NUSAMO Mock Olympiad Solutions

August 7th and 14th, 2015

1. In acute triangle ABC, M is the midpoint of AB, O is the circumcenter, and H isthe orthocenter. Let Oc be the reflection of O over line AB, and the circle centered atOc through B and the circle with diameter HM intersect at two points. Prove that oneof these points lies on the C-median.

B

C

AM

H T

Oc

H ′

F

Solution 1. Note that ∠AOcB = ∠AOB = 2∠C and ∠AHB = 180−(90−∠B+90−∠A) = ∠A+∠B = 180−∠C, so H lies on the circle centered at Oc through B and A. LetH ′ be the point such that ACBH ′ is a parallelogram. Since ∠AH ′B = ∠ACB = 180−∠AHB, AHBH ′ is cyclic. Then ∠HBH ′ = ∠HBA + ∠H ′BA = 90 −∠A + ∠A = 90

thus H ′ is the antipode of H on (AHB). Let T be the intersection of the C-median with(AHB). Then ∠HTH ′ = ∠HTM = 90 which implies that T lies on the circle withdiameter HM .

Solution 2. Note as before that H lies on the circle centered at Oc through B and A.Let CH ∩ AB = F and let T be the second intersection of the C-median with the

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circumcircle of HFM ; invert about C with radius√CH · CF . Remark that that the

circle AHB inverts to the nine-point circle of ABC, and the C-median and circumcircleof HFM invert to themselves. These three objects pass through M , so inverting backyields the desired conclusion.

Solution 3 (by blasterboy). Note as in Solution 1 that H lies on the circle centered atOc through B and A. Then if the other intersection of the two circles, apart from H, isT , then we wish to show C, T,M are collinear. Let F = CH ∩ AB. Then it suffices toshow ∠FTM + ∠FTC = 180.

Noting that FHTM is cyclic, ∠FTM = ∠FHM . Furthermore, ∠HMT = ∠TFH =⇒4CHM ∼ 4CTF =⇒ ∠FTC = ∠CHM

Now note that ∠FTM + ∠FTC = ∠FHM + ∠CHM = 180 since CF is an altitudepassing through orthocenter H.

Solution 4 (by djmathman). Remark that the circumradius of the circle centered at Oc

and passing through B is R. In addition, recall that the reflection of H over AB lies onthe circumcircle of 4ABC, so the circumradius of 4AHB is also R. Now let T be thefoot of the perpendicular from H to CM ; it suffices to show that AHTB is cyclic.

Let D = AH ∩ BC. Remark that since ∠CDH = ∠CTH = 90, quadrilateralCDTH is cyclic. This means that, among other things, ∠CTD = ∠CHD = ∠CBM ,so 4CDT ∼ 4CMB =⇒ BM2 = MT ·MC.

Now let T ′ be the reflection of T over M . Then the recently-discovered metric relationbecomes AM · BM = MC ·MT ′. By reverse Power of a Point, T ′ ∈ (ABC). Finally,since AM = MB and TM = MT ′, ATBT ′ is a parallelogram, so 4AT ′B = 4BTA.Hence the circumradius of 4ATB is also R, implying the desired cyclicity.

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2. Prove that for all n ∈ N,n2−1∑k=1

√k ≤ 4n3 − n− 3

6

Solution 1. Note that if r, a ∈ N, then√r2 + a ≤ r +

a

2r

(squaring both sides and subtracting gives(

a2r

)2 ≥ 0, which is true by the Trivialinequality). Then, note that

n2−1∑k=1

√k =

n2−1∑k=1

(b√kc+

√k)

For any integer k, with 1 ≤ k ≤ n − 1, there are obviously 2k + 1 integers m for whichb√mc = k (since (k + 1)2 − k2 = 2k + 1). Thus,

n2−1∑k=1

b√kc =

n−1∑k=1

k(2k + 1) =4n3 − 3n2 − n

6.

Now consider the fractional part. We concern ourselves with summing this from k2 to(k + 1)2 − 1, then summing across all values of k. Since

k ≤√k2 + a ≤ k +

a

2k

we know that √k2 + a ≤ a

2k . Then

(k+1)2−1∑m=k2

√m ≤

2k∑a=0

a

2k= k +

1

2.

Summing over all values of k gives

n2−1∑w=1

√w ≤

n−1∑k=1

(k +

1

2

)=

n2 − 1

2

which is an upper bound on the sum of the fractional parts. Adding this to the sum ofthe floors gives the desired bound.

