MIOTP WEEK-1 (SOLUTIONS)

1. a)Hint:-

Consider n=a+(a/2-1)²

Where a=126

 b)Hint:-

Choosing n = m² + 3m + 3, we have

m + n + 1 = m² + 4m + 4 = 

(m + 2)²

and mn+ 1 = m³+ 3m² + 3m + 1 = (m + 1)³.

2. Total number of triangle

19C3=969

Total number of isosceles triangle are 19

Rest are scalene 

P(scalene triangle)=969-19/969

=950/969

3.Consider x and y as squares of consecutive natural number 

Let x=a²  y=(a+1)² 

Then   √ (xy)=a(a+1)=a²+a

(x+y-1)/2=(2a²+2a)/2=a²+a

4.Let D be the intersection of PC with AB. Then one of the angle ADC andBDC is obtuse, say  ADC. Then in triangle ADC, AC>CD. Since AB ≥ AC, it follows that AB > CP. On the other hand, by the triangle inequality PA+PB > AB, it follows that PA+PB > PC.

5.

6.Let the line passing through B and parallel to AC meet the bisector of ∠C in point N (Fig.2). Since ∠BNC = ∠ACN = ∠BCN, triangle 4BCN

is isosceles and BM is its median. Thus S(AMC) = 1/2 S(ANC )=1/2=S(ABC) =1/2.

Where S denotes area

7) At a given perimeter an equilateral triangle has maximum possible area

So let the side of the equilateral triangle be a

    3a=126

=>a=42

Area of an equilateral triangle is 

(√3)/4•a² =  763.83(Ans)

8.We will use the 60° clockwise rotation around A. Let P be the image of P through this rotation. Since BP = CP = 5 and PP = AP = 3, PP'B is a right triangle (Figure 1.1.) with  BP'P = 90°. Also, in the equilateral triangle APP', the measure of the angle P

PA is 60°; hence  APB = 150°.

 Applying the law of cosines in the triangle APB, 

 We obtain AB^2 = 3² +4² +2 · 3 · 4 ·

9.

10.   Let P and Q be the midpoints of diagonals AC and BD. Then

AX² + CX² = 2PX ²+AC²/2 and BX² + DX² = 2QX² + BD²/2

and, therefore, in heading b) the locus to be found consists of points X such that PX² − QX²=1/4(BD²− AC²) and in heading a) 

P = Q and, therefore, the considered quantity is equal to 

1/2 (BD² − AC²).

11. Try to do it yourself, If any further problem tell us your doubt in the "Ask a question" in the sidebar.

12)Solution. In the rst part we prove CS = CE, afterwards we will prove ∠SCE = 60°.

Since C and D lie on the symmedian of BS, CBDS is a kite with

CS = CB and ∠CDB = ∠SDC = ∠EDC.

The line segments CB and CE therefore have equal length (equal inscribed angles at D). Hence, we

have shown that CS = CB = CE. Furthermore there exists a circle k1 with C containing E, S and B.

The triangle ESA is isosceles with base ES, because it is similar to the isosceles triangle BSD with

base BS (∠SEA = ∠DEA = ∠DBA = ∠DBS and ∠EAS = ∠EAB = ∠EDB = ∠SDB follow from

the inscribed angle theorem in k). We therefore have AE = AS = r, that is, the length of the chord AE

is equal to the radius r of the circle k. The central angle for the chord AE is therefore 60°, the inscribed

angle ∠ABE is 60/2 = 30°. But now ∠SBE = ∠ABE = 30° is an inscribed angle in k1 for the chord

SE, hence the central angle is ∠SCE = 2 · 30 = 60°.