Finding Isosceles Triangles

Mr. Isocel asked students in his class to find isosceles triangles whose vertices lie on a coordinate grid. For each student example below, explain why the triangle is isosceles.

Monika draws the following triangle:

 

Karl's picture is here:

Bob draws this picture:

Now, you find a general solution (a generic method)!

 





Answer:

1. We can see that Monika's triangle is isosceles in several ways. We will show that $|AC| = |BC|$. We can use the Pythagorean Theorem to check that these both have length $\sqrt{2^2 + 3^2}$. In the calculations below $M = (3,1)$ is the midpoint of $\overline{AB}$.
   
   
   $$\begin{align} |AC|^2 &= |AM|^2 + |MC|^2 = 2^2 + 3^2\\ |BC|^2 &= |BM|^2 + |MC|^2 = 2^2 + 3^2. \end{align}$$
   We can also observe that to get from $A$ to $C$ we go over (to the right) by two units and then up three units. To get from $B$ to $C$ we go over (to the left) by two units and then up three units. We travel the same distance to get from $A$ or $B$ to $C$ so $A$ and $B$ are equidistant from $C$. Finally, we can draw the horizontal line of symmetry for △ABC.
   
   
   
   Isosceles triangles are exactly those triangles with a line of symmetry so $\triangle ABC$must be isosceles.

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Next Domino

What would be the next domino in the following sequence?

m (+5)            >>> r
68 - (5) * 4      >>> 48
r - 48 == 18 - 48 >>> -30

Chameleons

On an island live 13 purple, 15 yellow and 17 maroon chameleons. When two chameleons of different colors meet, they both change into the third color. Is there a sequence of pairwise meetings after which all chameleons have the same color?

 

Let <p, y, m> denote a population of p purple, y yellow and m maroon chameleons. Can population <13, 15, 17> can be transformed into <45, 0, 0> or <0, 45, 0> or <0, 0, 45> through a series of pairwise meetings? Define function X(p, y, m) = (0p + 1y + 2m) mod 3. An interesting property of X is that its value does not change after any pairwise meeting because X(p, y, m) = X(p-1, y-1, m+2) = X(p-1, y+2, m-1) = X(p+2, y-1, m-1). Now X(13, 15, 17) equals 1. However, X(45, 0, 0) = X(0, 45, 0) = X(0, 0, 45) = 0. This means that there is no sequence of pairwise meetings after which all chameleons will have identical color.

Scales

Review the following picture of scale balances:

Based on the balances in the above picture, which of the following balances can be correct?

Answer is C.

Counterfeit Coin

One of twelve coins is counterfeit: it is either heavier or lighter than the rest. Is it possible to identify the counterfeit coin in three weighings on a beam balance?

Solution:

This solution is first presented by "BRIAN D. BUNDY":

We may label the coins 1, 2, ..., 12 so that we can distinguish between and identify them using these labels.

1.            1. Weigh 1, 2, 3, 4 against 5, 6, 7, 8.

2. They balance, so 9, 10, 11, 12 contain the odd coin. Weigh 6, 7, 8 against 9, 10, 11.
1. They balance, therefore 12 is the odd coin and so weigh 12 against any other to discover whether it is heavy or light.
2. 9, 10, 11 are heavy and so they contain an odd heavy coin. Weigh 9 against 10. If they balance, 11 is the odd heavy coin, otherwise the heavier of 9 and 10 is the odd coin.
3. If 9, 10, 11 are light, we use the same procedure to reach the same conclusion for the odd light coin.
3. 5, 6, 7, 8 are heavy and so either they contain an odd heavy coin or 1, 2, 3, 4 contain an odd light coin. Weigh 1, 2, 5 against 3, 6, 10.
1. They balance, so the odd coin is 4 (light) or 7 or 8 (heavy). Thus weigh 7 against 8. If they balance 4 is light, otherwise the heavier of 7 and 8 is the odd heavy coin.
2. 3, 6, 10 are heavy, so the odd coin can be 6 (heavy) or 1 or 2 (light). Thus weigh 1 against 2. If they balance 6 is heavy, otherwise the lighter of 1 and 2 is the odd light coin.
3. 3, 6, 10 are light, so the odd coin is 3 and light or 5 and heavy. We thus weigh 3 against 10. If they balance 5 is heavy, otherwise 3 is light.
4. If 5, 6, 7, 8 are light we use a similar procedure to that in 2.

Four Words

Four words related to basic gardening have been written in a row with neither spaces nor vowels. What are these four words?

PRNNGSLWDRSCP

Pruning
Soil
Weeder
Scoop

Shape and Size

With 10 sticks divide this figure into four parts that are all equal in shape and size!

1/3 Probability by One Coin

At a restaurant, how can John choose one out of three desserts with equal probability with the help of one coin?

He tosses the coin twice; where he supposes HT, TT and TT correspond to his three choices and if HH he repeats.

Which is the odd one out?

Which is the odd one out?

C.

C has three white and one black on the left and three black and one white on the right. The others are the opposite way round.

Which tile is missing?

Which tile is missing?

In each row and column, every ring is shaded black once.

Next Grid

What grid should come next?

What grid should come next?

(A)

(B)

(C)

(D)

C:

Which number is the odd one out?

Which number is the odd one out?

7768 9678 4572 5261 5133 4638 6895

4638: in the other numbers the sum of the first two digits is equal to the sum of the second two digits, for example in  5133 we have 5+1 = 3+3.