Collected Problems 2

I'm collecting various problems especially the short lived ones from twitter here for future use. My taste runs towards geometry as is probably evident. For ease of use I've hidden my notes. Click the button below to expose them.

Problem 1 [**]

There are 100 people in line to board a plane with 100 seats. The first person has lost their boarding pass, so they take a random seat. Everyone that follows takes their assigned seat if it's available, but otherwise takes a random unoccupied seat. What is the probability the last passenger ends up in their assigned seat? 


Provable via induction start with the case of  3 passengers.

Problem 2 [***]

Find the ratio of the area of the red triangle to the green quadrilateral.

Find the ratio of the radius of the incircle of a triangle to the its height in terms of its side lengths: a,b, and c.  Use Heron's formula. r/h=b/2s

Problem 3 [*]

[Simon Pampena/@solvemymaths]

How many squares of Area A can fit inside square of area B.

Problem 4 [*]

If you flip a particular unfair coin 6 times, P(3H & 3T) = 16%. If you flip the same coin 4x, what is P(2H & 2T)?

[Matt Enlow]

Don't forget about the number of combinations i..e 6!/(3!*3!)  = 20

Problem 5 [***]

[Henk Reuling]

This is quite tricky and requires calculus.
1. Derive the triangle ratio 2:1
2. Take derivative of ellipse ax^2 + y^2 = 100 and set to -1/2 at tangent to find pt (x,2ax)
3. Use similar triangles to find width is 4ax + x - 10 which gives a  2nd point
4. Plug both into the ellipse equation and solve for a to get 4/9
5. Plug the point (x,2ax) in to get (9,8) and calc the length.

Problem 6 [*]

ABCD is a 6x8 rectangle where the length of AB is 8, and length of BC is 6, and the 2 
subsegments in the cross are 2 and 3.

What's the area of EFGH?

Problem 7 [**]


In the  figure, triangle ABC is a right angle isosceles triangle with angle A = 90 degrees, and square ACDE is a parallelogram. When AE = 5 cm, BE = 8 cm, how many cm^2 is the area of ​​pentagon ABCDE?

Square off the figure by extending AE. That produces congruent triangles on both sides.
You can then find a formula for the total area 1/2(a^2+b^2) + 5a and a
Pythagorean equivalent (a+5)^2 + b^2 = 8^2.

Problem 8 [**]


1. The geometric approach is to extend chord and realize the max area is at its perpendicular
bisector. You can then easily angle chase and see this goes through the center and is also a 1:2.
2. If you have the linear alg. / calculus computing the area via vectors and differentiating is fairly
interesting as a secondary exercise.

Problem 9 [*]

"I have 7 sticks all of different lengths and all an integer length long. The longest one is shorter than a piece of paper and therefore under 30 centimeters.  Whenever I pick 3 sticks, I notice I can't form a triangle. What is the length of the shortest stick?"


Problem 10 [**]

Given the following areas, find the area X of the remaining piece of the triangle.


  • Use proportional thinking.
  • Draw a line from the top to the intersection and setup 2 proportional equations.

Problem 11 [*]



  • Extend ED to intersect with BC and an isosceles triangle will emerge.

Problem 12 [**]


  • Draw a picture to visualize and see symmetry.
  • You can square off the hexagon to find some 30-60-90 triangles.
  • From there its possible to get formulas for the area of the center triangle and exterior ones in terms of r.
  • Set a ratio between the 2 and solve using the quadratic equation \( \frac{r}{r^2 + r + 1} = \frac{1}{7} \)
  • Vieta gives a shortcut since the sum of roots is asked for.

Problem 13 [*]

Given a 4x6 cm  rectangle  (AB is the short side) what is the area of CFPG?


  • All the corners can be found directly.
  • That means EFP can be found via area addition.
  • Then calculate the ratio of HP:PF
  • That allows the remaining triangles to be found.

 Problem 14 [*]


1. All the implicit triangles are similar. You can use that and the fact that the 2 square sides are the same to find their ratios and then plug into the Pythagorean Theorem on one of them to solve.  (Its a 8:15:17) 

2. Its  much easier to calculate the diagonal of the square  which we can directly do with the distance formula since its 23 x 7.  From there its easy to get  the overall area.

