ISI MStat PSB 2006 Problem 2 | Cauchy & Schwarz come to rescue

This is a very subtle sample problem from ISI MStat PSB 2006 Problem 2. After seeing this problem, one may think of using Lagrange Multipliers, but one can just find easier and beautiful way, if one is really keen to find one. Can you!

Problem- ISI MStat PSB 2006 Problem 2


Maximize \(x+y\) subject to the condition that \(2x^2+3y^2 \le 1\).

Prerequisites


Cauchy-Schwarz Inequality

Tangent-Normal

Conic section

Solution :

This is a beautiful problem, but only if one notices the trick, or else things gets ugly.

Now we need to find the maximum of \(x+y\) when it is given that \(2x^2+3y^2 \le 1\). Seeing the given condition we always think of using Lagrange Multipliers, but I find that thing very nasty, and always find ways to avoid it.

So let's recall the famous Cauchy-Schwarz Inequality, \((ab+cd)^2 \le (a^2+c^2)(b^2+d^2)\).

Now, lets take \(a=\sqrt{2}x ; b=\frac{1}{\sqrt{2}} ; c= \sqrt{3}y ; d= \frac{1}{\sqrt{3}} \), and observe our inequality reduces to,

\((x+y)^2 \le (2x^2+3y^2)(\frac{1}{2}+\frac{1}{3}) \le (\frac{1}{2}+\frac{1}{3})=\frac{5}{6} \Rightarrow x+y \le \sqrt{\frac{5}{6}}\). Hence the maximum of \(x+y\) with respect to the given condition \(2x^2+3y^2 \le 1\) is \(\frac{5}{6}\). Hence we got what we want without even doing any nasty calculations.

Another nice approach for doing this problem is looking through the pictures. Given the condition \(2x^2+3y^2 \le 1\) represents a disc whose shape is elliptical, and \(x+y=k\) is a family of straight parallel lines passing passing through that disc.

The disc and the line with maximum intercept.

Hence the line with the maximum intercept among all the lines passing through the given disc represents the maximized value of \(x+y\). So, basically if a line of form \(x+y=k_o\) (say), is a tangent to the disc, then it will basically represent the line with maximum intercept from the mentioned family of line. So, we just need to find the point on the boundary of the disc, where the line of form \(x+y=k_o\) touches as a tangent. Can you finish the rest and verify weather the maximum intercept .i.e. \(k_o= \sqrt{\frac{5}{6}}\) or not.


Food For Thought

Can you show another alternate solution to this problem ? No, Lagrange Multiplier Please !! How would you like to find out the point of tangency if the disc was circular ? Show us the solution we will post them in the comment.

Keep thinking !!


ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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ISI MStat 2019 PSA Problem 14 | Reflection of a point

This is a problem from ISI MStat 2019 PSA Problem 14. First, try the problem yourself, then go through the sequential hints we provide.

Reflection of a point - ISI MStat Year 2019 PSA Question 14


The reflection of the point (1,2) with respect to the line \(x+2 y=15\) is

  • (3,6)
  • (6,3)
  • (10,5)
  • (5,10)

Key Concepts


Straight line

Check the Answer


Answer: is (5,10)

ISI MStat 2019 PSA Problem 14

Precollege Mathematics

Try with Hints


Find an algorithm to find the reflection,

Find the line perpendicular to \( x+2 y=15\) through (1,2).
Find the point of intersection.
Use Midpoint Segment Result.

The line perpendicular to \( x+2 y=15\) is of the form \(-2x+y=k \) .Now it passes through (1,2) . So, \( -2+2=k \Rightarrow k=0 \)

Hence the line perpendicular to \( x+2 y=15\) through (1,2) is y=2x.

Now we will find point of intersection (Foot of Perpendicular )

(3,6) is the point of intersection i.e the foot of perpendicular.

Use Mid-Point Formula (special case of Section formula) to get required point (Foot of perpendicular is mid-point of reflection and original point)

\( (3,6)=( \frac{x+1}{2} , \frac{y+2}{2} ) \) \( \Rightarrow x=5 , y=10 \)

Therefore the reflection point is (5,10) .

ISI MStat 2019 PSA Problem 14
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Coordinate Geometry - B.Stat. (Hons.) Admission Test 2005 – Objective Problem 5

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What are we learning ?

