Solutions to the 59th William Lowell Putnam Mathematical Competition Saturday, December 5, 1998 Manjul Bhargava, Kiran Kedlaya, and Lenny Ng A–1 Consider the plane containing both the axis of the cone and two opposite vertices of the cube’s bottom face The cross section of the cone and the cube in this plane consists of a rectangle of sides and ✂✁ ✄ inscribed in an isosceles triangle of base ✄ and height ☎ , where is the side-length of the cube (The ✂✁ ✄ side of the rectangle lies on the base of the triangle.) Similar triangles yield ✂✆ ☎✞✝✠✟☛✡✌☞ ✂✁ ✄✍✆✎✄✎✏✑✆ ✡ , or ✝✒✟✔✓ ✁ ✄ ☞✖✕ ✏✗✆✍✘✚✙ A–2 First solution: to fix notation, let ✛ be the area of region ✜✣✢✞✤✦✥ , and ✧ be the area of ✜✣✢✩★✍✪ ; further let ✫ denote the area of sector ✬✭✜✣✢ , which only depends on the arc length of If ✮ ✯✱✰✳✲✵✴ denotes the area of tri✫✹✸ angle ✮ ✯✶✰✳ ✌ ✲ ✴ , then we have ✷ ✛ ✝ ✮ ✭ ✬ ✞ ✢ ✥✺✴✻☞✹✮ ✬✭✜✣✤✺✴ ✫✽✸ ✮ ✬✭✜✣✪✾✴✿☞❀✮ ✬✭✢✞★❁✴ But clearly ✮ ✬✭✢✞✥✺ and ✧✼✝ ✸ ✫ ✴❂✝ ✮ ✬✭✢✞★✎✴ and ✮ ✬✭✜✣✤✺✴❃✝✠✮ ✬✭✜✣✪❄✴ , and so ✛ ✧❅✝ ✄ D H E I O F G Second solution: We may parametrize a point in by any of ❆ , ❇ , or ❈❉✝❋❊✑●❁❍❏■❃❑✂✟▲❇ ✆ ❆ ✏ Then ✛ and ✧ are just the integrals ✸ of ❇✵▼✎❆ and ❆◆▼✎❇ over the appropriate intervals; thus ✛ ✧ is the integral of ❆❖▼✍❇✺☞❇◗▼✎❆ (minus because the limits of integration are reversed) But ✸ ▼✎❈✩✝❘❆❖▼✎❇✩☞✶❇✵▼✎❆ , and so ✛ ✧❅✝❚❙✳❈ is precisely the radian measure of (Of course, one can perfectly well this problem by computing the two integrals separately But what’s the fun in that?) A-3 If at least one of ❯❱✟▲❲ ✏ , ❯❨❳❩✟▲❲ ✏ , ❯❨❳ ❳✔✟✔❲ ✏ , or ❯❨❳ ❳ ❳❩✟▲❲ ✏ vanishes at some point ❲ , then we are done Hence we may assume each of ❯❱✟▲❆ ✏ , ❯❨❳❩✟❬❆ ✏ , ❯❨❳ ❳❩✟❬❆ ✏ , and ❯❨❳ ❳ ❳❩✟❬❆ ✏ is either strictly positive or strictly negative on the real line