There are 3 hours to solve 6 questions of increasing difficulty (A1 to A6), followed by a 2 hour lunch break, followed by another 3 hours to solve 6 questions of increasing difficulty (B1 to B6). Each question is worth 10 points.

Even though the students taking the exam are skilled, in 1998 about 15 percent scored a 0, 35 percent scored between 1 and 10 points, and 21 percent scored between 11 and 20 points. So about 70 percent of participants did not fully solve even 2 questions on the exam. So even solving 1 problem can be seen as an accomplishment.

Today’s puzzle features problem B1 from 1998, which is a relatively easier but still fun question.

As usual, watch the video for a solution.

Easy Question From 1998 Putnam

Or keep reading.

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Answer To Easy Question From 1998 Putnam

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

First simplify the fraction:

Factor the 3 so the final expression is:

3(x + 1/x)

Recall the arithmetic mean geometric mean inequality for a, b > 0:

(a + b)/2 ≥ √(ab)
(a + b) ≥ 2√(ab)
Equality holds if and only if a = b

Since x > 0, let a = x and b = 1/x to get:

3(x + 1/x)
≥ 3(2)√[x(1/x)]
= 6

The minimum occurs when x = 1/x, which for positive x is x = 1.

Thus the minimum value is 6 and occurs when x = 1. And that’s the answer!

Special thanks this month to:

Kyle
Daniel Lewis
Lee Redden

Thanks to all supporters on Patreon and YouTube!

References

Putnam 1998
https://kskedlaya.org/putnam-archive/1998.pdf
https://prase.cz/kalva/putnam/psoln/psol987.html

Putnam 1998 stats
https://kskedlaya.org/putnam-archive/putnam1998results.html
https://kskedlaya.org/putnam-archive/putnam1998stats.html

Putnam 1998 B1
https://www.pinterest.com/pin/891009107547209844/
@LetsSolveMathProblems video solution
https://www.youtube.com/watch?v=MkRqe4x8f8E

University lecture
https://commons.wikimedia.org/wiki/File:Mathematics_lecture_at_the_Helsinki_University_of_Technology.jpg

Visual proof AM-GM inequality
https://www.youtube.com/watch?v=_gTClIbHreI

Which of these square numbers also happens to be the sum of two smaller square numbers?

A. 16
B. 25
C. 36
D. 49

The question was worth $16,000, but the contestant would drop down to just $1,000 for an incorrect answer. Can you figure it out?

As usual, watch the video for a solution.

When Not Knowing Math Could Cost You $15000

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Answer To When Not Knowing Math Could Cost You $15000

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Every student should instantly recognize the correct answer of B. 25 from the famous right triangle triple:

3² + 4² = 5² = 25

The 3-4-5 triangle was first known to the ancient Egyptians who used it for surveying land and constructing pyraminds. This is so ancient that it pre-dates Greek mathematicians like Archimedes, Pythagoras and Euclid by more than 1000 years!

So every student should know the answer instinctively, and any doubt would reflect a failure of the education system.

What happened on the show

Sadly, our education system is failing students.

The contestant was having pressure under the spotlight, so he used a lifeline and asked the audience. The studio audience answered in the following proportions:

A. 16 (50%)
B. 25 (30%)
C. 36 (19%)
D. 49 (1%)

The online AOL community also chose option A with the highest percentage of replies.

A. 16 (45%)
B. 25 (22%)
C. 36 (23%)
D. 49 (10%)

The contestant went with the audience answer of (A) 16 and was eliminated from the show, leaving with just $1000 instead of advancing to $16,000.

The lesson is: always pay attention in math class!

Verifying other answers are wrong

Suppose some square is equal to the sum of two smaller squares. Then we would have:

c² = a² + b²
c² – a² = b²

To see if a perfect square is the sum of two smaller squares, we can subtract smaller square numbers, one by one, and see if the remainder is also a perfect square.

Let’s test 4² = 16. We can subtract the smaller squares and see that none of the resulting differences is a perfect square.

4² – 1² = 15
4² – 2² = 12
4² – 3² = 7

Now let’s analyze 6² = 36 in the same fashion.

6² – 1² = 35
6² – 2² = 32
6² – 3² = 27
6² – 4² = 20
6² – 5² = 11

Now let’s analyze 7² = 49 in the same fashion.

7² – 1² = 48
7² – 2² = 45
7² – 3² = 40
7² – 4² = 33
7² – 5² = 24
7² – 6² = 13

References

Egypt
https://commons.wikimedia.org/wiki/File:All_Gizah_Pyramids.jpg
Ricardo Liberato, CC BY-SA 2.0 , via Wikimedia Commons
https://commons.wikimedia.org/wiki/File:Corda-a-12-nodi.svg

Video clip
https://www.youtube.com/watch?v=JpGya7hKtng

Reddit
https://www.reddit.com/r/facepalm/comments/rprz3s/when_not_knowing_high_school_level_math_costs_you/

As usual, watch the video for a solution.

Missing Area In Square

Or keep reading.

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Answer To Missing Area In Square

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Thanks Rolan for alerting me of several typos!

Triangles DHF and DAE are similar because each has a right angle, and they share a common angle ADE.

The ratio of their areas is the square of the ratio of corresponding sides:

[DHF]/[DAE] = 4/9
(HF)²/(AE)² = 4/9
(HF)/(AE)² = 2/3

Let HF = 2x so AF = 3x. Since we know the area of each right triangle and one leg, we can determine the value of the other leg in terms of x.

[DHF] = 4
(DH)(HF)/2 = 4
(DH)(2x)/2 = 4
DH = 4/x

[DAE] = 9
(AD)(AE)/2 = 4
(AD)(3x)/2 = 4
AD = 6/x

Since ABCD is a square CD = AD = 6/x.

Now construct GI parallel to CD.

Triangles GIF and DAE are congruent. They are similar because each has a right angle, and from the similarity of DAE and DHF the angles DFH and DEA are congruent. Since IG = AD the triangles GIF and DAE are congruent. Thus they have the same area of 9, and IF = AE = 3x.

We can now solve for the area of DIGC.

[DIGC]
= [DFGC] – [GIF]
= 4 + 11 – 9
= 6

But we have:

[DIGC] = (DC)(DI)
6 = (6/x)(DI)
x = DI

We now have the 3 sides of the right triangle DFH in terms of x, so we have:

(FH)² + (DH)² = (DF)²
(2x)² + (4/x)² = (4x)²
4x² + 16/x² = 16x²
16/x² = 12x²
4/3 = x⁴
2/√3 = x²

The area of the square ABCD is then:

(DC)²
= (6/x)²
= 36/x²
= 36/(2/√3)
= 18√3

But the area of the square is also the sum of 5, 4, 11, and the missing area. So we have:

5 + 4 + 11 + [missing] = 18√3
[missing] = 18√3 – 20

Thus the missing area is 18√3 – 20 ≈ 11.18 square meters.

Special thanks this month to:

Lee Redden
Kyle
Daniel Lewis

Thanks to all supporters on Patreon and YouTube!

Reference

Puzzle by Angel Silva Palacios, solution by Chew-Seong Cheong
https://www.facebook.com/photo/?fbid=1748313732795477&set=pb.100028506018835.-2207520000
https://www.pinterest.com/pin/713890978478397767/