Solution 2 (by blasterboy). We induct on n. If n = 1, then we have 0 = 0. Now assumefor some m we have

m2−1∑k=1

√k =√

1 + ... +√m2 − 1 ≤ 4m3 −m− 3

6

3

It suffices to show that

(m+1)2−1∑k=1

=m2+2m∑k=1

=√

1 + ... +√

m2 + 2m ≤ 4(m + 1)3 − (m + 1)− 3

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But note that √1 + ... +

√m2 + 2m

= [√

1+...+√m2 − 1]+[

√m2+...+

√m2 + 2m] ≤ 4m3 −m− 3

6+[√m2+...+

√m2 + 2m]

so it suffices to show that

√m2 + ... +

√m2 + 2m ≤ 4(m + 1)3 − (m + 1)− 3

6− 4m3 −m− 3

6=

(2m + 1)2

2

Note that by Cauchy-Schwarz, we have√(1 + ..... + 1)(m2 + .... + m2 + 2m) ≥

√m2 + ... +

√m2 + 2m

=⇒√

(2m + 1)2(m2 + m) ≥√m2 + ... +

√m2 + 2m

Furthermore, the following holds true:

0 ≤ 1 =⇒ 4m2+4m ≤ (2m+1)2 =⇒√m2 + m ≤ 2m + 1

2=⇒ (2m+1)

√m2 + m ≤ (2m + 1)2

2

Therefore we have

√m2 + ... +

√m2 + 2m ≤

√(2m + 1)2(m2 + m) ≤ (2m + 1)2

2

This concludes our argument.

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3. The points on a n × n lattice grid are colored one of k colors: A1, A2, . . . , Ak. Arectangle whose vertices lie on the grid and whose sides are parallel to those of the gridis called happy if for some r, all of its vertices are of color Ar.a) Show that if n > k(k + 1), there must exist a happy rectangle.

b) Show that if n > k(k + 1), there must exist at least(k+12

)2happy rectangles.

Solution. Let Ar be the most common color in the grid. By the pigeonhole principle,there are at least Nr = dn2

k e vertices of color Ar. Denote by f(i, r) the number of pointsof color Ar in the ith column of the grid. Then we know that

Nr =n∑

i=1

f(i, r)

Consider a graph Gr, whose vertices are the points of color Ar, and whose vertices areconnected only if they lie in the same column. Then

E(Gr) =n∑

i=1

(f(i, r)

2

)where E(Gr) denotes the number of edges in Gr. Note that(

n∑i=1

f(i, r)2

) n∑j=1

12

≥ ( n∑i=1

f(i, r)

)2

by Cauchy-Schwarz. Thenn∑

i=1

f(i, r)2 ≥ N2r

n.

We established that

E(Gr) =n∑

i=1

(f(i, r)

2

)=

1

2

n∑i=1

(f(i, r)2 − f(i, r)) ≥ 1

2

(N2

r

n−Nr

).

Call two edges in Gr equivalent if they lie in the same relative positions within theircolumns. Note that there are

(n2

)equivalence classes in Gr. We now follow the same

logic as before. Label these equivalences classes R1, R2, R3, . . . , R(n2). Then we know

that

E(Gr) =

(n2)∑i=1

|Ri|

Note that for any two elements in the same equivalence class, the four vertices of thetwo edges form a happy rectangle. Then there are

(n2)∑i=1

(|Ri|

2

)

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happy rectangles. Applying the similar manipulations as before using the Cauchy-Schwarz inequality, we see that there are at least

1

2

(E(Gr)

2(n2

) − E(Gr)

)

happy rectangles. We know that

E(Gr) ≥1

2

(N2

r

n−Nr

)=

1

2

⌈n2

k

⌉(dn2

k en− 1

)≥ 1

2

(n2

k

)(nk− 1)

=n2(n− k)

2k2

Let H denote the number of happy rectangles.

H ≥ 1

2

(E(Gr)

2(n2

) − E(Gr)

)

≥ 1

2

(n2(n−k)

2k2

)2(n2

) −(n2(n− k)

2k2

)=

n2(n− k)

4k2

(n2(n− k)

k2n(n− 1)− 1

)=

n2(n− k)

4k2

(n(n− k)

k2(n− 1)− 1

)≥ n2(n− k)

4k2

(n− k

k2− 1

)=

n2(n− k)(n− (k2 + k))

4k4

Note that if n > k2 + k, there is then at least one happy rectangle. Furthermore, lettingn ≥ k2 + k + 1, we see that

H ≥ (k2 + k + 1)2(k2 + 1)(1)

4k4≥ (k2 + k)2(k2)

4k4=

(k + 1)2

4

as desired.

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4. Call an n-sided nondegenerate convex polygon multi-Pythagorean if its n anglesa1, a2, . . . , an−1, an satisfy

a21 + a22 + · · ·+ a2n−1 = a2n

(here all ai are in degrees and n 6= 4). Find all multi-Pythagorean polygons whose anglesare all integers.

Solution. By Titu’s Lemma, we have that

a21 + a22 + · · ·+ a2n−1 ≥(a1 + a2 + · · ·+ an−1)

2

n− 1

=⇒ a2n(n− 1) ≥ (180(n− 2)− an)2

Since both sides are positive for n > 2, we can take the square root to obtain

an√n− 1 ≥ 180(n− 2)− an

=⇒ an ≥ 180(√n− 1− 1)

For n ≥ 5, an ≥ 180 which is impossible, so it suffices to check the case n = 3.

