**Note: I was a bit sloppy with crediting sources initially. I'm working on fixing that up.**

####

Problem 1.

JN is the angle bisector of corner KJM i.e. 45 degrees. What is side length LM?#### Problem 2.

- M is the median
- Area ABM = 8
- AC is parallel to DE

- Add diagonal AE.
- Altitudes (distance from DE to AC are the same for all the triangles since parallel
- triangle ACE has same area ADC since they have same base and same altitude.
- So area ABC + ADC = area ABC + ACE
- area ABM = area AME since its a median. and ABC + ACE is congruent to ABM + AME
- Therefore answer is 2 * 8 = 16

Note: this makes it seem easy what really happened which is probably about persistence is first

I identified the triangles formed by medians had the same area. ABM = AEM and ADM = DEM

Then I thought about parallel lines but went down the blind alley of saying the ADC was proportional to CDE based on one side. I was looking for a way to express the area of ABM in terms of the side and got stuck. I then drew all those lines in and looking at area addition / subtraction. I made a simple geogebra model and confirmed the goal was 16. I could even see which two areas had to be equal between ABE and ABCD but the soln wasn't apparent. A few days later when I stared back at the problem the equal area of AEC and ADC just popped out at me.

####

Problem 3.

Ratio Triangle AQH:APD is 4:9 since AH:AD 2:3

and area AFD = GCD since each is 1/3 S

That implies given the overlap that QPHD = NCPF

and therefore area GCN = AQH

Total area of ABCD - MNQP is therefore 9 * area (AQH)

Height of FD is 2/3 h from similar triangle.

Further area of AFD = 13/4 area AQH = 1/2 b * 2/3 h.

Solving for AFD you get AFD = 12/13 b * h = 12/13 S.

and area AFD = GCD since each is 1/3 S

That implies given the overlap that QPHD = NCPF

and therefore area GCN = AQH

Total area of ABCD - MNQP is therefore 9 * area (AQH)

Height of FD is 2/3 h from similar triangle.

Further area of AFD = 13/4 area AQH = 1/2 b * 2/3 h.

Solving for AFD you get AFD = 12/13 b * h = 12/13 S.

####

Problem 4.

Q,P split arc AB into thirds, find the shaded area.Fairly trivial.

####

Problem 5.

Triangle is a right isosceles with side length of 1. What is the radius of the inscribed semi-circle?

- Draw line from center of the circle to the hypotenuse.
- That forms a smaller right triangle with the side lengths of r and a hypotenuse of sqrt(2) * r.
- So the length of the side 1 = r + sqrt(2) * r

#### Problem 6.

What proportion of ABCD is DKL?

1. Split DKL into two triangles by adding a perpendicular at K. These are easily shown similar to CLN and BDN.

2. Further since they have the same base they are in exactly the same proportion.

3. CLN goes 1/2 the length of the rectangle and DBN the whole length. DKL goes 1/2.

4. So CLN and BDB must be scaled 1/3 for it to be the right length. (1 / 3 * (1/2 + 1)) = 1/2.

5. Therefore KL is 1/3 of LM and since DLM is 1/4 * 1/2 = 1/8 of the rectangle DKL is 1/3 * 1/8 = 1/24 of the rectangle.

#### Problem 7.

Find the ratio of the radii.

1. This is full of 1:2 triangles.

2. Easiest to draw median M for BC and call the cente intersection H and look at DCM and its reflection.

3. 2 right triangles that share 2 sides, also share the 3rd from Pythagorean theorem and are congruent SSS.

4. That shows HD is the same length as the radius of the red circle.

5. Repeat on the other side with AE = EM or 2r - x = r + x and you get the ED is 5/2.

6. Then notice ADE is a 3-4-5 triangle so DGM which is similar is also one.

7. Use the scaled incircles to find the ratios.

#### Problem 8.

AE:EC is 2:1 AD:DB is 1:2 EF:FD = 1:1 what fraction is the shaded area?

Easiest to do with proportional areas I started with letting AFD = A and went from there.

#### Problem 9.

Cute: What is the yellow area?

After angle chasing the two triangles are congruents 70-70-40's. Since they share an overlap that means the two non overlapping sections are equal in area. So you can swap the yellow piece with missing wedge in the 30 degree section and the area is equal to 1/3 of the quarter circle.

####

Problem 10.

Find the ratio of areas (or just the individual ones) Easiest via shape subtraction.

#### Problem 11.

Find angle X.

Unfold, find congruent angles and angle chase.

#### Problem 12.

Find the ratio of the 2 smaller circle's radii to the larger semicircle.

