Singapore Mathematical Society

About POTW

Dear mathematics enthusiast,

We will be posting one problem a week from the Singapore Mathematical Olympiad (SMO) archives. One problem will be posted on every Friday and its solution will be posted one week later to allow sufficient time for you to attempt it.

Your comments, including solutions are welcome. I may choose to not approve your comments right away so as not to give the game away.
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June 29, 2011 – 9:22 am Categories: About | Post a comment

POTW49

Let $f(x)$ be a function which satisfies $f(29+x)=f(29-x)$ for all values of $x$ . If $f(x)$ has exactly three real roots $\alpha, \beta$ and $\gamma$ , determine the value of $\alpha+\beta+\gamma$ .

This is Problem 1 from the 2002 SMO Open Part A.

June 1, 2012 – 12:42 pm Categories: POTW | Post a comment

Tagged 2002, functional equation, open |

POTW48 Solution

Answer to POTW48 is -1997001.

Using the identity $x^2 - y^2 = (x-y)(x+y)$ , then the sum becomes
$-(3+7+11+\cdots+3995) = -\frac{1998}{2} \times \frac{3+3995}{2} = -1997001.$

Note that we have used the formula for the sum of an arithmetic progression.

June 1, 2012 – 6:40 am Categories: POTW Solutions | Post a comment

POTW48

Find the value of $1^2 - 2^2 + 3^2 - 4^2 + \cdots + 1997^2 - 1998^2.$

This is problem 8 from the 1998 SMO Junior Round 1.

Next week is the first round of the 2012 SMO. Wishing all competitors the best of luck!

May 25, 2012 – 8:39 am Categories: POTW | Comments (3)

Tagged 1998, arithmetic, junior |

POTW47 Solution

Answer to POTW47 is $(a^2-b^2+c^2-d^2)(\tan \alpha)/4$ .

Let $AP=e, BP=f, CP=g, DP=h$ . Then the area of the quadrilateral can be calculated as the sum of the four triangles, which is
$\frac{1}{2} ( ef \sin \alpha + fg \sin \alpha + gh \sin \alpha + he \sin \alpha ). (*)$
To get rid of the extra variables, we apply the cosine rule to each of the four triangles.

$a^2 = e^2 + f^2 + 2ef \cos \alpha$ ,
$b^2 = f^2 + g^2 - 2fg \cos \alpha$ ,
$c^2 = g^2 + h^2 + 2gh \cos \alpha$ ,
$d^2 = h^2 + e^2 - 2he \cos \alpha$ .

So $ef + fg + gh + he = (a^2 - b^2 + c^2 - d^2)/2 \cos \alpha$ and substituting into (*) gives the answer. Note how we have actually derived a very general formula for the area of quadrilateral. A related formula is known as Bretschneider’s formula.

May 25, 2012 – 8:21 am Categories: POTW Solutions | Post a comment

POTW47

Let ABCD be a quadrilateral with each interior angle less than $180^\circ$ . Let P be the point of intersection of the two diagonals AC and BD. If AB=a, BC=b, CD=c, DA=d and $\angle BPC = \alpha$ , where $\alpha \ne 90^\circ$ , find the area of ABCD in terms of a, b, c, d and $\alpha$ .

This is problem 7 from the 1999 SMO Open Round A.

May 18, 2012 – 8:28 am Categories: POTW | Comments (1)

Tagged 1999, geometry, open |

POTW46 Solution

Answer to POTW46 is 350.

There are 3 subcases:
1) The packages contain 4,1,1,1 fruits. Total number of ways = $\binom{7}{4} = 35.$
2) The packages contain 3,2,1,1 fruits. Total number of ways = $\binom{7}{3}\binom{4}{2} = 35 \times 6 = 210.$
3) The packages contain 2,2,2,1 fruits. Total number of ways = $\binom{7}{2}\binom{5}{2}\binom{3}{2}/3! = \frac{7!}{3! 2^3} = 105.$

Total number of ways = 35+210+105 =350.

Remark: There is a mistake in the 2011 SMO solution book. The statement of the problem in the book is “How many ways are there to put 7 identical apples into 4 identical packages so that each package has at least one apple?” Clearly, there would only be 3 possibilities but the answer given in the book is essentially what we have written above. One either keep the question and change the answer as per the errata provided by the SMS or modify the question to fit the answer printed in the book.

