# New Zealand Mathematical Olympiad Committee  

# NZMO Round Two 2021 — Solutions  

1. Problem: Let \(ABCD\) be a convex quadrilateral such that \(AB + BC = 2021\) and \(AD = CD\) . We are also given that  

\[\angle ABC = \angle CDA = 90^{\circ}\]  

Determine the length of the diagonal \(BD\) .  

Solution: Since \(AD = DC\) and \(\angle ADC = 90^{\circ}\) , we can fit four copies of quadrilateral \(ABCD\) around vertex \(D\) as shown in the diagram.  

![md5:ed8c2107bcb244f3199667fd9acfdb08](ed8c2107bcb244f3199667fd9acfdb08.jpeg)
  

The outer shape is a quadrilateral because \(\angle DAB + \angle BCD = 180^{\circ}\) . Moreover it is a rectangle because \(\angle ABC = 90^{\circ}\) . In fact it is a square with side- length 2021 because of rotational symmetry and \(AB + BC = 2021\) . Also \(D\) is the centre of the square because it is the centre of the rotational symmetry. So \(BD\) is the distance from a vertex to the centre of the square, which is half the length of the diagonal of the square. Thus  

\[BD = \frac{1}{2}\left(2021\sqrt{2}\right) = \frac{2021}{\sqrt{2}}.\]



## Alternative Solution:  

First let \(x = AD = DC\) and \(a = BD\) and \(y = AB\) and \(z = BC\) . Now initially we can apply Pythagoras in triangles \(CDA\) and \(ABC\) to get \(x^{2} + x^{2} = AC^{2}\) and \(y^{2} + z^{2} = AC^{2}\) respectively. Putting this together gives us  

\[x^{2} = \frac{y^{2} + z^{2}}{2}.\]  

Now note that the opposite angles \(\angle ABC\) and \(\angle CDA\) (in quad \(ABCD\) ) are supplementary. Therefore \(ABCD\) is a cyclic quadrilateral. Equal chords subtend equal arcs (and chords \(AD = DC\) are equal) so \(\angle ABD = \angle DBC\) . Furthermore, since \(\angle ABC\) is a right angle, this means that \(\angle ABD = \angle DBC = 45^{\circ}\) .  

For any three points \(P\) , \(Q\) and \(R\) , let \(|PQR|\) denote the area of triangle \(PQR\) . Now consider the total area of quadrilateral \(ABCD\) calculated in two ways:  

\[|ABC| + |CDA| = |ABD| + |DBC|.\]  

We calculate the areas of the right- angled triangles using the \(\triangle = \frac{b h}{2}\) formula, and we calculate the area of the \(45^{\circ}\) - angled triangles using the \(\triangle = \frac{1}{2} ab\sin C\) formula.  

\[\frac{yz}{2} +\frac{x^{2}}{2} = \frac{1}{2} ay\sin (45^{\circ}) + \frac{1}{2} az\sin (45^{\circ})\]  

At this point we can substitute \(x^{2} = \frac{1}{2} (y^{2} + z^{2})\) into this equation, and rearrange:  

\[\frac{yz}{2} +\frac{x^{2}}{2} = \frac{ay\sin(45^{\circ})}{2} +\frac{az\sin(45^{\circ})}{2}\] \[\frac{yz}{2} +\frac{y^{2} + z^{2}}{4} = \frac{ay}{2\sqrt{2}} +\frac{az}{2\sqrt{2}}\] \[2yz + y^{2} + z^{2} = \frac{ay + az}{2\sqrt{2}}\] \[\frac{(y + z)^{2}}{4} = \frac{a(y + z)}{2\sqrt{2}}\] \[\frac{y + z}{\sqrt{2}} = a.\]  

Finally since \(y + z = 2021\) this gives our final answer of \(BD = a = \frac{2021}{\sqrt{2}}\) .



2. Problem: Prove that  

\[x^{2} + \frac{8}{xy} +y^{2}\geq 8.\]  

for all positive real numbers \(x\) and \(y\) .  

Solution: Since square numbers are always non- negative we have  

\[(x - y)^{2}\geq 0\qquad \mathrm{and}\qquad (x y - 2)^{2}\geq 0.\]  

Also since \(x\) and \(y\) are positive we have \(\frac{2}{x y} >0\) . Combining this all together gives us:  

\[(x - y)^{2} + \frac{2}{xy} (x y - 2)^{2}\geq 0.\]  

From here we expand and simplify:  

\[(x^{2} - 2x y + y^{2}) + \frac{2}{x y} (x^{2}y^{2} - 4x y + 4)\geq 0\] \[x^{2} - 2x y + y^{2} + 2x y - 8 + \frac{8}{x y}\geq 0\] \[x^{2} + \frac{8}{x y} +y^{2}\geq 8\]  

as required.  

