Test of Mathematics Solution Subjective 170 - Infinite Circles

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.

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 \(R_{n+1}\).

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

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

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

What is this topic: Coordinate Geometry

What are some of the associated concept: Intercept form of straight line equation.

Book Suggestions: Coordinate Geometry Volume I by S.L. Loney

Test of Mathematics Solution Subjective 63 - Pair of Straight Lines

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.

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.

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............

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

But try the problem first...

Answer: \(-8\)

Source

Suggested Reading

AMC-10A (2014) Problem 21

Pre College Mathematics

Try with Hints

First hint

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

But try the problem first...

Answer:\(18\)

Source

Suggested Reading

PRMO-2019, Problem 12

Pre College Mathematics

Try with Hints

First hint

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 ..........

Second Hint

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 \(k \geq 3\)

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

But try the problem first...

Answer:\(36\)

Source

Suggested Reading

PRMO-2018, Problem 13

Pre College Mathematics

Try with Hints

First hint

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

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

But try the problem first...

Answer: (-1,2)

Source

Suggested Reading

AMC 10B - 2019 - Problem No - 4

Challenges and Thrills in Pre-College Mathematics

Try with Hints

First Hint

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

Final Hint

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

[/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 be an equilateral triangle and in its interior. The distances from to the triangle's sides are denoted by respectively, where . Find the locus of the points for which 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 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 . 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 satisfy (why?). For any in the interior, Let be the foot of the perpendicular from to and be the foot of the perpendicular from to the XY plane. It is possible to show that (it is because the angle between the plane and the Z axis is ). Hence, . By symmetry, . This relates the two coordinate systems. For a triangle with sides , the square of the area is . For this to be positive, we must have (after simplification) . In cartesian coordinates, this translates to , which is equivalent to . As lies on the plane , this means that . 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 , which is precisely the interior of the circumcircle of .

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://www.cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark"][/et_pb_button][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="50px||50px" custom_padding="20px|20px|20px|20px"]