1. Note tangency of the upper circle means that line must also end at the center of the big circle and is parallel to the outer leg of the triangle.

2. Scale up the the similar triangle sides.

3. Apply the incircle properties to find the sides of the triangle in terms of r1 and r2. T

4. Using the Pythagorean theorem you get a relation between the 2.

5. Note: The triangle is i the 5-12-13 family!

####

Problem 13

Rectangle ABCD. E,F on AB so that AE=EF=FB. CE intersects line AD at P, PF intersects BC at Q.
BQ:QC=?

####

Problem 14

Show black and blue triangles have equal area.

Note: black = pentagon - 2 red isosceles and

blue pentagon - 1 red isosceles + 2 of the 1/2 side triangles. So the 1/2 side triangles must be half the area of the red isosceles triangles. This is easy to prove since you have another red isosceles triangle sharing the base with them that has the same height.

blue pentagon - 1 red isosceles + 2 of the 1/2 side triangles. So the 1/2 side triangles must be half the area of the red isosceles triangles. This is easy to prove since you have another red isosceles triangle sharing the base with them that has the same height.

####

Problem 15.

Draw a perpendicular down from E to F. Let BF = x then apply the pythagorean theorem twice to solve for x = 3/2. Then draw a parallel across for 3 and reapply the pythagorean theorem to get the other side. Finally scale by 4/3.

####

Problem 16.

Find the area of AEF.

Use the 2:1 triangle CEF and derive coordinates for Q and R in terms of x and 2x.

Then do the area subtraction and all the x's cancel out!

Then do the area subtraction and all the x's cancel out!

#### Problem 17.

1. Prove HC is congruent to CJ.

Use similar triangle CJ / GF = DC / DC + DC + CG and

HC / AB = EC / EC + BC.

2. Prove the 2 shaded regions are equal.

Let c = CH = CJ x = AB and Y = CG Triangle DEF has area y/2 ( y - c) + x/2(y -c ) = 1/2 ( y^2 - yc + xy - xc)

Its equal in area to CEF since same altitude and base = 1/2 y^2

So y^2 = y^2 - c(x + y) + xy or xy = c(x+y) That's the same as saying the whole triangle CDE is

equal in area to CDJ + CEH. Then just use area subtraction and you get the result.

Its equal in area to CEF since same altitude and base = 1/2 y^2

So y^2 = y^2 - c(x + y) + xy or xy = c(x+y) That's the same as saying the whole triangle CDE is

equal in area to CDJ + CEH. Then just use area subtraction and you get the result.

####

Problem 18.

Angle ABC is 60 degrees, what is the area of ABCD?

1. I experimented and discovered the BQ and QC are actually part of the diagonals AC and BD (which is an interesting thing to also prove.)

2. Using the formual for 15-17 triangles I could find BC and then using the similar triangles get a messy answer.

3. I went back added the point E on the top and filled in the full equilateral triangle BCE. This is

much easier to reason about.

4. Now draw a perpendicular line from BE to BQ . This creates a 45-45-90 and 30-60-90. From there

You can find the side length for the larger triangle BCE (2 + 2sqrt(3).

5. Then you can just subtract the 2 triangles BCE and ADE to find the area:

(sqrt(3)/4)[ (2 + 2sqrt(3)]^2 - 4^2] = 6.

6. I'd probably use similar triangles to extend and prove the extension of BQ has the same the same slope as QD. Note We can calculate the altitude of BQ by pythagorean + complete the square of the resulting square or using the square containing an equilateral triangle construction.

2. Using the formual for 15-17 triangles I could find BC and then using the similar triangles get a messy answer.

3. I went back added the point E on the top and filled in the full equilateral triangle BCE. This is

much easier to reason about.

4. Now draw a perpendicular line from BE to BQ . This creates a 45-45-90 and 30-60-90. From there

You can find the side length for the larger triangle BCE (2 + 2sqrt(3).

5. Then you can just subtract the 2 triangles BCE and ADE to find the area:

(sqrt(3)/4)[ (2 + 2sqrt(3)]^2 - 4^2] = 6.

6. I'd probably use similar triangles to extend and prove the extension of BQ has the same the same slope as QD. Note We can calculate the altitude of BQ by pythagorean + complete the square of the resulting square or using the square containing an equilateral triangle construction.

####

Problem 19.

Both triangles are equilateral, ABCD is a square. Find angle X.

