Test of Mathematics Solution Subjective 170 - Infinite Circles

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 170 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


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Problem

Let \({C_n}\) be an infinite sequence of circles lying in the positive quadrant of the \(XY\)-plane, with strictly decreasing radii and satisfying the following conditions. Each \(C_n\) touches both the \(X\)-axis and the \(Y\)-axis. Further, for all \(n\geq 1\), the circle \(C_{n+1}\) touches the circle \(C_n\) externally. If \(C_1\) has radius \(10\: cm\), then show that the sum of the areas of all these circles is \(\frac{25\pi}{3\sqrt{2}-4} \: cm^2\).


 

Solution

Consider the following diagram where the Green line segment is \(R_n\), the radius of the \(n^{th}\) circle, and the Yellow line segment is \(R_{n+1}\).

circle

 

As we are told about the symmetricity of the figure in the problem we can say that:

\(\sqrt{2}R_{n+1} + R_{n+1} + R_n = \sqrt{2} R_n\)

\(=> R_{n+1}(\sqrt{2}+1)=R_n(\sqrt{2}-1)\)

\(=> R_{n+1}= (3-2\sqrt{2})R_n\)

Let's say \(=> R_{n+1}= \alpha.R_n\).

Now the total sum of the areas of the circles is:

\((\pi R_1^2 + \pi R_2^2 + \cdots ) = \pi (R_1^2 + R_2^2 + R_3^2 + \cdots )\)

Now as \(R_{n+1}= \alpha.R_n\), we can say that:

\(\pi (R_1^2 + R_2^2 + R_3^2 + \cdots ) = \pi (R_1^2 + \alpha^2 R_1^2 + \alpha^4 R_1^2 + \cdots ) = \pi \frac{R_1^2}{1-\alpha^2}\) as \(\alpha^2 < 1\).

Substituting the value of \(\alpha = 3-2\sqrt{2}\) and \(R_1 = 10 \: cm\)we have,

Sum = \(\frac{25\pi}{3\sqrt{2}-4} \: cm^2\).

Hence Proved.

Test of Mathematics Solution Subjective 113 - Vertices of a Triangle

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 113 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


Problem:

Find the vertices of the two right angles triangles, each having area 18 and such that the point (2, 4) lies on the hypotenuse, and the other two sides are formed by the \(x\) and \(y\) axes.


Solution:

Suppose the vertices are (a, 0) and (0, b). Clearly $ \frac{1}{2} ab = 18 $ or ab = 36.
Also the equation of the line through (0,a), (b,0) is $ \displaystyle{ \frac{x}{a} + \frac{y}{b} = 1 } $. Since we know that (2, 4) is on that line, there fore $ \displaystyle { \frac{2}{a} + \frac{4}{b} = 1 } $.
In this equation, lets replace $ a $ by $ \frac{36}{b} $. Hence we get $ \displaystyle { \frac{2}{\frac{36}{b}} + \frac{4}{b} = 1 } $ or $ \displaystyle { \frac{b}{18} + \frac{4}{b} = 1 } $.
Therefore we get a quadratic in b.
$ \displaystyle { b^2 -18b + 72 = 0 } $. We can simply middle term factorize this to find $ \displaystyle {(b-12)(b-6) = 0 } $. Thus b = 12 or 6 implying a = 3 or 6.
The other two vertices are: (0,12) and (3, 0) OR (0,6) and (6,0).


Chatuspathi: Test of Mathematics Solution Subjective 113

Test of Mathematics Solution Subjective 63 - Pair of Straight Lines

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 63 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

If any one pair among the straight lines
ax + by = a + b, bx –(a + b)y = - a, (a + b)x –ay = b
intersect, then show that the three straight lines are concurrent.


Solution:

Three lines are concurrent if each of them is linear combination of other two & they are not parallel.
Now given one pair intersect that is they are not parallel.
Now
ax + by = a + b … (i)
bx – (a + b)y = - a …(ii)
(a +b)x –ay = b …(iii)
(i) + (ii) - (iii) = 0
So they are concurrent.

Graph Coordinates | AMC 10A, 2015 | Question 12

Try this beautiful Problem on Graph Coordinates from coordinate geometry from AMC 10A, 2015.