For n = 3, we have that

a21 + a22 = (180− (a1 + a2))2

which reduces to(180− a1)(180− a2) = 16200

We want 0 ≤ a1, a2 ≤ 180 which is equivalent to 0 ≤ 180 − a1, 180 − a2 ≤ 180, so weseek numbers m and n such that mn = 16200 and m,n ≤ 180. Looking through thefactors of 16200 = 180(90), we see that the only such pairs are (100, 162), (135, 120),(150, 108) (and permutations). Then if WLOG a1 ≤ a2, we see that the only orderedtriples (a1, a2, a3) which are multi-Pythagorean are (18, 80, 82), (30, 72, 78), (45, 60, 75).

In summary, the only multi-Pythagorean polygons are those with angles (18, 80, 82),(30, 72, 78), (45, 60, 75) and appropriate permutations.

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5. Let ABC be a triangle, and define X to be the intersection of the external anglebisectors of ∠B and ∠C and I to be the intersection of the internal angle bisectors of∠B and ∠C. Let M be the circumcenter of 4BIC, and let G be the point on BC suchthat XG ⊥ BC. Construct a circle Ω with diameter AX. If Ω and the circumcircle of4ABC intersect at a second point P , and the altitude from A onto BC intersects Ω atH, prove that H, M , G, and P are collinear.

B

A

H

C

I

X

G

M

P

DK

R

S

Solution 1. Let R be the tangency point of the A-excircle on AB and S the tangencypoint on AC. By Miquel’s Theorem, there is a spiral similarity that carries 4PBR to4PCS, so PB

PC = BRCS = BG

CG . By the Angle Bisector Theorem, this implies that G lieson PM , solving half the problem.

To prove that H lies on this line, it suffices to prove that ∠APM = ∠APH. Byangle chasing, we get ∠APM = ∠ACM = ∠C + ∠BCM = ∠C + ∠BAM = ∠C + ∠A

2 .Similarly, ∠APH = ∠AXH. Let AH ∩ BC = K; it is clear that 4AKD ∼ 4AHX,where D is AX ∩ BC; we get that ∠AXH = ∠ADK = 180 − ∠B − ∠A

2 = ∠C + ∠A2 ,

and we’re done.

Solution 2. To prove the collinearity of M,G and P , invert about M with radius MB.It is well-known that M is the midpoint of minor arc BC in the circumcircle of 4ABCand that A, I,M and X are collinear. Note that Ω and the circumcircle of 4BIC passthrough X by Fact 5 and their centers are collinear through this point, so they are

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tangent. This inversion sends line MG to itself, the circumcircle of 4ABC to line BC,and Ω to the circle with diameter DX. Remark that ∠DGX = 90, so G lies on theinverted circle and all three objects pass through this point, implying the conclusion.Now proceed as in Solution 1.

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6. A sequence of positive integers a1, a2, a3, . . . , an has the property that every term isa distinct power of 2. Let

S =

(j∏

k=i

ak

)− 1

∣∣∣∣∣i ≤ j

.

Suppose S does not contain a multiple of n + 2. Find all possible values of n.

Solution. Consider when n is even. Then given a sequence 〈ai〉 not containing 1, allelements of S are odd so it is impossible that S contains a multiple of n + 2 which iseven. Now we consider when n is odd. Let f(k) = ordk+2(2); evidently, there are f(n)possible values of ai modulo n+ 2. Suppose n ≥ f(n). We claim that in this case, everysequence ai gives a set S containing a multiple of 2. Denote

Pi,j =

j∏k=i

ak.

There are at most n different possible values of P1,j . If they are all pairwise not congruentmodulo n+2, then since n ≥ f(n), by the Pigeonhole Principle one of them is congruentto 1 (mod n + 2) and we’re done. Otherwise assume for some j1 6= j2, it is true thatP1,j1 ≡ P1,j2 (mod n + 2). Then

j1∏k=1

ak ≡j2∏k=1

ak (mod n + 2)

WLOG let j1 > j2. n + 2 is odd and the LHS and RHS are both powers of 2, so we cansafely say

j1∏k=j2

ak ≡ 1 (mod n + 2)

so then Pj2,j1 − 1 is a multiple of n + 2 and we are done. Thus, if n ≥ f(n), wehave shown that regardless of the sequence 〈ai〉, S contains a multiple of n + 2. Nowwe consider n < f(n). Since f(n) = ordn+2(2), by Euler’s Theorem we know thatf(n) ≤ ϕ(n + 2) ≤ n + 1 since the maximal value would occur at primes. If n < f(n),then n < f(n) ≤ n + 1 =⇒ f(n) = n + 1. Thus n + 2 is prime and 2 is a primitiveroot. Let p = n + 2 then be a prime. We must show that for these values of p thereexists a sequence 〈ai〉 for which S contains no multiples of n+ 2. Consider the sequenceak = 2k(p−1)+1 = 2kn+k+1. These are all equivalent to 2 (mod n + 2), and since 2 is aprimitive root modulo n+2, it is impossible that multiplying any consecutive subsequenceof these will be congruent to 1 (mod n + 2). Then our final solution set is

n ∈ 2k | k ∈ N ∪ p− 2 | p is a prime for which 2 is a primitive root.

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