# Calvin Testing 6

## Requires high school knowledge only

Problems that any high schooler should be able to solve. It only uses concepts that they are aware of, but may not know how to apply in this scenario.

**1) Idea: Pythagoras Theorem**

Problem: [Henry Dudeney] Find the area of a triangle with side lengths \( \sqrt{61}, \sqrt{153}, \sqrt{388}\).

What is hard about this problem: (Assuming that they do not know Area by sine rule or Heron's formula). Middle school kids are not used to finding the area just given the sides. They are used to seeing a base and height.

Solution: Using pythagoreas theorem, we can show that this is the triangle. The area can then be calculated from the large rectangle and subtracting off the other areas.

**2) Idea: Representing information as a (x-y) graph**

Problem:

Cars A, B, C, and D are on a straight road (not necessarily at the same point). They each drive at a constant speed (which is possibly different from each other).

Car A meets

- car B at 10 am
- car C at 1 pm
- car D at 4 pm.

Car D meets

- car B at 12 pm
- car C at 2 pm
- car A at 4 pm (as stated earlier).

What time do car B and car C meet?

\(\)

**Hint:** There is a nice way to represent this problem.

What is hard about this problem: It's not clear how to get started. There seems to be a lot of information, but hard to present it clearly.

Solution: Drawing the location-time chart tells us that the graphs for B and C are parallel, hence they never meet.

**3) Idea: Given 4 lines \( l_1, l_2, l_3, l_4\) in \(n-\)dimensional space, if lines \(l_1, l_2, l_3 \) lie in the same plane and lines \( l_1, l_2, l_4 \) lie in the same plane, then all 4 lines lie in the same plane.**

Problem: (Sorry for the horrendous presentation)

4 ships are sailing on a flat earth, which we can model as \( \mathbb{R}^2 \).

Each ship moves in a straight line, at a constant velocity.

Each pair of ships will travel in paths that intersect in the future.

If two ships are at the point of intersection **at the same time**, then they will collide but can continue on their journey with no time lost. (Obviously, if the ships are at the point of intersection at distinct times, then no collision will occur.)

No three ships collide at the same time and spot. However, pairs of ships could end up colliding on the same spot (but different times), or at the same time (but different spots).

Each ship can only survive two collisions--on the third collision, the ship would sink.

There are \( { 4 \choose 2 } = 6 \) possible collisions that can occur. After some time, 5 collisions have taken place, which implies that 2 ships have sunk and there are the other 2 left. What is the fate of these 2 remaining ships?

Specifically, you can model the ship's movement as \( S_i (t) = ( x_i + t v_{xi} , y_i + t v_{yi}) \). Then, you know that there exist \( t_{12}, t_{13}, t_{14}, t_{23}, t_{24} \) which make \( S_i (t) = S_j (t) \) at \( t_{ij} \). The question then becomes "Is there a \( t_{34} \) that makes \( S_3(t) = S_4(t) \)?"

The condition "Each pair of ships will travel in paths that *intersect in the future*." can be ignored if we allow the ships to travel in negative time (or equivalently, backtrack and start the experiment early enough). The ships have to take non-parallel paths, though.

What is hard about this problem: It's not clear how to get started. There seems to be a lot of information, but hard to present it clearly.

Solution: Represent the ships as \( l_i = (t, x_i, y_i) \), which is a straight line. Suppose that we don't know if \(l_3, l_4 \) intersect. We know that pairwise \( l_1, l_2, l_3 \) intersect, which means that these lines lie on the plane determined by the triangle of their intersection points. Likewise, \( l_1, l_2, l_4 \) lie on a plane. Hence, all 4 lines lie on a plane, so lines \(l_3, l_4 \) must also intersect.

**4) Idea: Given a specific point (the focus) and a specific line (the directrix), the parabola is the locus of all points such that its distance from the focus is equal to its perpendicular distance from the directrix, provided the focus doesn't lie on the directrix.**

Problem:

Consider an infinite sequence of triangles with odd side lengths laid out as above. Let \(P_1, P_2, \ldots \) be the apex of the equilateral triangles. Does there exist a point \(P\) such that \( |P P_i | \) is always an integer?

What is hard about this problem: It's not clear that such a point exists. We could draw concentric circles of integer radius about the points \(P_i\), but why must all of them intersect at one point?

It's not clear how to prove such a point does not exist. There could be some weird combination (esp given how the lengths were chosen) that made this possible.

Solution: Suppose that the lower left corner of the diagram is \( (0,0) \). Let's try to describe the coordinates of the apex. The height increases linearly, while the width increases quadratically, since it's the sum of a linear increase. Thus, we have a parabola. Now, a natural point \(P\) to consider is the focus of this parabola. In particular, since \(PP_i\) is the distance from \(P_i\) to the focus, it is also equal to the distance from \(P_i\) to the directx. In particular, since horizontal distance between 2 apexes is an integer, as long as \(PP_1\) is an integer, we can conclude that all of these distances are integers.

It remains to verify that this claim holds true. (Sorry, I'm too lazy to fill in the details.)

**5) Idea: Cosine rule / Acute triangles**

Problem: Given any 8 real numbers \( a_1, a_2, \ldots, a_8 \), show that there exists \( (i,j) \) such that \( a_i a_j + a_{i+1} a_{j+1} \) is non negative.

What is hard about this problem: It seems almost obvious, has a lot of degrees of freedom, but it's hard to make sense of the expression.

Solution: Let \(O\) be the origin. Consider the 4 points \( P_i = (a_{2i-1}, a_{2i} ) \). There exists \(i,j\) such that \( \angle P_i O P_j \) is at most \( 90 ^ \circ \). This tells us that

\[ | P_i O | ^2 + |O P_j|^2 \geq |P_i P_j|^2 \Rightarrow ( a_{2i-1} ^2 + a_{2i} ^2 ) + ( a_{2j -1}^2 + a_{2j}^2 ) \geq ( a_{2i-1} - a_{2j-1} ) ^2 + (a_{2i} + a_{2j} )^2 \Rightarrow a_{2i-1} a_{2j-1} + a_{2i} a_{2j} \geq 0 . \]

## Requires some Olympiad/Non-curricular knowledge

**Cite as:**Calvin Testing 6.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/calvin-testing-6/