By replacing ❯❱✟❬❆ ✏ by ☞❖❯❱✟❬❆ ✏ if necessary, we may assume ❯❨❳ ❳❩✟▲❆ ✏✠❭❫❪ ; by replacing ❯❱✟❬❆ ✏ by ❯❱✟❴☞✌❆ ✏ if necessary, we may assume ❯❨❳ ❳ ❳❩✟❬❆ ✏✦❭❚❪ (Notice that these substitutions not change the sign of ❯❱✟❬❆ ✏ ❯❨❳✔✟▲❆ ✏ ❯❨❳❵❳✔✟▲❆ ✏ ❯❨❳❵❳ ❳❩✟❬❆ ✏ ) Now ❯❨❳ ❳❩✟❬❆ ✏✩❭✠❪ implies that that ❯ ❳ ✟❬❆ ✏ ❯ ❳ ✟❬❆ ✏ is increasing, and✸ ❯ ❳ ❳ ❳ ✟❬❆ ✏✦❭✒❪ implies ✸ ✦ ✏ ❭ ✏ is convex, so that ❯❨❳✔✟▲❆ ❲ ❯❨❳✔✟▲❆ ❲❛❯❨❳ ❳❩✟▲❆ ✏ for all ❆ and ❲ By letting ✸ ❲ ✏ increase in the latter inequality, ❲ must be positive for sufficiently we see that ❯❨❳✔✟▲❆ large ❲ ; it follows that ❯❨❳✔✟▲❆ ✏❜❭❝❪ for all ❆ Similarly, ❯❨❳✔✟▲❆ ✏✩❭✒❪ and ❯❨❳ ❳❩✟❬❆ ✏✞❭✒❪ imply that ❯❱✟❬❆ ✏✦❭✒❪ for all ❆ Therefore ❯❱✟▲❆ ✏ ❯❨❳✔✟▲❆ ✏ ❯❨❳❵❳❩✟❬❆ ✏ ❯❨❳❵❳ ❳❩✟❬❆ ✏◆❭❘❪ for all ❆ , and we are done A-4 The number of digits in the decimal expansion of ✛◆❞ is the Fibonacci number ✤❡❞ , where ✤ ✝❢✡ , ✤❤❣❄✝✐✡ , ✸ ❑ and ✤❤❞❥✝❢✤❤❞ ✤❤❞ ❣ for ❦ ❭❧✄ It follows that ■❃❑ ■ the sequence ♠♥✛ ❞♣♦ , modulo 11, satisfies the recursion ✸ ❴ ✏ ✎ q ❁ r ✎ s t ✛ ❞ ✝❝✟❴☞✦✡ ✛ ❞ ✛ ❞ ❣ (Notice that the recur■❃❑ ■ sion for ✛ ❞ depends only on the value of ✤ ❞ ❣ modulo ■ 2.) Using these recursions, we find that ✛◆✉✖✈ ❪ and ✛❖✇✞✈①✡ modulo 11, and that ✤ ✉ ✈✼✡ and ✤❤✇✞✈①✡ modulo It follows that ✛❖❞✼✈②✛◆❞❁③❏④ (mod 11) for all ❦⑥⑤⑦✡ We find that among ✛ ✛❖❣ ✛❖⑨ ✛❖⑩ ✛◆❶ , and ❑✂⑧ ⑧ ⑧ ⑧ vanishes✸ modulo 11 Thus 11 divides ✛◆❞ ✛❖④ , only ✛ ❑ if and only if ❦✹✝❚✕❸❷ ✡ for some nonnegative integer ❷ A–5 Define the sequence ✜❜❹ by the following greedy algorithm: let ✜ be the disc of largest radius (breaking ties ❑ arbitrarily), let ✜ ❣ be the disc of largest radius not meeting ✜ , let ✜ ⑨ be the disc of largest radius not meeting ❑ ✜ or ✜ ❣ , and so on, up to some final disc ✜ ❞ To see ❑ that ✢❧❺①❻ ❼✗❞ ❽ ☎✍✜ ❼ , consider a