Extremal Principle for Counting - AMC 10

Extremal Principle is used in a variety of problems in Math Olympiad. The following problem from AMC 10 is a very nice example of this idea.

AMC 10 Problem 4 (2019)- Based on Extremal Principle

A box contains 28 red balls, 20 green balls, 19 yellow balls, 13 blue balls, 11 white balls, and 9 black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 15 balls of a single color will be drawn?

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Try this problem. Send answers to helpdesk@cheenta.com

Suppose there 5 blue balls, 7 white balls, and 9 green balls. At least how many balls should you pick up (without looking a and without replacement) to be sure that you have picked up at least 4 balls of the same color?

You can also try these problems related to spiral similarity.

You may also click the link to learn its application:- https://www.youtube.com/watch?v=8o8AAWt960o

Spiral Similarity

Geometric Transformation is a powerful tool in Geometry. We look at one such transformation: Spiral Similarity.

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Try these Spiral Similarity problems. Send answers to helpdesk@cheenta.com

Problem 1: Suppose O = (0, 0), A = (2, 0) and B = (0, 2). Let T be the spiral similarity that sends A to B (center of T is O). What is the angle of spiral similarity? What is the dilation coefficient?

Problem 2: Can you rigorously prove the claim made at the end of the video?

Problem 3: Revisit the spiral similarity T described in problem 1. This can be realized by multiplying a complex number to A. What is that complex number?

Also try the problems related to Cyclic Quadrilaterals.

You may also click the link to learn its application:- https://www.youtube.com/watch?v=8o8AAWt960o

Spiral Similarity of cyclic quadrilaterals

ABC be any triangle. P is any point inside the triangle ABC. ( PA_1, PB_1, PC_1 ) be the perpendiculars dropped from P on the sides BC, CA and AB respectively. ( A_1 B_1 C_1) constitutes a pedal triangle.

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Advanced Math Olympiad Program

Drop perpendiculars from P on ( A_1 B_1, , B_1 C_1, C_1 A_1 ) at ( C_2, A_2, B_2 ) respectively. ( A_2 B_2 C_2 ) known as the second pedal triangle.

Finally, repeat the process to have the third pedal triangle (A_3 B_3 C_3 ).

Proposition (easy angle chasing): The third pedal triangle is similar to the original triangle ( ( \Delta ABC \sim \Delta A_3 B_3 C_3 ) )

Spiral Similarity Cyclic Quadrilateral Pedal triangle

Spiral Similarity

Notice that quadrilateral ( q_1 = P A_1 B C_1 ) is cyclic (why?). Rotate ( q_1 ) by ( 180^\circ ) and dilate it by a factor of ( \frac {1}{8} ). This spiral similarity sends the vertex B to ( B_3 ).

Exercise 1: Proof this using complex bashing or otherwise.

Exercise 2: Normalize by recreating the process in an equilateral triangle.

Remark: It is interesting to note that ( P A_1 B C_1 ) appears to be spirally similar to ( PB_2 C_1 A_2 ) and ( PC_3 A_2 B_3 ) but that does not happen.

 

Spiral Similarity - Complex Number 1

What is spiral similarity?

Spiral similarity, for our purpose, will be a beautiful motion in the plane. It is a combination of rotation and dilation (expansion or contraction by a constant factor.

Suppose AB is a (finite) segment in the plane. CD be another segment. Assume AB and CD to be disjoint.

It is possible to send AB onto CD using a combination of rotation and dilation.

Spiral Similarity and Complex Number

To accomplish this, we need to find:

How to find center of rotation (and dilation)

Join AC and BD. Suppose they intersect at X. Draw the circles CXD and AXB. Let them intersect at O.

O could be same as X or could be a different point.

Claim: O is the center of rotation and dilation, that sends AB to CD.

Spiral Similarity and Complex Number

Join OA, OX, OB, OC, OD.

Pause and think: Can you find some cyclic quadrilaterals in this picture.

Spiral Similarity and Complex Number

Clearly ABXO is cyclic. This implies \( \angle OBX = \angle OAX \). This is because they are subtended on the circumference by the same segment OX. (Angles subtended by a chord at the circumference are equal, by property of cyclic quadrilaterals).

Similarly CDXO is cyclic implying \( \angle ODX = \angle OCX \)

Pause and think: Can you find a pair of similar triangles?

Hint: Equiangular triangles are similar.

\( \Delta OBD \sim \Delta OAC \)

This is because we have shown pair of corresponding angles to be equal. Hence the two triangles are equiangular.

Pause and imagine:

This implies \( \angle AOC = \angle BOD \). We may imagine this as following:

Since the angles (of rotation) \( \angle AOC, \angle BOD \) are equal, hence this works out nicely.

Moreover, OA grows into OC and OB grows into OD by same constant factor. This is true because \( \frac {OA}{OC} = \frac{OB}{OD} \). This follows from the similarity of triangles.

Hence A goes to C and B goes to D, via rotation and dilation, centered at O.

Rotation and dilation is known spiral similarity.

Complex numbers are algebraic tools to encode this motion in the plane.

(This is a supplemental note to live lecture session of Complex Number module of Cheenta I.S.I., C.M.I. Entrance Program and Math Olympiad Program)