Graph Coordinates - AMC-10A, 2015- Problem 12


Points $(\sqrt{\pi}, a)$ and $(\sqrt{\pi}, b)$ are distinct points on the graph of $y^{2}+x^{4}=2 x^{2} y+1 .$ What is $|a-b| ?$

,

  • $0$
  • $1$
  • $2$
  • $3$
  • \(4\)

Key Concepts


Co-ordinate geometry

graph

Distance Formula

Check the Answer


Answer: $2$

AMC-10A (2015) Problem 12

Pre College Mathematics

Try with Hints


The given points are $(\sqrt{\pi}, a)$ and $(\sqrt{\pi}, b)$ which are satisfying the equation $y^{2}+x^{4}=2 x^{2} y+1$.

So we can write $y^{2}+\sqrt{\pi}^{4}=2 \sqrt{\pi}^{2} y+1$

Can you now finish the problem ..........

Therefore

$y^{2}+\pi^{2}=2 \pi y+1$
$y^{2}-2 \pi y+\pi^{2}=1$
$(y-\pi)^{2}=1$
$y-\pi=\pm 1$
$y=\pi+1$
$y=\pi-1$

can you finish the problem........

There are only two solutions to the equation, so one of them is the value of $a$ and the other is $b$. our requirement is $|a-b|$ so between a and b which is greater is not importent............

So, $|(\pi+1)-(\pi-1)|=2$

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Sum of Co-ordinates | AMC-10A, 2014 | Problem 21

Try this beautiful problem from Algebra based on Sum of Co-ordinates

Sum of Co-ordinates - AMC-10A, 2014- Problem 21


Positive integers $a$ and $b$ are such that the graphs of $y=ax+5$ and $y=3x+b$ intersect the $x$-axis at the same point. What is the sum of all possible $x$-coordinates of these points of intersection?

  • \(16\)
  • \(12\)
  • \(-8\)
  • \(-4\)

Key Concepts


Geometry

Co-ordinate

Check the Answer


Answer: \(-8\)

AMC-10A (2014) Problem 21

Pre College Mathematics

Try with Hints


The given equations are $y=ax+5$\(\Rightarrow x=\frac{-5}{a}\).....(1)

$y=3x+b$\(\Rightarrow x=\frac{-b}{3}\)............(2)

Since two lines intersect the $x$-axis at the same point,then at first we have to find out the common point on x -axis.........

Now the intercept form of the given two equations will be

\(\frac{x}{(-5/a)} +\frac{y}{5}=1\) ,Therefore the straight line intersect x-axis at the point (\(\frac{-5}{a},0\))

\(\frac{x}{(-b/3)}+\frac{y}{b}=1\) Therefore the straight line intersect x-axis at the point (\(\frac{-b}{3},0)\).

Can you now finish the problem ..........

Since two lines intersect the $x$-axis at the same point so we may say that

(\(\frac{-5}{a},0\))=(\(\frac{-b}{3},0)\)

\(\Rightarrow ab=15\)

The only possible pair (a,b) will be \((1,15),(3,5),(5,3),(15,1)\)

can you finish the problem........

Now if we put the values \((1,15),(3,5),(5,3),(15,1)\) in (1) & (2) we will get \(-5\),\(\frac{-5}{3}\),\(-1\),\(\frac{1}{3}\)

Therefore the sun will be \(-8\)

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Good numbers Problem | PRMO-2019 | Problem 12

Try this beautiful problem from PRMO, 2019 based on Good numbers.

Good numbers Problem | PRMO | Problem-12


A natural number \(k > \) is called good if there exist natural numbers
\(a_1 < a_2 < ………. < a_k\)

\(\frac{1}{\sqrt a_1} +\frac{1}{\sqrt a_2}+................... +\frac{1}{\sqrt a_k}=1\)

Let \(f(n)\) be the sum of the first \(n\) good numbers, \(n \geq 1\). Find the sum of all values of \(n\) for which
\(f(n + 5)/f(n)\) is an integer.

  • $20$
  • $18$
  • $13$

Key Concepts


Number theory

Good number

Integer

Check the Answer


Answer:\(18\)

PRMO-2019, Problem 12

Pre College Mathematics

Try with Hints


A number n is called a good number if It is a square free number.