point in ✢ ; if it lies in ❑ one of the ✜❜❹ , we are done Otherwise, it lies in a disc ✜ of radius ❾ , which meets one of the ✜❜❹ having radius ⑤❢❾ (this is the only reason a disc can be skipped in our ✸ algorithm) Thus the centers lie at a distance ❿✾➀ than ❾ ❾ , and so every point at distance less ✸ ❿➁ ➀ ☎ from the center of ✜ lies at distance at most ❾ from the center of the corresponding ✜ ❹ ✫ ❣ ✫ ❣ ✸ ➂ ✧ ✫ ➂ ⑤ ✄ ➂ ✛❖✧➂➄➂ ✧ ➂ A–6 Recall the inequalities ➂ ✫ ✛◆✧➃➂ ➂❱⑤ ✄ ✮ ✛❖✧ ✴ (Law of Sines) (AM-GM) and ➂ ✛❖✧➂➄➂ ✧ Also recall that the area of a triangle with integer coordinates is half an integer (if its vertices lie at ✏ ✟▲❾ ✻✏ , the area is ➂ ➅ ☞ ✟ ❪ ❪✍✏ ✟➆➅ ❾❛➂ ✆❁✄ ), and that ⑧ ⑧ ⑧✗➇ ⑧ ⑧ ➇ if ✛ and ✧ have integer coordinates, then ➂ ✛❖✧➂ ❣ is an integer (Pythagoras) Now observe that ✫ ✫ ❣ ✸➋➊ ✫ ➈ ❣ ✸ ✮ ✛◆✧ ✴❂➉✒➂ ✛❖✧➃➂ ➂✧ ➂ ✮ ✛◆✧ ✴ ✫ ❣ ✸ ✄ ✫ ❣ ✸ ➉✒➂ ✛❖✧➃➂ ➂✧ ➂ ➂ ✛◆✧➃➂➌➂ ✧ ➂ ✫ ✸ ➀ ➈ ✮ ✛❖✧ ✴ ✡ ⑧ and that the first and second expressions are✸ both✫ inte✫ ➈ ❣ ❣ ✸ gers We conclude that ✮ ✛❖✧ ✴➍✝➎➂ ✛❖✧➂ ➂✧ ➂ ➊ ✫ ✫ ❣ ✫ ❣ ✸ ➂✧ ➂ ✝ ✄ ➂ ✛◆✧➃➂➌➂ ✧ ➂➏✫ ✝ ➊ ✮ ✛❖✧ ✫ ✴ , and so ➂ ✛❖✧➃➂ ✮ ✛❖✧ ✴ ; that is, ✧ is a right angle and ✛❖✧➐✝⑦✧ , as desired B–1 Notice that ✸ ④ ④ ✸ ❬✟ ❆ ✡ ✆ ❆ ✏ ☞❀✟❬❆ ✡ ✸ ✸ ✸ ✆ ✏ ⑨ ⑨ ✟❬❆ ✡ ❆ ✟▲❆ ✸ ⑨ ⑨ ✸ ⑨ ✟▲❆ ✡ ✆ ❆ ✏ ☞❀✟❬❆ ✡ ✆ ❆ Piecing our various cases together, we easily deduce that ➺✌✟▲➻ ❦ ✏ ✝ ❪ if and only if the highest powers of ⑧ dividing ➻ and ❦ are different B–5 Write Ơ➙✝✠✟☛✡ ❪ ❑☛Õ✑Õ ✇ ☞Ư✡ ✑✏ ✆ ✓ Then ✆ ❆ ④ ✏ ☞ ✄ ✝ ✡ ✆ ❆ ⑨ ✏ ✏ ✝✷☎➑✟❬❆ ✸ ✡ ✆ ❆ ✏ ✁ Ô➎✝ ✝ (difference of squares) The latter is easily seen (e.g., by AM-GM) to have minimum value (achieved at ❆➃✝✒✡ ) ✡ ❪ ✗Õ Õ✑Õ ➛ ✇ ✡✌☞×✡ ❪ ❃ ■ ❑☛Õ✑Õ ☎ ✡ ❪ Õ✗Õ✑Õ ✡ ✇ ✸ ■ ❑❴Õ✑Õ ✟❴✡❖☞ ✄ ✡ ❪ ❏ ☎ ❾ ✏ ⑧ ➀ ❣ ✡ ❪ ■ ➞✑➞✑➞ Now the digits after the deciwhere ❾ ❪ mal point of ✡ Õ✗Õ✑Õ ✆ ☎ are given by ✙ ☎✎☎✎☎✍☎ ✙➤✙➠✙ , while the digits after the decimal point of ④❑ ✡ ❪ ■♣Õ✑Õ✑Õ are given by ✙ ❪✍❪✎❪✍❪✎❪✌✙➠✙➤✙ ✡➠✕✍✕✎✕✎✕✍✕✎✕ ✙➠✙➤✙ It follows that the first 1000 digits of ✁ Ô are given by ✙ ☎✎☎✎☎✍☎✎☎ ✙➠✙➤✙ ☎✎☎✍☎➝✡ ; in particular, the thousandth digit is ✡ ✸❉Ø ⑨ ✸ ❣ ✸ B–6 First solution: Write Note ➅❡✟❬❦ ✏ ✝❚❦ ❲❸❦ ❦ ✸ ➓ ✄✍✏ have the same parity, and recall that ➅❡✟▲❦ ✏ and ➅❡✟▲❦ B–2 Consider a triangle as✫ described by the problem; label its vertices ✛ ✫ ✧ so that ✛❧✝➒✟✔❲ ✏ , ✧ lies on ⑧ ⑧ ⑧✑➓ the ❆ -axis, and lies on the line ❇➔✝❫❆ Further ✏ let ✜→✝➣✟✔❲ ☞ be the reflection of ✛ in the ❆ -axis, ⑧ ➓ and let ✢↔✝↕✟ ❲ ✏ be the reflection ✫ of ✛ ✫ in the line ➓✻⑧ ❇✒✝➣❆ Then ✛❖✧❫✫ ✝②✜✣✧ and ✛ ✢ , and so ✸ ✫❚✸✽✝ ✫ the of ✛❖✧ is ✜✣✧ ✧ ✢➙⑤✒✜✣✢➎✝ ➛ perimeter ✸ ✸ ✏❣ ✸ ✄ ❣ It is clear that ✟✔❲✞☞ ✏ ❣ ✟▲❲ ✝ ✁ ✄ ❲ ❣ ✫ ➓ ➓ ➓ this lower bound can be achieved; just set ✧ (resp ) to be the intersection between the segment ✜✣✢ and the ❆ -axis (resp line ✝✒❇ ); thus the minimum perimeter ✸ ❆✶ ✄ ❣ is in fact ✁ ✄ ❲ ❣ ➓ B–3 We use the well-known result ✸ that the surface area of ❣ ✸ ❣ ✏ ➂❴❆ ❣ the “sphere cap” ♠❸✟▲❆ ❇ ❇ ✝✠✡ ⑤ ♦ ⑧✑➜ ➜ ⑧➝➜ ➜♥➞ ✏ ⑧ (This is simply ✄✂➟ ✟❴✡✿☞ result is easily verified using ➜♥➞ calculus; we omit the derivation here.) Now the desired surface area is just ✄✂➟ minus the surface areas of five identical halves of sphere caps; these caps, up to isometry, correspond to being the distance from the center ➜➠➞ ✝➒➡➤➢✍➥➧➦ ❶ of the pentagon to any of its sides, i.e., ➜ ➞ ✄✂➟ ☞ ❶ ✄✂➟ ✟☛✡✌☞✖ Thus the desired area is ➤ ➡ ✍ ➢ ➥➧➦ ❶ ✏✗➩ ✝ ✩ ❣ ➨ ➫ ➟ ➟ ❡ ➟ ✎ ✆ ✄ ) ➡➭➢✍➥ ➦ ❶ ☞➋☎ (i.e., ✧❅✝ ❹ ✸ ❹ B–4 For convenience, define ❯✂➯◆➲ ❞❨✟❬➳ ✏ ✝➔➵ ➯✳➸ ➵ ❞❃➸ , so that ➯✵❞ the given sum is ➺✌✟❬➻ ❦ ✏ ✝➣➼ ❹ ❽ ■❏❑ ✟☛☞✦✡ ✏☛➽☛➾❱➚ r❸➪ ❹➌➶ If ⑧ ➻ and ❦ are both