Mathematics Olympiad Sample Problems

Figure 1

I. The Figure 1 shown is a square with quarter circles drawn from two adjacent corners. If the side of the square is 16, find the area of the largest circle that can be inscribed in the blue region.

Solution

Let O be the center of the circle and r be the radius of the circle.
The circle must touch the arc AC and BD. Let extended DO touch the arc AC at P.
DP= 16 as it is the radius of the quarter circle ACD.
OP=r,    DO=10-r.
Radius from center O is perpendicular bisector of DC.
So by Pythagoras theorem we can write,
r²+8²=(10-r)²
20r=36
r=1.8
So area of the circle=π(1.8)²

II. There are 20 questions in an exam and you must answer 12 questions including at least 5 questions from the first 6. How many different combinations of questions can you choose to answer?

Solution

Let's imagine two scenarios. In first scenario you chose to answer all of the first 6 questions and in second scenario you chose 5 questions out of the first 6 questions.

If you choose to answer the first 6 questions, there are 14ℂ6=3003 ways to answer the remaining 6.
If you choose to answer 5 questions from the first 6 questions, there are 6ℂ5=6 ways to choose them.
Remaining 7 questions can be chosen in 14ℂ7=3432 ways.
So, 3003+3432×6=23595 ways of choosing 12 questions.

III. Consider a right-angled triangle ABC, where AC=15, CB=20 and AĈB= 90’. Let CD be a perpendicular on AB. Let P and Q be in centre of triangle ACD and BCD. Find the length of PQ.

Solution

AB=25 (By Pythagoras Theorem)
sin B=15/25= 3/5
3/5=CD/20
CD=15
DB=16
Area of ΔBCD= 0.5*12*16= semi perimeter* in-radius= 0.5*(12+16+20)*r
r=4
AD=9
Area of ΔACD= 0.5*9*12= 0.5*(15+12+9)*r̅
r̅=3
r+r̅=7
PQ=7.

IV. ABCD is a square. P is a point inside the square.∠PAB=∠PBA=15°. What is the measure of ∠CPD in degrees?

Solution

Let length of square is x
Triangle PAB is an isosceles triangle.Draw a perpendicular bisector PZ from P on AB.
tan 15°= PZ/(x/2) = 2PZ/x
PZ= xtan15°/2
Notice triangle PDC is also an isosceles triangle.
The perpendicular bisector from P to DC= x-(xtan15/2)
tan (PDC)= (2x-xtan15)/x
2-tan(60°-45°)=tan (PDC)
tan(PDC)= 2- (tan 60°-tan 45°)/(1+tan 45°tan 60°)
tan (PDC)=√3
angle PDC=60°=angle PCD
So angle CPD= 180°-120°=60°

V. Given that, (TWO)²=(THREE) where TWO is a 3 digit number where different letters represent different digits. Find the five digit number THREE.

Solution
TWO is a 3-digit number less than 317 because 317² is more than 10000, but THREE is a 5-digit number.
T can be 1, 2 or 3.
Suppose that T = 2; observing that 200*200 = 40000, whose first digit is 4 and not 2 we can say that T=1.
Now O²= E. So E can be 1, 4 or 9, but as T=1 E can only be 4 or 9.
If we multiply TWO*TWO we'll notice that O²=2OW. As O² is even, E is even, so E=4.
Since (TWO)² < 20000, we must have W ≤ 4. And since T = 1, E = 4, we get W = 2 or 3.
Since the units digit of (TWO)² is 4, the units digit of TWO must be 2 or 8. That leaves us with TWO= 128 or 132 or 138.
For only 138² ones and tens digit are same. So THREE=19044.

  

Figure 2

VI.  In the Figure 2, area of DQBP is 18unit². Given that BP=PC and AQ=QB. Find the area of ABCD. 

Solution 

Let AQ be x and BP be y.
We can write Area of ABCD- Area of AQD- Area of PCD= Area of DQBP
So 4xy- 0.5*x*2y- 0.5*2x*y=18
2xy=18
4xy= 36
So area of ABCD is 36unit²

VII. Let a, b, c and d be distinct positive integers such that ab+cd=60, ac+bd=75, ad+bc=68. Find the value of a+b+c+d.

Solution 

ac+bd+ad+bc=75+68=143
a(c+d)+b(c+d)=143
(a+b)(c+d)=11*13
As 11 and 13 are prime numbers and (a+b) and (c+d) are integers, we can write (a+b)=11 and (c+d)=13
So a+b+c+d=24.

VIII. Looking at pascal's triangle (figure 3) from left to right, find the 2015th number on the 2017th row of pascal's triangle. Consider the very top row as the 1st row. 

Figure 3

Solution

From 5th row if you look at the 3rd number from right for every row, you'll notice a pattern. The 3rd number from right for each consecutive rows are : 6, 10, 15, 21, 28, 36, 45 and so on.
If you look closely you will notice that the 3rd number from right in nth row is the summation of all integers from 1 to (n-2)
So the 3rd number from right in the 2017th row is:
2015*2016/2= 2031120

IX. How many two digit numbers are there so that the sum of the number and its reverse is a perfect square?

Solution

Let one number be 10x+y. So its reverse is 10y+x.
So 10x+y+x+10y=a*a
11x+11y= 11(x+y) = a*a
So (x+y) needs to be 11 or 11b*b.
As 1<x,y <=9, (x+y) <=18
So (x,y) can be (2,9), (3,8), (4,7), (5,6), (6,5), (7,4), (8,3), (9,2)
So there are 8 such numbers.

X. Find all integer values of N for which N²+ 25N+158 is a perfect square.

Solution
Let’s assume N²+25N+158=y²
Or, N²+25N+(158-y²) =0
We know, N= {-b ±√( b²-4ac)}/2a

Let, D= b²-4ac and D needs to have an integer root for N to be an integer.
25²-4×(158-y²)= 625-632+4y²= -7+4y²
So, (-7+4y²) needs to have an integer root
Let, -7+4y² = x²
x²- (2y)² =-7
(x+2y)(x-2y)=-7
As both “a” and “x” are integers, (x+2y) can be 1 (x-2y) can be -7 or (x+2y) can be 7 and (x-2y) can be -1.
Solving these equations for “a” we get a=-2 or 2.
So D= 9
N= (-25+3)/2 or (-25-3)/2
So N= -11 or -14

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Neblina Sikta.

Mathematics and Informatics Intern.

Science Olympiad Blog.