aime 2015 2

download aime 2015 2

of 19

Transcript of aime 2015 2

  • 7/25/2019 aime 2015 2

    1/19

    Problem 1

    Let be the least positive integer that is both percent less than one integer and

    percent greater than another integer. Find the remainder when is divided by .

    Solution

    If is percent less than one integer , then . In addition, is

    percent greater than another integer , so . Therefore, is divisible

    by ! and is divisible by ". Setting these two e#ual, we have . $ultiplying

    by on both sides, we get .

    The smallest integers and that satisfy this are and , so .

    The answer is .

    Problem "

    In a new school, percent of the students are freshmen, percent are sophomores,

    percent are %uniors, and percent are seniors. &ll freshmen are re#uired to ta'e Latin,

    and percent of sophomores, percent of the %uniors, and percent of the seniorselect to ta'e Latin. The probability that a randomly chosen Latin student is a sophomore

    is , where and are relatively prime positive integers. Find .

    Solution

    (e see that of students

    are learning Latin. In addition, of students are sophomores learningLatin. Thus, our desired probability is and our answer is

    Problem )

    Let be the least positive integer divisible by whose digits sum to . Find .

  • 7/25/2019 aime 2015 2

    2/19

    Solution "

    The digit sum of a base integer is %ust . In this problem, we 'now ,

    or for a positive integer .

    &lso, we 'now that , or .

    *bviously is a solution. This means in general, is a solution for non+

    negative integer .

    hec'ing the first few possible solutions, we find that is the first solution that

    has , and we-re done.

    Problem

    In an isosceles trape/oid, the parallel bases have lengths and , and the

    altitude to these bases has length . The perimeter of the trape/oid can be written in

    the form , where and are positive integers. Find .

    Solution

    all the trape/oid with as the smaller base and as the longer. The point

    where an altitude intersects the larger base be where is closer to .

    Subtract the two bases and divide to find that is . The altitude can be e0pressed

    as . Therefore, the two legs are , or .

    The perimeter is thus which is .

    So

    Problem

    Two unit s#uares are selected at random without replacement from an grid of unit

    s#uares. Find the least positive integer such that the probability that the two selected unit

    s#uares are hori/ontally or vertically ad%acent is less than .

    Solution

  • 7/25/2019 aime 2015 2

    3/19

    all the given grid 2rid &. onsider 2rid 3, where the vertices of 2rid 3 fall in the centers

    of the s#uares of 2rid &4 thus, 2rid 3 has dimensions . There is a one+

    to+one correspondence between the edges of 2rid 3 and the number of ad%acent pairs of

    unit s#uares in 2rid &. The number of edges in 2rid 3 is , and the number of

    ways to pic' two s#uares out of 2rid & is . So, the probability that the two chosen

    s#uares are ad%acent is . (e wish to find the

    smallest positive integer such that , and by inspection the first such

    is .

    Problem 5

    Steve says to 6on, I am thin'ing of a polynomial whose roots are all positive integers. The

    polynomial has the form for some positive

    integers and . an you tell me the values of and 7

    &fter some calculations, 6on says, There is more than one such polynomial.

    Steve says, 8ou-re right. 9ere is the value of . 9e writes down a positive integer and

    as's, an you tell me the value of 7

    6on says, There are still two possible values of .

    Find the sum of the two possible values of .

    Solution

    (e call the three roots :some may be e#ual to one another; , , and .

  • 7/25/2019 aime 2015 2

    4/19

    (e can then subtract twice our second e#uation to get .

    Simplifying the right side>

    So, we 'now .

    (e can then list out all the triples of positive integers whose s#uares sum to >

    (e get , , and .

    These triples give values of , , and , respectively, and values of , ,

    and , respectively.

    (e 'now that 6on still found two possible values of when Steve told him the value, so

    the value must be . Thus, the two values are and , which sum to .

    Problem ?

    Triangle has side lengths , , and .

    @ectangle has verte0 on , verte0 on , and vertices and

    on . In terms of the side length , the area of can be e0pressed as the

    #uadratic polynomial

    .

    Then the coefficient , where and are relatively prime positive integers.

    Find .

    Solution 1

  • 7/25/2019 aime 2015 2

    5/19

    If , the area of rectangle is , so

    and . If , we can reflect over PA, over , and

    over to completely cover rectangle , so the area of is half the area of

    the triangle.

  • 7/25/2019 aime 2015 2

    6/19

    Similar triangles can also solve the problem.