1. Draw in BE

2. ABE is isosceles so by angle chasing <ABE is 15

3. Let G be the intersection of BE with DF. By angle chasing BFG is a 30-60-90.

4. So FG is 1/2 of FB and FD and the midpoint of FD.

5. Triangle DEG is a right isosceles so EG = FB = FD.

6. Therefore triangle EFG is also a right isosceles and angle X is 45.

Much Simpler

DEF is congruent to ABD via SAS.

####

Problem 20.

1. The sides of the squares are 2sqrt(2)

2. Extend BC to form a right triangle with E.

3. That triangle is a 45-45-90 and has a side of sqrt(2)

4. So the triangle BCE has an area of 1/2 * sqrt(2) * 2sqrt(2) = 2

5. Divide in half and we just need the height between the 2 lines (HI).

6. Use Pythagorean Theorem to find BE sqrt(3sqrt(2)^2 + sqrt(2)^2) = sqrt(20)

7. Now we can find the height using the Pythagorean theorem again on the right triangle BEL with the base of 2 = 4.

8. The area of half trapezoid BEHI is 1/2 * 4 (1 + 3) = 8.

9. So the whole area is 2 (2 + 8) = 20.

2. Extend BC to form a right triangle with E.

3. That triangle is a 45-45-90 and has a side of sqrt(2)

4. So the triangle BCE has an area of 1/2 * sqrt(2) * 2sqrt(2) = 2

5. Divide in half and we just need the height between the 2 lines (HI).

6. Use Pythagorean Theorem to find BE sqrt(3sqrt(2)^2 + sqrt(2)^2) = sqrt(20)

7. Now we can find the height using the Pythagorean theorem again on the right triangle BEL with the base of 2 = 4.

8. The area of half trapezoid BEHI is 1/2 * 4 (1 + 3) = 8.

9. So the whole area is 2 (2 + 8) = 20.

#### Problem 21.

The triangle is equilateral. Ratio of radii between the blue circle and green semicircle?

####

Problem 22.

Radii and legs of isosceles triangle are the same. What is x?

####

Problem 23.

Given 2 right triangle. ABD is right isosceleses. AC = 8, BD = 10. what is AE:EC?

Approach 1:

1. Think circularly as above. Once the radii are added several 3:4:5 triangles start to appear and its

easy to find sub-lengths of the AC.

2. "Complete the square" This is cool but uses a couple of additional similarity/proportionality arguments. It also misses the interior structure with the 3:4:5's.

The total interior square is 100. The total exterior square square is defined by the diagonals (AC) and 128. You can then subtract and calculate the area ratios which are the same as the segment ratios.

####

Problem 25.

Find the area of the total square.

1. Use the fact that any triangle from an entire edge to the other side is half the area of the square.

Let E be the point on AB. The CDE is 1/2 s^2.

2. Let F be the point on BC and M be the intersection OF DF and CE. Then CDM is s^2 - 30.

3. CDF is s^2 - 20 and ABF is 2s. ABF + CDF = 1/2 s^2 or s^2 - 20 + 2s = 1/2 s^2. Solve.

Let E be the point on AB. The CDE is 1/2 s^2.

2. Let F be the point on BC and M be the intersection OF DF and CE. Then CDM is s^2 - 30.

3. CDF is s^2 - 20 and ABF is 2s. ABF + CDF = 1/2 s^2 or s^2 - 20 + 2s = 1/2 s^2. Solve.

#### Problem 26.

1. Interestingly independent of the larger square size. Area subtraction is the easiest way to find.

Treat as 3 triangles. Orientation is the key here. Treat BC like base then height is always constant because the diagonals are parallel.

Treat as 3 triangles. Orientation is the key here. Treat BC like base then height is always constant because the diagonals are parallel.

#### Problem 27.

In the Footprints Cafe each table has three legs, each chair has four legs and all the customers and the three members of staff have two legs each. There are four chairs at each table. At a certain time, three-quarters of the chairs are occupied by customers and there are 206 legs altogether in the cafe.

**How many chairs are there in the cafe?**

**Problem 28.**

#### Problem 29.

Given the parallelogram ABCD where E is the midpoint of BD, BD is perpendicular to CB,

EF is the angle bisector of CFD, DF length is 4 and CF length is 13, find the length of BD.

- E must be the midpoint of AC as well (use similar triangles to verify) so since EF is the angle bisector and it intersects AC at its midpoint that angle AEF is a right angle.
- That means AD must be 9 since triangle AEF is congruent to CEF.
- At this point most of the triangles are provably similar including EFD and AEF.
- So FD / ED = ED / AD or ED^2 = 36.

#### Problem 30

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?

[@brilliantorg]

[@brilliantorg]

Provable via induction start with the case of 3 passengers.

#### Problem 31

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

[@fayzshafloot5]

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