odd, then ➺➹✟❬➻ ❦ ➞ ✏ is the sum of an ⑧ odd number of ➘✳✡ ’s, and thus cannot be zero Now ➻ and ❦ have opposite parconsider the case where ❹ ✸ ❹➌③ ➵✔❷➋☞ ity Note that ➵ ➯ ➸ ➯ ❑ ➸ ✝➴❷➷☞➬✡ for all ✸ ❹ ➯✵❞ ❹ integers ➳ ❷ ➻ Thus ➵ ➯ ➸ ➵ ➯ ■ ■❃❑ ➸ ✝❫❦✽☞⑥✡ ❹ ✸ ⑧ ⑧ ➯✵❞ ❹ ■ ❞ ■❏❑ ➸ ✝➮➻➱☞✃✸ ✡ ; this implies that ➵ and ➵ ❞ ➸ ✸ ❯✂➯◆➲ ❞♣✟▲➳ ✏ ❯✂➯◆➲ ❞♣✟▲➻➃❦✭☞❜➳❸☞✡ ✏ ✝❀➻ ❦✭☞ ✄ is odd, and so ❹➌➶ ➯✵❞ ❹ ➶ ❸ r ➪ ❸ r ➪ ✗ ✏ ❴ ➽ ❱ ➾ ➚ ✗ ✏ ❴ ➽ ❱ ➾ ➚ ✟❴☞✦✡ ✝✒☞✞✟☛☞✦✡ ■ ■❏❑ for all ➳ It follows that ➺✌✟❬➻ ❦ ✏ ✝ ❪ if ➻ and ❦ have opposite parity ⑧ Now suppose that❼ ➻➣✝ ✄ ❷ and ❦✽✝ ✄❁❐ are both even ❼❣ ❣ ③ Then ➵ ❣✑➯ ➸ ✝❢➵ ❣✗➯ ❑ ➸ for all ❒ , so ➺ can be computed as twice the sum over only even indices: ❣❰Ð ❮❰Ï ■❏❑ ✸ ❮Ó③❏Ï ✏Ó✙ ➽☛Đ ➚ Ị ➪ ❹➌➶ ✄ ✎ ✄ ✔ ❐ ✏ ✄ ➺✌✟ ❷ ✝ ✟☛☞✦✡ ✏ ✝✷➺✌✟❩❷ ❐❩✏ ✟☛✡ ✟❴☞✦✡ ✏ ⑧ ⑧ ❽❹ ➞ ✄ ❁ ✄ ❩ ❐ ✏ Thus✸ ➺➹✟ ❷ vanishes if and only if ➺✌✟❩❷ ❐❩✏ vanishes ⑧ ✏ ❮Ó③❏Ï ⑧ (if ✡ ✟❴☞✦✡ ✝ ❪ , then ❷ and ❐ have opposite parity ✔ ❐ ✏ and so ➺✌✟✔❷ also vanishes) ⑧ that any perfect square✸ is congruent to or (mod 4) ✄✍✏ are perfect squares, they are Thus if ➅❡✟▲❦ ✏ and ➅❡✟▲❦ ✸ ✄✎✏ ❣ ✸ ✄ congruent mod But ➅❡✟❬❦ ☞➹➅❡✟▲❦ ✏ ✈ ✄ ❦ (mod ➓ 4), which is not divisible by if ❦ and have opposite ➓ parity Second solution: We prove more generally that for any polynomial Ù➏✟ ✏ with integer coefficients which is not ➜ a perfect square, there exists a positive integer ❦ such that Ù➏✟❬❦ ✏ is not a perfect square Of course it suffices to assume Ù➏✟ ✏ has no repeated factors, which is to say ➜ Ù➏✟ ✏ and its derivative Ù✦❳❩✟ ✏ are relatively prime ➜ ➜ In particular, if we carry out the Euclidean algorithm on Ù➏✟ ✏ and Ù✦❳✔✟ ✏ without dividing, we get an in➜ ➜ teger ✜ (the discriminant of Ù ) such that the greatest common divisor of Ù➏✟▲❦ ✏ and Ù✦❳❩✟❬❦ ✏ divides ✜ for any ❦ Now there