Problem 15 [**]


  • By checking the intercepts of the line AB, you get side lengths and its a 30-60-90
  • This means all the other triangles are either 30-60-90 or 15-75-90's
  • I like subdividing big one: Let F be the (0,2sqrt(3)   Draw a perpendicular for the origin to DB
  • From there its possible to get an expression for length of the original triangle 6 all in terms of OD.

Problem 16 [*]


Triangles ABE and BCD are congruent by SAS.

Problem 17 [*]


1. BF is 8 from the area, EF is 17 by Pythagorean Theorem.
2. Let FH = x and  EH = 17 - x, then x + 8 = 17 - x + 15. Solve for the radius.

Problem 18 [**]


If DQ is 5cm and the area of ADQP = area of BCQP what is the length of AP?

Note: this diagram is deceptive the actual figure looks like:

So if one starts to calculate the altitude via the Pythagorean Theorem odd negative numbers will pop out.

  • First since DQ = CD = 5 i.e. its a median the area of triangle DQP = CPQ. 
  • Then using area subtraction area of triangle BCP =  APQ.
  • Drop altitudes which are parallel and alt 1 = 3/2 of alt2 to have the same area.
  • That also creates similar triangle with the side AP so AP : PB  = 3 : 2 

Problem 19 [*]

[Fred Hardwood]

* Note don't forget about the 3 different forms to check 1^n  n^0 and -1^n

Problem 20 [**]


  • Isolate y,z and substitute into the 3rd equation i.e. standard techniques work fine here.

Problem 21 [**]


The key is the 2 triangles on the left and top are congruent. You can then setup an area equation and
show the green areas are equal to middle triangle.

Problem 22[*]

Find the cube root of \(x^6  - 9x^5 + 33x^4 - 63x^3 + 66x^2 - 36x + 8 \).

[This is actually from an old book by Wallace  Boyden 1895]

Assume that a cube root exists then it must be of the form \((ax^2 + bx +c)^3\)
Expand this out and correlate with various terms. a and c fall out fairly easily.

Problem 23[*]

The 2 circles have the same radius and are tangent to each other and the sides of the 16x18 rectangle. Find the radius.


Use the Pythagorean theorem on the interior right triangle formed by the 2 centers  of the circles.

Problem 24[**]

If (x-a)(x-1) + 1 = (x+b)(x+c) and a,b and c are all integers, find the sum of all values of x.

[From a friend's son's homework.]
  • Key is this is a diophantine equation.
  • After expanding -bc = b + c  or  b = -c/1+c.
  • Do some case work mostly this is just a fraction exception around the boundaries where the denominator is 1 or -1.

Problem 25[**]

p(x) is a quadratic equation with real coefficients. If the following is true for all reals 
and p(2) = 34, find p(x)

$$x^2 + 4x + 14 <= p(x) <= 2x^2 + 6x + 18$$

  • p(x) = ax^2 + bx + c.   Substitute 2 into get one equation in terms of a, b, and c
  • Find where the 2 bounds equal each other to get another equation.
  • Check the minima, All 3 must align which gives the final piece needed to solve.

Problem 26[***]

Given the range of \(f(x)=\frac{ax^2 + 8x + b}{x^2+1}\)  is \( 1<=f(x)<=9 \)
Find a and b.

[Hint: find a+b first]


Compute f(1) and f(-1). The value of a + b falls out
Then look at either limit and check when the discriminant is 0 i.e. one root or vertex there.

Problem 27[*]

For what right triangles (if any) are the average of the length of two of the sides equal to the thirds side? [James Tanton]

Set one side to 1 and solve the equation.

Problem 28[*]

When is the following equation an integer?

$$ \large \frac{6^{r+s} \cdot 12^{r-s}}{8^r \cdot 9^{r+2s}}$$

[@solvemymaths via MAT]

Once simplified, don't forget about negative exponents.

Problem 29[*]

Equilateral △ABC with sides 12. E is the midpoint of AC. Point A is folded over DE to point F. Given that FG=2, determine FH.


There are 3 similar triangle and there is enough info to determine the ratios between 2 of them and them correlate with the overall side length of 12.

Problem 30 [*]


  • Add AM angle CAM is half CAB i.e. 30 degrees
  • Therefore angle QAC is a - 30 

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