# How deep is my garden pond MKII

#### Hymie

Joined Mar 30, 2018
985
Following on from the simplicity of my last attempt; I have another garden pond with a water-lily within it - but his time I want to know how long the stem of the water lily is.

Within the middle of the pond is a water-lily, standing at a slight angle (as shown in the diagram), the length of stem from the water to the base of the flower head is exactly 250mm. If I pull the flower towards me, when the base of the flower head sinks to the level of the pond water, I notice that the distance between the original location of the flower stem, to where I have pulled it, is exactly 600mm. Conversely when I push the flower directly away, when the base of the flower head sinks to the level of the pond water, its distance from the original location is exactly 200mm – how long is the water-lily flower stem (in total).

For clarity, below is a dimensioned diagram of the above problem (not to scale).

As ever, along with your answer – show your working.

#### Attachments

• 57.9 KB Views: 25

#### WBahn

Joined Mar 31, 2012
26,398
No one has responded?

I was on the road for 15 hours yesterday and just got up.

Did you intentionally give a problem that has no solution?

If you bend the part of the stem that is out of the water down to the water it is long enough to go past the point where the straight stem intersects the surface. But that means that your system violates the triangle inequality theorem, which requires that

(S - 250 mm) + 200 mm > S

#### Hymie

Joined Mar 30, 2018
985
No one has responded?

I was on the road for 15 hours yesterday and just got up.

Did you intentionally give a problem that has no solution?

If you bend the part of the stem that is out of the water down to the water it is long enough to go past the point where the straight stem intersects the surface. But that means that your system violates the triangle inequality theorem, which requires that

(S - 250 mm) + 200 mm > S
Wbahn, your maths skills are too good for me – you can spot the ‘deliberate’ flaws in my questions.

Correction: The length of lily stem above the water should be 150mm (and not 250mm).

Hymie.

#### jpanhalt

Joined Jan 18, 2008
11,088
Between 256.1 mm and 256.2 mm

I am curious as to why the problem wasn't corrected earlier.

#### Hymie

Joined Mar 30, 2018
985
Between 256.1 mm and 256.2 mm

View attachment 153341

I am curious as to why the problem wasn't corrected earlier.
Looks like you are on the right track – show your working before WBahn beats you with the correct answer.

#### jpanhalt

Joined Jan 18, 2008
11,088
I did show my work. It was something Archimedes discovered, but I can't find the original right now.

Here:

#### WBahn

Joined Mar 31, 2012
26,398
Wbahn, your maths skills are too good for me – you can spot the ‘deliberate’ flaws in my questions.

Correction: The length of lily stem above the water should be 150mm (and not 250mm).

Hymie.
We know that the distance between the two extremes is 200 mm + 600 mm = 800 mm. So we know that the displacement of the original stem/surface intersection is 200 mm from the vertical.

If S is the length of the stem and D is the depth of the pond.

From the original position:

(S - 150 mm)² = D² + (200 mm)²

From the second position:

S² = D² + (400 mm)²

Combining them so as to eliminate D:

-(300 mm)S + (150 mm)² = - (400 mm)² + (200 mm)²

Solving for S:

S = [(400 mm)² + (150 mm)² - (200 mm)²] / (300 mm)

S = 4.75 dm = 475 mm

S = [(4 dm)² + (1.5 dm)² - (2 dm)²] / (3 dm)

S = [(4² + 1.5² - 2²) / 3 ] dm

S = [(16 + 2.25 - 4) / 3 ] dm

S = (14.25 / 3) dm

=====================
S = 4.75 dm = 475 mm
=====================

Check:

D from first position:

D² = (475 mm - 150 mm)² - (200 mm)² => D = ~256.174 mm

D from second position:

D² = (475 mm)² - (400 mm)² => D = ~256.174 mm

#### WBahn

Joined Mar 31, 2012
26,398

#### Hymie

Joined Mar 30, 2018
985
We know that the distance between the two extremes is 200 mm + 600 mm = 800 mm. So we know that the displacement of the original stem/surface intersection is 200 mm from the vertical.

If S is the length of the stem and D is the depth of the pond.

From the original position:

(S - 150 mm)² = D² + (200 mm)²

From the second position:

S² = D² + (400 mm)²

Combining them so as to eliminate D:

-(300 mm)S + (150 mm)² = - (400 mm)² + (200 mm)²

Solving for S:

S = [(400 mm)² + (150 mm)² - (200 mm)²] / (300 mm)

S = 4.75 dm = 475 mm

S = [(4 dm)² + (1.5 dm)² - (2 dm)²] / (3 dm)

S = [(4² + 1.5² - 2²) / 3 ] dm

S = [(16 + 2.25 - 4) / 3 ] dm

S = (14.25 / 3) dm

=====================
S = 4.75 dm = 475 mm
=====================

Check:

D from first position:

D² = (475 mm - 150 mm)² - (200 mm)² => D = ~256.174 mm

D from second position:

D² = (475 mm)² - (400 mm)² => D = ~256.174 mm

I take my hat off to jpanhalt & WBahn for solving this problem.

I was going to post my answer/solution – but jpanhalt & WBahn have shown that my answer is wrong.

#### WBahn

Joined Mar 31, 2012
26,398
I take my hat off to jpanhalt & WBahn for solving this problem.

I was going to post my answer/solution – but jpanhalt & WBahn have shown that my answer is wrong.
But do you know WHY your answer is wrong? Was it an error in your fundamental approach, or was it just a stupid algebra mistake, or something else?

#### Hymie

Joined Mar 30, 2018
985
But do you know WHY your answer is wrong? Was it an error in your fundamental approach, or was it just a stupid algebra mistake, or something else?
The attached (figure 1) shows the solution to the original problem using Pythagoras – soon after posting the question I realised this solution was much simpler than the one I had (shown in figure 2).

I then attempted to modify the question such that the method shown in figure 2 was simplest – but broke the rules for it to be applicable (hence obtaining the wrong answer).

Analysing the original problem shown by figure 2; (Xmm + Ymm) x 200mm = 600mm x 600mm (†)

Therefore (Xmm + Ymm) = 1800mm

The diameter of the circle = 1800mm + 200mm = 2000mm
The circle radius = 1000mm

Therefore the pond depth = 1000mm – 200mm = 800mm (without using Pythagoras)

(†) This equation is derived from Euclid discovery that when two chords of an arc intersect within a circle, the product of the parts of one will be equal to the product of the parts of the other.

#### Attachments

• 98.2 KB Views: 8