May 18, 2012 – 6:10 am Categories: POTW Solutions | Post a comment

POTW46

How many ways are there to put 7 distinct fruits into 4 identical packages so that each package has at least one fruit?

This is modified from Problem 26 of the 2011 SMO Junior Round 1.

May 11, 2012 – 9:07 am Categories: POTW | Comments (2)

Tagged 2011, combinatorics, junior |

POTW45 Solution

Answer to POTW45 is (E).

Based on Theresa’s and Ann’s statements, Jack has a daughter and at least two sons. We also know that Marie and Frank are siblings and Frank’s father has the least children. This eliminates the possibility of Frank being Jack’s son as Jack would then have at least 2 girls and 2 boys out of 8.

So Jack is father to Theresa and the remaining two boys John and Ivan. Based on John’s statement, these three are Jack’s only children. With five remaining children, Marie and Frank must have different fathers from the other three because of Frank’s statement. Since Catherine is Peter’s child, she must then be siblings with Ann and Isabelle, leaving Marie and Frank as Paul’s only children.

May 11, 2012 – 6:22 am Categories: POTW Solutions | Post a comment

POTW45

Three brothers, Peter, Paul and Jack, meet with their children in the family house. The children announce the following in turn:
Isabelle: “I’m three years older than John”
Theresa: “My father’s name is Jack.”
Ivan: “I’m two years older than Isabelle.”
Marie: “I’d rather play with one of my cousins that with my brother Frank.”
Catherine: “My father’s name is Peter.”
Ann: “I get along best with Uncle Jack’s sons.”
John: “My father has fewer than four children.”
Frank: “My father is the one with the least children.”

Which of the following statement is NOT correct?
(A) Peter has three daughters.
(B) Jack has two sons.
(C) Marie and Frank have no other siblings.
(D) Catherine and Isabelle are sisters.
(E) Ann is Paul’s daughter.

This is problem 20 from the 2003 SMO Junior Round 1.

May 4, 2012 – 8:21 am Categories: POTW | Post a comment

Tagged 2003, junior, logical reasoning |

POTW44 Solution

Answer to POTW44 is 309 (or 32.)

Perhaps the difficulty in this problem lies with interpreting the condition $a_{i+1}$ does not exceed $a_i$ by 1. Does it mean $a_{i+1} \ne a_i + 1$ ? But the word exceed seems to imply an order relation, so could it be that $a_{i+1} \le a_{i} + 1$ ?

Using the first interpretation, the problem can be solved by the principle of inclusion and exclusion. (This is the solution from the SMO solution book.) Let $A_i$ be the subset of permutations where $i+1$ follows $i$ . Then $A_1$ means that 12 occurs as a block together and should be permuted with 3, 4, 5 and 6, hence a total of $5!$ ways. Then $A_{1} \cap A_{2}$ means that 12 occurs as a block as 23 occurs as a block (i.e. 123 is a block), so there are $4!$ ways. Note that this is the same number as $A_{1} \cap A_{3}$ .

Since
$\sum |A_{i}| = \binom{5}{1} 5!,$
$\sum |A_{i} \cap A_{j}| = \binom{5}{2} 4!,$
$\sum |A_{i} \cap A_{j} \cap A_{k}| = \binom{5}{3} 3!,$
$\sum |A_{i} \cap A_{j} \cap A_{k} \cap A_{l}| = \binom{5}{4} 2!,$
$|A_{1} \cap A_{2} \cap A_{3} \cap A_4 \cap A_5| = \binom{5}{5} 1!.$
By the principle of inclusion and exclusion, the answer is
$\displaystyle \sum_{i=0}^5 (-1)^i \binom{5}{i}(6-i)! = 309$ .

If we use the second interpretation, $a_{i+1} \le a_{i} + 1$ , the problem is quite interesting. We have for example, 123456 being an admissible choice. But 234561 is also possible. In total there would be 32 possibilities, can you see why?

May 4, 2012 – 6:26 am Categories: POTW Solutions | Post a comment