    First, solve for the area of the triangle. . This can be done by 9eron-s Formula

    or placing an right triangle on and solving. :The side would be collinear

    with line ;

    &fter finding the area, solve for the altitude to . Let be the intersection of the altitude

    from and side . Then . Solving for using the Pythagorean Formula, we

    get . (e then 'now that .

    Bow consider the rectangle . Since is collinear with and parallel to

    , is parallel to meaning is similar to .

    Let be the intersection between and . 3y the similar triangles, we 'now

    that . Since . (e can solve for and in

    terms of . (e get that and .

    Let-s wor' with . (e 'now that is parallel to so is similar to .

    (e can set up the proportion>

    . Solving for , .

    (e can solve for then since we 'now that

    and .

    Therefore, .

    This means that .

    Solution )

    9eron-s Formula gives so the altitude from to

    has length

  • 7/25/2019 aime 2015 2

    7/19

    Bow, draw a parallel to from , intersecting at . Then in

    parallelogram , and so . learly, and are similar

    triangles, and so their altitudes have lengths proportional to their corresponding base sides,

    and so Solving gives , so theanswer is .

    Problem C

    Let and be positive integers satisfying . The ma0imum possible value

    of is , where and are relatively prime positive integers. Find .Solution

    Let us call the #uantity as for convenience. Dnowing that and are positive

    integers, we can legitimately rearrange the given ine#uality so that is by itself, which

    ma'es it easier to determine the pairs of that wor'. Eoing so, we have

    Bow, observe that

    if we have that , regardless of the value of . If , we have the

    same result> that , regardless of the value of . 9ence, we want to find

    pairs of positive integers e0isting such that neither nor is e#ual to , and that the

    conditions given in the problem are satisfied in order to chec' that the ma0imum value

    for is not .

    To avoid the possibility that , we want to find values of such that . If we

    do this, we will have that , where is greater than , and this allows us to

    choose values of greater than . &gain, since is a positive integer, and we want ,

    we can legitimately multiply both sides of by to

    get . For , we have that , so the only possibility

  • 7/25/2019 aime 2015 2

    8/19

  • 7/25/2019 aime 2015 2

    9/19

    *ur aim is to find the volume of the part of the cube submerged in the cylinder. In the

    problem, since three edges emanate from each verte0, the boundary of the cylinder touches

    the cube at three points. 3ecause the space diagonal of the cube is vertical, by the

    symmetry of the cube, the three points form an e#uilateral triangle. 3ecause the radius of

    the circle is , by the Law of osines, the side length s of the e#uilateral triangle is

    so . &gain by the symmetry of the cube, the volume we want to find is the volume

    of a tetrahedron with right angles on all faces at the submerged verte0, so since the lengths

    of the legs of the tetrahedron are :the three triangular faces touching the

    submerged verte0 are all triangles; so

    so

    Problem 1!

    all a permutation of the integers quasi-

    increasingif for each . For e0ample, and are

    #uasi+increasing permutations of the integers , but is not. Find the

    number of #uasi+increasing permutations of the integers .

    Solution

    The simple recurrence can be found.

    (hen inserting an integer n into a string with n+1 integers, we notice that the integer n has )

    spots where it can go> before n+1, before n+", and at the very end.

    GH&$PLG> Putting into the string 1")> can go before the "> 1"), 3efore the )> 1"),

    &nd at the very end> 1").

    Thus the number of permutations with n elements is three times the number of permutations

    with elements.

  • 7/25/2019 aime 2015 2

    10/19

    9owever, for , there-s an e0ception> there-s only " places the " can go :before or after

    the 1;.

    For , there are permutations. Thus for there are

    permutations.

    Problem 11

    The circumcircle of acute has center . The line passing through point

    perpendicular to intersects lines and at and , respectively.

    &lso , , , and , where and are relatively prime

    positive integers. Find .

    Solution

    all the and foot of the altitudes from to and , respectively. Let

    and let . Botice that because both are right triangles,

    and . Then, . 9owever, since is the

    circumcenter of triangle , is a perpendicular bisector by the definition of a

  • 7/25/2019 aime 2015 2

    11/19

    circumcenter. 9ence, . (e can use the Pythagorean theorem to

    find , so we have

    Li'ewise,because both are right triangles, and . 9ence,

    since as well, we have that . It follows

    that . (e add this to to get ,

    so . *ur answer is .

    Solution "

    Botice that , so . From this we get

    that . So , plugging in the given values we

    get , so , and .

    Problem 1"

    There are possible +letter strings in which each letter is either an & or a 3.

    Find the number of such strings that do not have more than ad%acent letters that are

    identical.