Let \(a_1 ={A_1}^2\),\(a_2={A_2}^2\),..................\(a_k={A_k}^2\)
we have to check if it is possible for distinct natural number \(A_1, A_2………….A_k\) to satisfy,
\(\frac{1}{A_1}+\frac{1}{A_2}+...........+\frac{1}{A_k}=1\)

Can you now finish the problem ..........

For \(k = 2\); it is obvious that there do not exist distinct\( A_1, A_2\), such that \(\frac{1}{A_1}+\frac{1}{A_2}=1 \Rightarrow 2\) is not a good number

For \(k = 3\); we have \(\frac{1}{2} +\frac{1}{3}+\frac{1}{6}=1 \Rightarrow 3\) is a good number.

\(\frac{1}{2}+\frac{1}{2}\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1\) \(\Rightarrow 4\) is a good number

Let \(k\) wil be a good numbers for all \(k \geq 3\)

\(f(n) = 3 + 4 +… n\) terms =\(\frac{n(n + 5)}{2}\)
\(f(n + 5) =\frac{(n + 5)(n +10)}{2}\)

\(\frac{f(n+5}{f(n)}=\frac{n+10}{n}=1+\frac{10}{n}\)

Can you finish the problem........

Therefore the integer for n = \(1\), \(2\), \(5\) and \(10\). so sum=\(1 + 2 + 5 + 10 = 18\).

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Direction & Angles | PRMO-2019 | Problem 4

Try this beautiful Geometry problem from PRMO, 2019 based on Direction & Angles.

Direction & Angles | PRMO | Problem-4


An ant leaves the anthill for its morning exercise. It walks 4 feet east and then makes a 160° turn to the right and walks 4 more feet. It then makes another 160° turn to the right and walks 4 more feet. If the ant continues this pattern until it reaches the anthill again, what is the distance in feet it would have walked?

  • $20$
  • $36$
  • $13$

Key Concepts


Geometry

Co-ordinate geometry

Trigonometry

Check the Answer


Answer:\(36\)

PRMO-2018, Problem 13

Pre College Mathematics

Try with Hints


Direction & Angles - figure

According to the problem we draw the figure and try to solve using co-ordinate method...

Can you now finish the problem ..........

Let \(A_0(0,0)\)

Therefore \(A_1(4 cos 0,4sin 0)\)

\(A_2(4cos 0+4 cos 160, 4 sin 0+4 sin 160)\)

\(\Rightarrow A_n =(0,0)\)

and \(4(cos 0+cos 160 +...+sin 160(n-1)=0\)

\(\Rightarrow n =9\)

Can you finish the problem........

Therefore distance covered =\(4 \times 9\)=\(36\)

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Circles and semi-circles| AMC 8, 2010|Problem 23

Try this beautiful problem from Geometry based on Ratio of the area of circle and semi-circles.

Area of circles and semi-circles - AMC-8, 2010 - Problem 23


Semicircles POQ and ROS pass through the center O. What is the ratio of the combined areas of the two semicircles to the area of circle O?

circles and semi-circles
  • $\frac{1}{2}$
  • $\frac{2}{\pi}$
  • $ \frac{3}{2} $

Key Concepts


Geometry

Circle

co-ordinate geometry

Check the Answer


Answer:$\frac{1}{2}$

AMC-8 (2010) Problem 23

Pre College Mathematics

Try with Hints


Find the radius of the circle

Can you now finish the problem ..........

ratio of the areas

Join O and Q

can you finish the problem........

ratio of the areas of circles and semi-circles

The co-ordinate of Q is (1,1), So OB=1 and BQ=1

By the Pythagorean Theorem, the radius of the larger circle i.e OQ=\(\sqrt{1^2+1^2}\)=\(\sqrt 2\).

