On a tropical island, 34 mischievous elves live. At the beginning, there are 7 blue, 10 red, and 17 yellow ones. When two elves of different colors meet, they decide to play a trick on their fellows and simultaneously adopt the third color. One day, the intrepid Maryam arrives on the island and observes that all the elves are of the same color. What is this color?

## Exercise Solutions

## Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Let's give one point to each blue elf, two points to each red elf, and three points to each yellow elf. Let's study how the sum of these scores varies when two elves of different colors meet:

- if they are blue and red, they had accumulated three points, and now accumulate six;
- if they are red and yellow, they had accumulated five points, and now accumulate two;
- if they are yellow and blue, they had accumulated four points, and still accumulate four.

Therefore, the sum of the elves' scores varies by -3, 0, or +3 at each meeting. In particular, its congruence modulo 3 never changes.

Initially, this sum is $7 + 10 \times 2 + 17 \times 3 \equiv 27 \equiv 0 (\bmod 3)$. Consequently, if our 34 elves take on the same color, they will each have a score $s \equiv 34s \equiv 0 (\bmod 3)$. This means that $s=3$, and therefore that our 34 elves will be yellow.

## Alternative Solution

Let $b, r$, and $j$ be the number of blue, red, and yellow elves at a given time. When two elves of different colors meet, two of our three numbers decrease by 1, and the third increases by 2. We would like to avoid distinguishing these two cases, which means considering our numbers modulo 3: in this case, each of the numbers $b, r$, and $j$ increases by 2 at each step.

When Maryam arrives, two of these numbers, corresponding to the two disappeared colors, are equal, as they are 0. Moreover, initially, we know that $b \equiv r \equiv 1 (\bmod 3)$, while $j \equiv 2 (\bmod 3)$. The two disappeared colors are therefore necessarily blue and red, which means that our 34 elves will be yellow.

## 4 Tangent Angle (Benoît)

We have reviewed the various properties from the previous course and established the theorem of the tangent angle (with its converse).

Theorem 1 (Tangent Angle).

Let $ABC$ be a triangle inscribed in a circle $\Gamma$ and $t$ a line passing through $A$. Let $D$ be a point on $t$ on the side of $B$.

The line $t$ is tangent to $\Gamma$ if and only if

$$
\widehat{DAB} = \widehat{ACB}
$$

![](https://cdn.mathpix.com/cropped/2024_05_10_a954a7760d797f08770eg-102.jpg?height=557&width=651&top_left_y=241&top_left_x=702)

## Example 2.

![](https://cdn.mathpix.com/cropped/2024_05_10_a954a7760d797f08770eg-102.jpg?height=557&width=985&top_left_y=949&top_left_x=547)

Let $\Gamma$ be a circle and $P$ a point outside the circle.

The tangents to $\Gamma$ passing through $P$ intersect the circle at $A$ and $B$.

Then $PAB$ is isosceles at $P$.

Indeed, by considering a point $C$ on the circle, we can use the tangent angle theorem twice and obtain:

$$
\widehat{PAB} = \widehat{ACB} = \widehat{ABP}
$$

## Exercises