    Solution

  • 7/25/2019 aime 2015 2

    12/19

    all the and foot of the altitudes from to and , respectively. Let

    and let . Botice that because both are right triangles,

    and . Then, . 9owever, since is the

    circumcenter of triangle , is a perpendicular bisector by the definition of a

    circumcenter. 9ence, . (e can use the Pythagorean theorem to

    find , so we have

    Li'ewise,

    because both are right triangles, and . 9ence,

    since as well, we have that . It follows

    that . (e add this to to get ,

    so . *ur answer is .

    Solution "

  • 7/25/2019 aime 2015 2

    13/19

    Botice that , so . From this we get

    that . So , plugging in the given values we

    get , so , and .

    Problem 1)

    Eefine the se#uence by , where represents radian

    measure. Find the inde0 of the 1!!th term for which .

    Solution 1

    If , . Then if satisfies , , and

    Since is positive, it does not affect the sign of .

    Let . Bow

    since an

    d , is negative if and only

    if , or when . Since is irrational, there is

    always only one integer in the range, so there are values of such that

    at . Then the hundredth such value will be when

    and .

    Solution "

    Botice that is the imaginary part of , by Guler-s formula.

  • 7/25/2019 aime 2015 2

    14/19

    (e only need to loo' at the imaginary part, which is

    Since , , so the denominator is positive. Thus, in order for the

    whole fraction to be negative, we must

    have . This only holds

    when is between and for integer continuity proof hereJ, and since this has

    e0actly one integer solution for every such interval, the th such is .

    Problem 1

    Let and be real numbers satisfying and .

    Gvaluate .

    Solution

    The e0pression we want to find is .

    Factor the given e#uations as and ,

    respectively. Eividing the latter by the former e#uation yields .

    &dding ) to both sides and simplifying yields . Solving for and

    substituting this e0pression into the first e#uation yields . Solving for ,

    we find that , so . Substituting this into the second e#uation and

    solving for yields . So, the e0pression to evaluate is e#ual

    to .

    Solution "

    Factor the given e#uations as

    and , respectively. 3y the first

    e#uation, . Plugging this in to the second e#uation and simplifying

  • 7/25/2019 aime 2015 2

    15/19

    yields . Bow substitute . Solving the #uadratic in , we

    get or &s both of the original e#uations were symmetric in and ,

    (L*2, let , so . Bow plugging this in to either one of the e#uations, we get

    the solutions , . Bow plugging into what we want, we

    get

    Solution )

    &dd three times the first e#uation to the second e#uation and factor to

    get . Ta'ing the cube root

    yields . Boting that the first e#uation is , we

    find that . Plugging this into the second e#uation and dividing

    yields . Thus the sum re#uired, as noted in Solution 1,

    is .

    Problem 1

    ircles and have radii and , respectively, and are e0ternally tangent at point .

    Point is on and point is on such that is a common e0ternal tangent of the

    two circles. & line through intersects again at and intersects again at .

    Points and lie on the same side of , and the areas of and are

    e#ual. This common area is , where and are relatively prime positive integers.

    Find .

  • 7/25/2019 aime 2015 2

    16/19

    9int

    This is a K1 on an &I$G, so it must be difficult. Indeed, there are two possible approaches

    :both of them very computational;> coordinate geometry, or regular Guclidean geometry

    combined with a bit of trigonometry.

    Solution 1

  • 7/25/2019 aime 2015 2

    17/19

    all and the centers of circles and , respectively, and call and the feet ofthe altitudes from to and to , respectively. G0tend and to

    meet at point .

  • 7/25/2019 aime 2015 2

    18/19

    Eiscarding ,

    the +coordinate of is . The distance from to is

    then The perpendicular distance from to or the

    height of is Finally, the common area

    is , and .

    Solution "

    3y homothety, we deduce that . :The proof can also be e0ecuted by similar

    triangles formed from dropping perpendiculars from the centers of and to .; Therefore,

    our e#uality of area condition, or the e#uality of base times height condition, reduces to the

    fact that the distance from to is four times that from to . Let the distance from

    be and the distance from be .

    Let and be the centers of their respective circles. Then dropping a perpendicular

    from to creates a right triangle, from which and, if ,

    that . Then , and the Law of osines on triangles

    and gives and

    Bow, using the Pythagorean Theorem to e0press the length of the pro%ection of onto

    line gives S#uaring and simplifying gives

    and s#uaring and solving gives

    3y the Law of Sines on triangle , we have 3ut we 'now , and

    so a small computation gives The Pythagorean Theorem now gives

  • 7/25/2019 aime 2015 2

    19/19

    and so the common area

    is The answer is