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Competency in Focus: Coordinate Geometry

This problem from B.Stat. (Hons.) based on coordinate geometry Admission Test 2005 – Objective Problem 5  is based nature of curve.

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First look at the knowledge graph:-

[/et_pb_text][et_pb_image src="https://www.cheenta.com/wp-content/uploads/2020/02/drawit-diagram-9.png" alt="calculation of mean and median- AMC 8 2013 Problem" title_text=" mean and median- AMC 8 2013 Problem" align="center" force_fullwidth="on" _builder_version="4.3.1" min_height="429px" height="189px" max_height="198px" custom_padding="10px|10px|10px|10px|false|false"][/et_pb_image][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Next understand the problem

[/et_pb_text][et_pb_text _builder_version="4.2.2" text_font="Raleway||||||||" text_font_size="20px" text_letter_spacing="1px" text_line_height="1.5em" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]

The equation $x(x+3)=y(y-1)-2$ represents
(A) a hyperbola

(B) a pair of straight lines
(C) a point

(D) none of the foregoing curves

[/et_pb_text][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.3.1" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" _builder_version="4.3.1" open="on"]B.Stat. (Hons.) Admission Test 2005 – Objective problem 5[/et_pb_accordion_item][et_pb_accordion_item title="Key Competency" open="off" _builder_version="4.3.1" inline_fonts="Abhaya Libre"]

Coordinate Geometry

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" open="off"]4/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.3.1" open="off"]Challenges and Thrills in Pre College Mathematics

Excursion Of Mathematics[/et_pb_accordion_item][/et_pb_accordion][/et_pb_column][/et_pb_row][et_pb_row _builder_version="4.0"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="4.0.9" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|0px|20px||" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Start with hints 

[/et_pb_text][et_pb_tabs _builder_version="4.2.2"][et_pb_tab title="HINT 0" _builder_version="4.0.9"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="HINT 1" _builder_version="4.2.2"]

It cannot be a straight line because Straight line are linear equation of the form. \(ax+by+c=0\).

[/et_pb_tab][et_pb_tab title="HINT 2" _builder_version="4.2.2"]

So it may be a circle or a point if radius is zero, But when we generalized it to the standard form of circle we get negative radius. so it cant be either of this one.

Hint: Stanrd form of circle is $(x-h)^{2}+(y-k)^{2}=r^{2}$.

[/et_pb_tab][et_pb_tab title="HINT 3" _builder_version="4.2.2"]

Now lets try to factorize to find the product of two linear equations so as we can verify the pair of straight line.

$a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0$
This equation represents two straight lines, if $\Delta=a b c+2 f g h-a t^{2}-b g^{2}-c h^{2}=0$
or $\left|\begin{array}{lll}{a} & {h} & {g} \\ {h} & {b} & {f} \\ {g} & {f} & {c}\end{array}\right|=0$.

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The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

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Slope of straight line - AMC 10B, 2012 Problem 3

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What are we learning ?

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Competency in Focus: Slope of straight line.

This problem is based on slope of straight line  from American Mathematics contest (AMC 10B, 2012). It includes image formation due to reflection from a line of A point.

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First look at the knowledge graph.

[/et_pb_text][et_pb_image src="https://www.cheenta.com/wp-content/uploads/2020/02/drawit-diagram-10.png" align="center" force_fullwidth="on" _builder_version="4.3.1" min_height="388px" height="198px" max_height="207px"][/et_pb_image][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Next understand the problem

[/et_pb_text][et_pb_text _builder_version="4.3.1" text_font="Raleway||||||||" text_font_size="20px" text_letter_spacing="1px" text_line_height="1.5em" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]The point in the xy-plane with coordinates $(1000, 2012)$ is reflected across the line $y = 2000$. What are the coordinates of the reflected point? $\textbf{(A)}\ (998,2012)\qquad\textbf{(B)}\ (1000,1988)\qquad\textbf{(C)}\ (1000,2024)\qquad\textbf{(D)}\ (1000,4012)\qquad\textbf{(E)}\ (1012,2012)$  [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="4.0"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.3.1" 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="4.3.1"]American Mathematical Contest 2012, AMC 10B  Problem 3[/et_pb_accordion_item][et_pb_accordion_item title="Key Competency" _builder_version="4.3.1" open="off"]Slope of the Sriaght line[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" open="off"]4/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.2.2" open="off"]Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="4.0.9" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|0px|20px||" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Start with hints 