## Alternative Solution:  

Consider the AM- GM inequality applied to \(\left\{x^{2},\frac{4}{x y},\frac{4}{x y},y^{2}\right\}\)  

\[\frac{x^{2} + \frac{4}{x y} + \frac{4}{x y} + y^{2}}{4}\geq \sqrt[4]{x^{2}\times\frac{4}{x y}\times\frac{4}{x y}}\times y^{2}\] \[\frac{x^{2} + \frac{8}{x y} + y^{2}}{4}\geq 2\] \[x^{2} + \frac{8}{x y} + y^{2}\geq 8.\]



3. Problem: Let \(\{x_{1},x_{2},x_{3},\ldots ,x_{n}\}\) be a set of \(n\) distinct positive integers, such that the sum of any 3 of them is a prime number. What is the maximum value of \(n\) ?  

Solution: First we show that \(n = 4\) is possible with an example. The example \(\{x_{1},x_{2},x_{3},x_{4}\} = \{1,3,7,9\}\) satisfies the problem because:  

\(\cdot 1 + 3 + 7 = 11\) is prime,  

\(\cdot 1 + 3 + 9 = 13\) is prime,  

\(\cdot 1 + 7 + 9 = 17\) is prime, and  

\(\cdot 3 + 7 + 9 = 19\) is prime.  

We still have to prove that \(n\geq 5\) is impossible.  

Consider any set \(\{x_{1},x_{2},x_{3},\ldots ,x_{n}\}\) such that the sum of any 3 of them is a prime number. Also consider the three "pigeonholes" modulo 3; the residue classes 0, 1 and 2. If all three pigeonholes were non- empty, then it would be possible to choose three numbers – one from each pigeonhole. This would result in a sum which is \(0 + 1 + 2\equiv 0\) (mod 3), and since the numbers are distinct positive integers, this sum would be \(>3\) . Thus the sum would not be prime which is a contradiction. Hence at least one of the pigeonholes must be empty. i.e.  

The numbers \(\{x_{1},\ldots ,x_{n}\}\) are distributed amongst (at most) two different residue classes modulo 3.  

Now assume for the sake of contradiction that \(n\geq 5\) . By the pigeonhole principle at least one residue class contains at least 3 of the numbers. The sum of any three numbers from the same residue class is always a multiple of 3 and so this is a contradiction.  

Therefore \(n< 5\) as required.  

Comment: The example \(\{1,3,7,9\}\) is not the only example that satisfies the problem with \(n = 4\) . Here are many other examples: \(\{1,5,7,11\}\) , \(\{3,5,11,15\}\) , \(\{1,3,13,15\}\) , \(\{3,9,11,17\}\) , \(\{5,9,15,17\}\) , ... Searching for an example when \(n = 4\) is much easier if you conjecture that all the \(x_{i}\) must be odd.



4. Problem: Let \(AB\) be a chord of circle \(\Gamma\) . Let \(O\) be the centre of a circle which is tangent to \(AB\) at \(C\) and internally tangent to \(\Gamma\) at \(P\) . Point \(C\) lies between \(A\) and \(B\) . Let the circumcircle of triangle \(POC\) intersect \(\Gamma\) at distinct points \(P\) and \(Q\) . Prove that \(\angle AQP = \angle CQB\) .  

Solution: Construct the tangent line to \(\Gamma\) at \(P\) . Note that this line is also tangent to the circle through points \(C\) and \(P\) with centre \(O\) . Also construct point \(E\) on this tangent line to the right of \(P\) . Note that \(\angle EPO = 90^{\circ}\) and \(\angle OCA = 90^{\circ}\) because the radii and tangents are perpendicular.  

![md5:d5576ee95456eaef99d3c209d49a4200](d5576ee95456eaef99d3c209d49a4200.jpeg)
  

Let \(x = \angle PBQ\)  

\(\angle EPQ = x\) (by alternate segment theorem) \(\angle OPQ = x - 90^{\circ}\) (because \(\angle EPO = 90^{\circ}\) ) \(\angle QCO = 180^{\circ} - \angle OPQ\) (because opposite angles in a cyclic quad \(= 270^{\circ} - x\) are supplementary) \(\angle ACQ = 360^{\circ} - \angle QCO - \angle OCA\) (angles around point \(C\) are \(360^{\circ}\) ) \(= 360^{\circ} - (270^{\circ} - x) - 90^{\circ}\) \(= x\) .  

We also have \(\angle QPB = \angle QAB\) (angles subtended by chord \(QB\) ) in cyclic quad \(QAPB\) . Therefore we have similar triangles  

\[\triangle QBP\sim \triangle QCA\qquad (\angle PBQ = \angle ACQ\mathrm{~and~}\angle QPB = \angle QAC)\]  

Hence \(\angle AQC = \angle PQB\) . Therefore  

\[\angle AQP = \angle AQC + \angle CQP = \angle PQB + \angle CQP = \angle CQB.\]



5. Problem: Find all pairs of integers \(x, y\) such that  

\[y^{5} + 2x y = x^{2} + 2y^{4}.\]  

Solution: Rearrange and factorize to get  

\[y^{2}(y - 1)(y^{2} - y - 1) = (x - y)^{2}.\]  

