sice 00 (mod 0, this implies that,, 80, AB, 8, 080 is cogruet to ( 8 ( 7 ( 8( 5 ( (AB( ( 80( 80( 0 8 AB 80 80 AB 0 0 (mod 0. Thus, AB 0 9 (mod 0, ad sice A ad B are digits, this implies that AB 9. Hece, 0A B 9.
8th Grade Solutios Joe Holbrook Memorial Math Competitio October 5, 07 7. We epress the degree polyomial P ( as ( ( ( (, so P (5 0. 8. First, you wat to fid slowest the eastboud trai ca be goig for it to ot crash with the orthboud oe. I order for that to happe, the eastboud trai has to hit the tail of the orthboud oe. The orthboud trai will be completely clear of the itersectio i: (00 5/50 9/ hours. This meas that the head of the eastboud trai has to be there at this time, which leads to the miimum speed beig: 00/(9/ 00/. For the maimum speed, the orthboud trai has to hit the tail of the eastboud trai. The time it will take for the tail of the eastboud trai to reach the itersectio is (00 50/s, where s is the speed. We wat this to equal 00/50, the time it takes for the orthboud trai to reach the itersectio. Therefore, solve, ad get that s is 75/. Therefore, take their differece, ad you get 5/. 9. Let E be the epected value of P. The epected value of ay roll of the die is 5 so a b c d 7. It follows that E( 7, E( 9, E( 7, E( 9. 7, Sice P has degree at most, E does as well, so its third fiite differeces are costat. its first fiite differeces are Note that E( E(, E( E(, E( E(, E(5 E(,..., its secod fiite differeces are E( E( E(, E( E( E(, E(5 E( E(,..., so its third fiite differeces are E( E( E( E(, E(5 E( E( E(,... Thus, E( E( E( E( E(5 E( E( E(, E(5 E( E( E( E( 9 9 E(5 (7 7 9 9 7 9. 0. Defie the fuctio f( to retur the umber of iversios i the sequece whe the sequece has terms. Parsig the recurrece, we ca see that the terms a, a,, a is formed by takig the first terms of the sequece, reversig their order, ad addig. By uiqueess of biary represetatio, all umbers are distict. We shall ow epress f( i terms of f(. Amog the first half of the sequece {a i } i, by defiitio, there are f( iversios. Now, ote that if (a i, a j was a iversio i the first half, (a i, a j will ot be a iversio (sice the order was reversed, ad vice versa. Therefore, amog the secod half, there are eactly f( iversios. Fially, it ca be see that the umbers i the first half of the sequece are betwee 0 ad iclusive, so each umber i the secod half will be strictly greater tha ay umber i the first half, yieldig o iversios across the two halves of the sequece. We may thus coclude from the previous paragraph ( that f( f(. f( (, so we may coclude that f( ( 5