exist infinitely many primes ➅ such that ➅ divides Ù➏✟❬❦ ✏ for some ❦ : if there were ✙➤✙➠✙ ➅ ❮ , then for any ❦ dionly finitely many, say, ➅ ⑧ ❑ ⑧ ✍ ❪ ✏ ✝ÚÙ➏✟ ➅ ➅ ❣ÜÛ➠Û➤Û ➅ ❮ , we have Ù➏✟❬❦ ✏ ✈ visible by ➻ ❑ ❸ ❪ ✏ ✏ ✑ ✏ Ù➏✟ ✟▲Ý❜➢✚Þ➻ , that is, Ù➏✟❬❦ ✆ Ù➏✟ ❪✍✏ is not divisible ✙➠✙➤✙ ➅♣❮ , so must be ➘✳✡ , but then Ù takes some by ➅ ❑✻⑧ ⑧ value infinitely many times, contradiction In particular, we can choose some such ➅ not dividing✸ ✜ , and choose✸ ❦ such that ➅ divides Ù➏✟▲❦ ✏ Then Ù➏✟❬❦ ❷✂➅ ✏ ✈ ✏ íòịĩ (write out the Taylor series of the left side); in particular, since ➅ does✸ not divide Ù✦❳✔✟▲❦ ✏ , we can find some ❷ such that Ù➏✟▲❦ ❷✂➅ ✏ is di❣ visible by ➅ but not by ➅ , and so is not a perfect square Third solution: (from David Rusin, David Savitt, and ⑨ ✸ ❣ ✸ ❦ ❸ ❲ ❦ Richard Stanley independently) Assume that ✸❀Ø is a square for all ❦ ❭✒❪ For sufficiently large ❦ ➓ ❦ , ✟❬❦ ⑨❰à✑❣ ✸ ✡ à✑❣ ❣ ➀ ❲ ❦ ❑ ☞Ö✡ ✏ ✄ ❸ ➀ ⑨ ✸ ❦ ✟▲❦ ❲❸❦ ⑨✑à✑❣ ✸ ❣ ✸ ➓ ❦ ✸❉Ø ✡ à✑❣ ✸ ❲ ❦ ❑ ✄ ❸ ❣✎á ✡ ✏ ⑨ ✸ thus✸ if ❦ ✸✺isØ a large even perfect square, we have ❦ ✸ ❣ ⑨❰à✑❣ à✑❣ ✏ ❣ ❲✍❦ ❦ ✝✒✟❬❦ We conclude this is an ❣❑ ❲❸❦ ❑ ➓ equality of polynomials, but the right-hand side is not a perfect square for ❦ an even non-square, contradiction (The reader might try generalizing this approach to arbitrary polynomials A✸ related✸ argument, due to Greg ✸❉Ø Kuperberg: write ✁ ❦ ⑨ ❲❸❦ ❣ ❦ as ❦ ⑨✑à✑❣ times a ➓ power series in ✡ ✆ ❦ and take two finite differences to get an expression which tends to as ❦❉â↔ã , contradiction.) ✸Ừ ⑨ ✸ ❣ ✸ ❲❸❦ ❦ has no repeated facNote: in case ❦ ➓ tors, it is a square for only finitely many ❦ , by a theorem of Siegel; work of Baker gives an explicit (but large) bound on such ❦ (I don’t know whether the graders will accept this as a solution, though.)