Note that \(y\) and \((y - 1)\) are coprime (their greatest common divisor is 1) because they are consecutive integers. Note since \(y(y - 1)\) and \((y^{2} - y - 1)\) are consecutive integers, we see that \((y^{2} - y - 1)\) is coprime to both \(y\) and \((y - 1)\) . Therefore the three factors  

\[y^{2},(y - 1)\mathrm{~and~}(y^{2} - y - 1)\mathrm{~are~pairwise~coprime}.\]  

Since their product is a perfect square it follows that either: one of \(y^{2}\) , \((y - 1)\) and \((y^{2} - y - 1)\) is zero, or all three of them are perfect squares. So we have four cases:  

Case A: \(y = 0\)  

Substituting this into the original equation yields \((0)^{5} + 2x(0) = x^{2} + 2(0)^{4}\) . Solving this quadratic yields \(x = 1\) and so \((x,y) = (0,0)\) is the only solution in this case.  

Case B: \(y = 1\)  

Substituting this into the original equation yields \((1)^{5} + 2x(1) = x^{2} + 2(1)^{4}\) . Solving this quadratic yields \(x = 1\) and so \((x,y) = (1,1)\) is the only solution in this case.  

Case C: \(y^{2} - y - 1< 0\)  

This rearranges to give us \((2y - 1)^{2}< 5\) . But the only odd square less than 5 is 1, and so we would have \((2y - 1) = \pm 1\) . which leads to \(y = 0,1\) (but we have already covered this in Cases A and B).  

Case D: \(y^{2} - y - 1 = k^{2}\)  

If \(y > 2\) then \((y - 1)^{2} = y^{2} - 2y + 1< y^{2} - y - 1< y^{2}\) . This would give us \((y - 1)^{2}< k^{2}< y^{2}\) which is a contradiction because \((y - 1)^{2}\) and \(y^{2}\) are consecutive squares.  

If \(y< - 1\) then \((- y)^{2}< y^{2} - y - 1< y^{2} - 2y + 1 = (- y + 1)^{2}\) . This would give us \(y^{2}< k^{2}< (y - 1)^{2}\) which is a contradiction because \((- y)^{2}\) and \((- y + 1)^{2}\) are consecutive squares.  

Therefore we must have \(- 1\leq y\leq 2\) . We have already covered \(y = 0\) and \(y = 1\) in cases A and B respectively. So it suffices now only to consider \(y = - 1\) and \(y = 2\) .  

If \(y = - 1\) then \((- 1)^{5} + 2x(- 1) = x^{2} + 2(- 1)^{4}\) . This rearranges into \((x + 1)^{2} = - 2\) which has no real solutions.  

If \(y = 2\) then \(2^{5} + 4x = x^{2} + 2\times 2^{4}\) . This rearranges to give \(x^{2} - 4x = 0\) which has solutions \(x = 0\) and \(x = 4\) .  

Hence in this case we have solutions \((x,y) = (0,2)\) and \((4,2)\) .  

In summary we have four distinct solutions for \((x,y)\) being:  

\[(x,y) = (0,0),(1,1),(0,2)\mathrm{~and~}(4,2).\]



## Alternative Solution:  

In a similar manner to above get to Case D:  

\[y^{2} - y - 1 = k^{2}.\]  

Then rearrange it to give \((2y - 1)^{2} - 4k^{2} = 5\) . This then factorizes as a difference between two squares as  

\[(2y - 1 + 2k)(2y - 1 - 2k) = 5\]  

Since 5 is prime it can only be factored in two ways: \(5 = 5\times 1 = (- 5)\times (- 1)\) . The sum of these two factors is \((2y - 1 + 2k) + (2y - 1 - 2k) = 4y - 2\) . Therefore:  

\[4y - 2 = 5 + 1\qquad \mathrm{or}\qquad 4y - 2 = (-5) + (-1).\]  

From \((4y - 2) = 6\) we get \(y = 2\) , and from \((4y - 2) = - 6\) we get \(y = - 1\) .  

Thus we have ruled out all possibilities except for \(y = - 1,0,1,2\) . Checking these each individually yields the four answers.  

- \(y = 2\) yields \((2)^{5} + 2x(2) = x^{2} + 2(2)^{4}\) which simplifies to become \(x^{2} - 4x = 0\) , and has solutions \(x = 0\) and \(x = 4\) .- \(y = 1\) yields \((1)^{5} + 2x(1) = x^{2} + 2(1)^{4}\) which simplifies to become \(x^{2} - 2x + 1 = 0\) , and has solution \(x = 1\) only.- \(y = 0\) yields \((0)^{5} + 2x(0) = x^{2} + 2(0)^{4}\) which simplifies to become \(x^{2} = 0\) , and has solution \(x = 0\) only.- \(y = -1\) yields \((-1)^{5} + 2x(-1) = x^{2} + 2(-1)^{4}\) which simplifies to \(x^{2} + 2x + 3 = 0\) , but this has no real solutions.  

Thus all solutions are  

\[(x,y) = (0,2),(4,2),(1,1)\mathrm{~and~}(0,0).\]