A ship carpenter has a problem of plugging a 12in square hole by cutting a single 9in x 16in wood block into two pieces and joining the two. Is it possible?

### The problem of plugging a square hole puzzle

A ship carpenter discovered a hole in his ship of size 144 sq in and exactly square in shape. His problem was - he had with him only a single piece of wood of size 9in by 16in. How could he cut the wood block into two pieces and join the two to plug the ship's hole?

**Time to solve:** 30 minutes.

You may go on to make many trials. Extra time is available.

Pun aside, this is well worth your time both for solving as well as for knowing how to solve by using step by step reasoning.

### Systematic solution to ship carpenter's problem of plugging a square hole puzzle

For convenience, the rectangular wood block and the square hole shape superimposed on it shown below.

Concern is only with cutting the front face of the rectangular block as shown. The block will be cut in a direction perpendicular to the face so that both blocks after cut **would be of same thickness.**

**No need to consider thickness of the block then.** The problem is transformed to a two dimensional one.

At the start, with a quick try be convinced that the **rectangular wood block can't be cut in any way along a single straight line cut.**

For example, in the figure the area B of the block and area C of the square hole both are of same size of 36 sq inch. Can you cut out the extra 36 sq in and reshape it to fit on top to make a 12 inch sq shape? It is not possible in any way.

Upshot is,

Conclusion 1:Solutioncan't be a curved line cut as well as it can't be a single straight line cut.

What could be the option left then?

The conclusion is easy to make,

Conclusion 2:The cut must have at least two straight lines.

Given condition says, there must be two pieces of wood after the cut. Thus,

Conclusion 3:The line segments of cut must be joined together so that after the complete cut only two pieces are there.

**Observation:** The 36 sq in extra area on the right of the rectangle (area B) must move up on top even if it is in parts (I don't mean multiple pieces here). So the natural decision taken is,

Conclusion 4:Thefirst cut would be made vertically along the line demarcating 36 sq inch extra area B and common area A four inches from the right side.Theline of cut can't be all the way up through 9 inchesto split the wood piece into two,it would move up for a certain length and stop.

It follows,

Conclusion 5:As the line of cut can't move up all the way through 9 inches height,it must turn left horizontally after moving vertically for a certain distance.It would bein the form of a step.

The following schematic gives an idea about the nature of the line of cut thought out till now.

The line of cut would move vertically upwards from point B to point C, turn left and move horizontally left to point D.

When more mysteries are uncovered soon, these unknown lengths of BC and CD would also be known. The **only known distance known at this time is length AB as 4 inches.**

What more can be discovered at this time?

**Analyze.** Consider cut face at the bottom AB as the Base of the second piece being cut out. Realize that this base would sit on top of exposed face CD after the cut.

Question is:Can length CD be more than the length AB (it is obvious that CD cannot be less than AB)?

Imagine such a situation where CD is larger than AB. That means the piece cut out cannot be put on face CD and pushed left to fit snugly leaving no space between the two vertical cut faces when the two pieces are joined.

Only **one certain conclusion** can then be made,

Conclusion 6:Length CD must also be 4 inches equal to length AB.

What would be the next step of the cut?** It must turn right and move vertically up again.** At this point of time 8 inches leftwards and an unknown length upwards are left. Length BC yet to be known.

Impatiently just move up through the rest of 9 inches to complete the cut. Following is the schematic of the result,

It is clear that to fit without a gap along the joining line, length of BC must be equal to DE and that would then be half of 9 inches, that is, 4.5 inches. It is possible to measure 4.5 inches and make the cut this way.

**Realize now** that first cut at 4 inches left of right face won't produce the desired result. The two pieces to fit without a gap between the cut faces that are to be joined, total horizontal length of 16 inches has to be divided into three equal parts.

Dividing 16 by 3 and measure each divided portion **EXACTLY** **is not possible**. **But ignore this roadblock.**

Assume that it is possible to cut the second line segment of cut CD with AB = CD = EF and BC = DE = 4.5 inches. In this case, after the second piece sits on top of the remaining first piece on the left, the height becomes 13.5 inches, more than 12 inches, and length becomes 14.67777... inches.

**The result is not a square.**

It is crystal clear now that,

Conclusion 7:thelast vertical line segment of the cut must be 3 inches so that the height becomes 9 + 3 = 12 inches.

This is a **critical conclusion.**

And thus the final solution become very clear,

Conclusion 8:from point D the line of cut must then stop at 3 inches at point E, turn left and move horizontally for 4 inches to point F.

Horizontal distance left is 4 inches and vertical 3 inches. **A perfect situation for the final cut.**

Conclusion 9:turn right and cut through rest 3 inches vertically up to point G. Two pieces are formed and separated. Push the right piece up and left against the left sitting piece. The two would form a perfect square of side length 12in.

The final line of cut is shown.

And the final joined two pieces is shown.

