# Thread: a logic teaser!

1. ## a logic teaser!

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2. Originally Posted by soyeahiknow
AThey say "walk to the North shore and find a lone oak, a lone pine, and a gallows pole. Stand at the gallows and walk to the oak, counting your steps.
Just pretend the gallows are anywhere...as long as same number of steps is taken,
it matters not where the gallows are.

3. Hello, soyeahiknow!

Wilmer is absolutely correct!

A pirate has been marooned on a deserted island.
He finds instructions to locate a buried treasure on the island.

It says: "Walk to the North shore and find an oak, a pine, and a gallows.
Stand at the gallows and walk to the oak, counting your steps.

"Turn 90 degrees to your right and walk the same number of steps.
Put a spike in the ground there.

"Return to the gallows and walk to the pine, counting your steps.
Turn 90 degrees to your left and walk the same number of steps.
Put a second spike in the ground there.

Dig halfway between spikes."

He finds the oak and the pine, but no trace of gallows.
Can he still find the treasure? .Is so, where is it?

If the gallows were still there, the pirate would do the following:

He starts at the gallows $G$ and walks to the oak $O.$
He turns to the right and walks to $A,$
. . so that: . $\angle AOG = 90^o,\;OA = OG.$
He plants a spike at $A.$

He returns to the gallows $G$ and walks to the pine $P.$
He turns to the left and walks to $B,$
. . so that: . $\angle BPG = 90^o,\;PB = PG.$
He plants a spike at $B.$

The treasure is found at $X$, the midpoint of $AB.$

Code:
                                    B
o
*  |
X  *     |*
o        |
*  |        | *
A  *     |        |
o        |        |  *
* |        |        |
*   |        |        |   *
*     |        |        |
*       |        |        |    *
*         |   D    |        |F
O o - - - - - + - + - -+- - - - + - - o P
*       C   |    E         *
*       |         *
*   |    *
o
G

Draw altitudes $AC,\:GD,\:XE,\:BF\,\text{ to }OP.$

$\text{Since }\Delta ACO \cong \Delta ODG\!:\;\begin{Bmatrix}(1)\;OC\:=\:DG \\ (2)\;AC\:=\:OD \end{Bmatrix}$

$\text{Since }\Delta BFP \cong \Delta PDG\!:\;\begin{Bmatrix}(3)\;FP \:=\:DG \\ (4)\;BF\:=\:DP \end{Bmatrix}$

From (1) and (3): . $OC = FP$ and $X$ is the midpoint of $AB.$
. . Hence: $E$ is the midpoint of $OP:\;OE \,=\,\frac{1}{2}OP$

$X\!E$ is the average of $AC$ and $BF.$. That is: . $X\!E \:=\:\dfrac{AC + BF}{2}$
From (2) and (4): . $X\!E\:=\:\dfrac{OD+DP}{2} \:=\:\frac{1}{2}OP \:=\:OE$

He can locate the treasure without the gallows.

(a) Locate $E$, the point halfway between the oak and the pine.

(b) From $E$, move directly north the distance $OE.$
. . . [And dig there.]