Therefore the area of the larger circle be \(\pi (\sqrt 2)^2=2\pi\)

Now for the semicircles, radius OB=OC=1(as co-ordinate of P=(1,1) and S=(1,-1))

So, the area of the two semicircles is  \(2\times\frac{\pi(1)^2}{2}=\pi\)

 Finally, the ratio of the combined areas of the two semicircles to the area of circle O is

\(\frac{\pi}{2\pi}\)=\(\frac{1}{2}\)

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Co-ordinate Geometry - AMC 10B - 2019 - Problem No - 4

Co-ordinate Geometry - AMC 10B - 2019 - Problem No - 4


This problem on co-ordinate geometry is from AMC 10B, 2019. First try it yourself.

All lines with equation ax+by= c such that a,b,c form an arithmetic progression pass through a common point . What are the coordinates of that point?

  • (-1,2)
  • (0,1)
  • (1,-2)
  • (1,0)

Key Concepts


Arithmetic Progression

2D - Co- ordinate Geometry

Cartesian System of Points

Check the Answer


Answer: (-1,2)

AMC 10B - 2019 - Problem No - 4

Challenges and Thrills in Pre-College Mathematics

Try with Hints


If all lines satisfy the condition, then we can just plug in values for a,b,c that form an arithmetic progression. Let's use a =1,b= 2,c=3 and a= 1,b=3,c =5. Then the two lines we get are:

x+2y =3

x+3y = 5 so

y= 2

Now plug the value of y into one of the previous equations. We get :

x+4=3

x = -1

Ans is (-1,2)

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Trilinear coordinates and locus

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.22.4"][et_pb_column type="4_4" _builder_version="3.22.4"][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_padding="20px|20px|20px|20px"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]Let $ ABC$ be an equilateral triangle and $ P$ in its interior. The distances from $ P$ to the triangle's sides are denoted by $ a^2, b^2,c^2$respectively, where $ a,b,c>0$. Find the locus of the points $ P$ for which $ a,b,c$ can be the sides of a non-degenerate triangle.

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.22.4"][et_pb_column type="4_4" _builder_version="3.22.4"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.23.3" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.23.3" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]Romanian Master in Mathematics, 2008[/et_pb_accordion_item][et_pb_accordion_item title="Topic" open="off" _builder_version="3.23.3" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]Geometry[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" open="off" _builder_version="3.23.3" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]Medium[/et_pb_accordion_item][et_pb_accordion_item title="Comments" open="off" _builder_version="3.23.3" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]This problem obviously points towards an application of trilinear coordinates (see more here). [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.23.3" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.23.3"]If there exists a non-degenerate triangle with sides a,b,c then the area of the triangle must be positive. Make use of this fact employing Heron's formula.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.23.3"]Hint 1 will give you a locus in terms of trilinear coordinates. However, it is easier to work in cartesian coordinates as we are already familiar with many curves in them.[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.23.3"]Take A=(1,0,0), B= (0,1,0), C= (0,0,1). There is a simple relationship between the cartesian and trilinear coordinates for this choice.[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.23.3"]

.Note that all points in the plane of ABC satisfy x+y+z=1 (why?). For any P in the interior, Let Q be the foot of the perpendicular from P to AB and R be the foot of the perpendicular from P to the XY plane. It is possible to show that \frac{PR}{PQ}= \sqrt{\frac{2}{3}} (it is because the angle between the plane and the Z axis is \arccos \sqrt{\frac{2}{3}}). Hence, \frac{z}{c^2}=\sqrt{\frac{3}{2}}. By symmetry, \frac{x}{a^2}=\sqrt{\frac{2}{3}}=\frac{y}{b^2}. This relates the two coordinate systems. For a triangle with sides a,b,c, the square of the area is \frac{1}{16}(a+b+c)(-a+b+c)(a-b+c)(a+b-c). For this to be positive, we must have (after simplification) a^4+b^4+c^4< 2(a^2b^2+b^2c^2+c^2a^2). In cartesian coordinates, this translates to \frac{3}{2}(x^2+y^2+z^2)<3(xy+yz+zx), which is equivalent to (x+y+z)^2>2(x^2+y^2+z^2). As P lies on the plane x+y+z=1, this means that x^2+y^2+z^2<\frac{1}{2}. This last equation is that of the interior of a solid sphere. Hence, our desired locus is the intersection of this solid sphere with the plane x+y+z=1, which is precisely the interior of the circumcircle of ABC.  

 

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Similar Problems

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