**Note that the three vertical cut segments actually divided the horizontal side length of 16in into 4 segments of 4in each** and the **two horizontal cut segments actually divided the 9in vertical side into 3 segments of 3in each.**

Line of cut was along BCDEFG. The piece at the top was on the right making up the second part of the 9in x 16in rectangular block.

Okay, you have solved the main part of the puzzle. But what about the second part of the puzzle?

### Solution to second part of Ship Carpenter's problem of plugging a square hole puzzle: More rectangles convertible to squares

To find more such rectangles, **reasonable approach is to understand clearly why could you convert the 9in x 16in rectangle to a square of 12in side.** The **mechanism of conversion must be made clear** and **a model is to be formed.**

From the model that would describe clearly

what are the properties of such a rectangle,other such rectangles can be deduced.

In the problem solved just now, the dimensions were 9 inches vertically and 16 inches horizontally. Line of cut had to move by 4 inches horizontally three times and vertically by 3 inches two times.

In effect,

16 had to be divided equally in four parts and 9 equally in three parts.

This has been possible simply because, **16 is 4 square and 9 is 3 square.**

First conclusion about candidate rectangle would be easy to form,

Conclusion A.Thetwo dimensions must be squares of integers.

The math supports the transformation of the rectangle to a square without showing the line of cut explicitly,

$\textbf{Square Hole: }12^2 = 3 \times{ 4\times{3 \times{ 4}}}$

$= 3 \times{ 3 \times{ 4 \times{ 4}}}$

$= 9 \times{ 16} \textbf{ :Rectangular wood block.}$

The repeated factor of 3 and 4 in $12^2$ are simply rearranged and transformed into the dimensions of the rectangular wood block.

Question:Would this be enoughfor the candidate rectangle to be converted to a square by a single cut and joining two pieces?

Can a rectangle with dimension $3^2$ by $5^2$ be a feasible candidate for such a transformation by a single cut?

With the steps now known, it is easy to check.

**For a single line of cut:** $5^2$ to be divided into 4 equal vertical moves 5 units separated and $3^2$ to be divided into 2 equal horizontal moves 3 units separated. But after three vertical cuts 5 units separated and two horizontal cuts 3 units separated, the cut will be completed and the two pieces formed cannot be joined to make the desired square. The **rectangle dimensions are invalid.**

Recall that for the 9 x 16 rectangle (with 16 along horizontal direction), **3 vertical moves separated 4 units each** and *2 vertical moves separated 3 units each*** formed the perfect cut and two mutually fitting pieces.**

Question:Any specialrelationship between number of horizontal and vertical moves?

Yes. And that is the **second critical condition** that must be fulfilled by the dimensions of a feasible candidate rectangle,

Conclusion B:Thenumber of horizontal moves along the line of cut in the direction of longer dimension must be one less than the number of vertical moves along the smaller dimension.

In essence, **line of cut must move horizontally 1 less time than vertically along the shorter dimension**.

**Note:** The vertical cuts divide longer dimension and horizontal cuts divide shorter dimension.

**Then only** the right side piece can be **moved up and left on top of the left side piece** to form a perfect square. All the cut faces of the two pieces would fit flush against each other having no awkward gap or overflow between them.

It means mathematically, for a candidate rectangle piece to be convertible to a square by a single cut and joining the two pieces,

Final condition:The two dimensions must each be a square of two integers that are consecutive with difference unity.

The first few such feasible rectangles are,

Height: $1^2=1 \qquad$ Length: $2^2=4 \qquad$ Square side: 2 $\qquad$ Area: 4

Height: $2^2=4 \qquad$ Length: $3^3=9 \qquad$ Square side: 6 $\qquad$ Area: 36

Height: $3^2=9 \qquad$ Length: $4^2=16 \qquad$ Square side: 12 $\qquad$ Area: 144

Height: $4^2=16 \qquad$ Length: $5^2=25 \qquad$ Square side: 20 $\qquad$ Area: 400

Height: $5^2=25 \qquad$ Length: $6^2=36 \qquad$ Square side: 30 $\qquad$ Area: 900

There would be **infinite numbers of such feasible rectangles.**

**Is that all?** What about the rectangle with dimensions as, **each of the two dimensions of a feasible rectangle multiplied by an integer?** That would in other words be **multiples of dimensions of a feasible rectangle.**

For example can the following rectangle be a feasible one?

$\text{Height: }3\times{4^2}, \qquad \text{ Length: }3 \times {5^2}$.

Does it seem that number of segments would be 3 x 4 = 12 and 3 x 5 = 15? It cannot be so as the difference between the two must be 1.

Feasible line of cut:Number of steps would be4 vertical separated 15 units each and 3 horizontal separated 12 units each. In effect there would be 4 vertical cut segments each 12 units long and 3 horizontal cut segments each 15 units long.

Following shows the figure. Moves are again 4 vertical (dividing the horizontal length) and 3 horizontal (dividing the vertical shorter height).

Add the number of feasible rectangles as another infinity to the earlier infinity!

Under this second group of feasible rectangles, consider using fractional multiples of dimensions of an originally feasible rectangle and that should be all that can be there.

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