[/et_pb_text][et_pb_tabs _builder_version="4.3.1"][et_pb_tab title="HINT 0" _builder_version="4.3.1"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="HINT 1" _builder_version="4.3.1"]

The line $y = 2000$ is a horizontal line located $12$ units beneath the point $(1000, 2012)$. When a point is reflected about a horizontal line, only the $y$ - coordinate will change.[/et_pb_tab][et_pb_tab title="HINT 2" _builder_version="4.3.1"]The $x$ - coordinate remains the same. Since the $y$-coordinate of the point is $12$ units above the line of reflection, the new $y$ - coordinate will be $2000 - 12 = 1988$. Thus, the coordinates of the reflected point are $(1000, 1988)$.[/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built="1" fullwidth="on" _builder_version="4.2.2" global_module="50833"][et_pb_fullwidth_header title="AMC - AIME Program" button_one_text="Learn More" button_one_url="https://www.cheenta.com/amc-aime-usamo-math-olympiad-program/" header_image_url="https://www.cheenta.com/wp-content/uploads/2018/03/matholympiad.png" _builder_version="4.2.2" title_level="h2" background_color="#00457a" custom_button_one="on" button_one_text_color="#44580e" button_one_bg_color="#ffffff" button_one_border_color="#ffffff" button_one_border_radius="5px"]

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Beautiful problems from Coordinate Geometry

The following problems are collected from a variety of Math Olympiads and mathematics contests like I.S.I. and C.M.I. Entrances. They can be solved using elementary coordinate geometry and a bit of ingenuity.

  1. The equation \( x^2 y - 3xy + 2y = 3 \) represents:
    • (A) a straight line;
    • (B) a circle;
    • (C) a hyperbola
    • (D) none of the foregoing curves;
  2. The equation \( r = 2a \cos \theta + 2b \sin \theta \) in polar coordinates represents:
    • (A) a circle passing through the origin;
    • (B) a circle with the origin lying outside it;
    • (C) a circle with radius \( 2 \sqrt {a^2 + b^2 } \) ;
    • (D) a circle with the center at the origin;
  3. The curve whose equation in polar coordinates is \( r \sin^2 \theta - \sin \theta - r = 0 \), is
    • (A) an ellipse;
    • (B) a parabola;
    • (C) a hyperbola;
    • (D) none of the foregoing curves;
  4. A point P on the line 3x + 5y = 15 is equidistant from the coordinate axes can lie in
    • (A) quadrant I only;
    • (B) quadrant I or quadrant II;
    • (C) quadrant I or quadrant III;
    • (D) any quadrant;
  5. The set of all points (x, y) in the plane satisfying the equation \( 5x^2 y - xy + y = 0 \) forms:
    • (A) A straight line;
    • (B) a parabola;
    • (C) a circle;
    • (D) none of the foregoing curves;
  6. The equation of the line through the intersection of the lines $$ 2x + 3y + 4 = 0 \textrm{and} 3x + 4y - 5 = 0 $$ and perpendicular to \( 7x - 5y+ 8 = 0 \) is:
    • (A) 5x + 7y - 1 = 0;
    • (B) 7x + 5y + 1 = 0;
    • (C) 5x - 7y + 1 = 0;
    • (D) 7x - 5y - 1 = 0;
  7. The two equal sides of an isosceles triangle are given by the equations y = 7x and y = -x and its third side passes through (1, -10). Then the equation of the third side is
    • (A) 3x + y + 7 = 0 or x - 3y - 31 = 0
    • (B) x + 3 y + 29 = 0 or -3x + y + 13 = 0
    • (C) 3x + y + 7 = 0 or x + 3y + 29 = 0
    • (D) x - 3y - 31 = 0 or - 3x + y + 13 = 0
  8. The equations of two adjacent sides of a rhombus are given by y = -x and y = 7x. The diagonals of the rhombus intersect each other at the point (1, 2). The area of the rhombus is:
    • (A) \( \frac{10}{3} \)
    • (B) \( \frac{20}{3} \)
    • (C) \( \frac{50}{3} \)
    • (D) none of the foregoing quantities.

More problems are in the Cheenta student portal. You may send answers to support@cheenta.com.

We will keep on adding more problems in this list as well.