The time has come to issue a warning: this entire series of caches will be archived between May 1st and October 31st, 2022.
It is difficult to line up the threads on this container - especially now (as of 2020-08-15) that we have added tape to help prevent leakage. Be patient and don't force it. It helps sometimes in aligning the threads to screw the top in reverse until you can feel that the threads are "connecting." Thanks!
Caches in the Series
The series consists of: Archimedes (GC3TT8C), Bernoulli (GC3V9MW), Cayley (GC3W898), Descartes (GC3X0W7), Euclid (GC3XGN0), Fermat (GC413V5), Gauss (GC4219Z), Heron (GC4VGBX), Ingham (GC4WC72), Jarnik (GC4Z9PK), Kepler (GC4ZMRG), Lovelace (GC50BN6), Merryman (GC51D34) and Newton (GC5AJ04).
Cayley the Mathematician
Arthur Cayley was a British mathematician of the C19th. He had a talent for languages and practised law for fourteen years but it was in mathematics that he excelled. He showed his mathematics ability as a child when he amused himself by solving complex mathematical problems. Cayley’s name precedes too many theorems, numbers, processes and the like to mention. We can sum it up by saying that he was one of the founders of the modern British school of pure mathematics. Some might be interested in checking out a game he created: “Cayley’s Mousetrap.” We should mention too, that the Cayley Mathematics Contest run by the University of Waterloo (in Waterloo, Ontario) for capable students in Grade 10 or below bears his name.
Cayley the Cache
Many will find it unnecessary to go through all of this material. It is provided chiefly for those who, for some reason, missed this "when will we ever use this stuff anyway" material in school and we hope that the next few steps will enable them to work their way through this "puzzle" satisfactorily. Others may wish to go straight to "4. We Now Know!" But, hey, if you run into a problem, back up and "have a go!"
1. Similar Triangles
Put simply, similar triangles are of different sizes but with the same shape. But let’s be a little more specific - and this is intended to help, not hinder! Please use a sharp pencil (I just sharpened my own!) and be as accurate as you can; otherwise, the earth-shaking illustration we are trying to provide won’t work!!
Draw a triangle with sides of three different lengths. Call it △ABC - i.e. one letter at each vertex. If convenient outline this triangle in a distinctive way - black or colour. Extend side AC (which means from A to C and beyond) until the line from A is about twice the length of AC itself. Extend AB in similar fashion. Place your ruler with one edge (edge1) along side BC and the other edge of the ruler (edge2) further away from A. Draw a line along edge2 taking it far enough to “cut” (cross) the extension of AC at a point we’ll call D and to “cut” the extension of AB at a point we’ll call E. This process will ensure that side ED is parallel to side BC.
We now have a second - larger - triangle called △AED. Please outline its sides in some distinctive way so that you can see it clearly as distinct from △ABC. In so doing, please try to ensure that you can still see the lines with which you marked the sides of △ABC. We now have two triangles with the smaller nesting in a “corner” of the larger. We don't usually "nest" similar triangles; we are doing so here only for the purpose of illustrating certain properties.
The two triangles are obviously different sizes but one might say that they are the same shape. What that implies is that the three angles at A, B and C in the smaller triangle are the same size as the three angles at A E and D in the larger triangle. At A, the angles are, in fact, the same angle. The way we drew line ED ensured that it would be parallel to side BC and, although we won’t prove it here, that means that ∡ABC = ∡AED and that ∡ACB = ∡ADE. Such triangles - triangles with three angles in one equal to the three angles in the other - are said to be "similar." (Thanks, kirkey4, for the correction to the error in this paragraph submitted on 2016-01-03.)
2. Side Relationships in Similar Triangles
We are going to measure the lengths of the six sides in centimetres to two decimal places as accurately as possible - not easy but let’s see how this turns out! Then, in pairs, we will use a calculator to divide the longer by the smaller and compare the values - rounded off to two decimal places, please.
Side AE = _____cm; side AB = _____cm; AE⁄AB = _____.
Side AD = _____cm; side AC = _____cm; AD⁄AC = _____.
Side ED = _____cm; side BC = _____cm; ED⁄BC = _____.
These three fractions (side over side) are called "ratios" and, with any luck, you will find that they are pretty close to being equal. How close did you come? Mine came to: 1.428, 1.418 and 1.420 and these round off to 1.43, 1.42 and 1.42 respectively . . . and that’s pretty good for drawings of this accuracy! (Yours will almost certainly not be 1.42 but the three numbers you have should be close to each other.)
In similar triangles, the sides opposite the equal sides are called "corresponding sides" and it can be proven that the ratios of these three pairs of corresponding sides are equal.
A statement that two ratios are equal is called a "proportion" so for the triangles we used we can write the proportion: AE⁄AB = AD⁄AC = ED⁄BC; any two of these equated would also be a proportion. Let’s look at how one can work with proportions.
3. Proportions
Let us write a simple proportion. You will find it easier to print them on paper using a horizontal fraction line but it is difficult (for me!) to do it here so we will use the slash for that line.
We can all accept that this is true: 3/5 = 9/15. (This could also be written: 3:5 = 9:15 and read “3 is to 5 as 9 is to 15 but is easiest to work with in fractional form.)
You can do a lot of things with the numbers in a proportion and still get a true statement:
Exchange the numerator (top) of the first ratio with the denominator of the second ratio or the other way around or both and still the proportion holds; eg.: 15/5 = 9/3; 3/9 = 5/15; 15/9 = 5/3.
Move the denominator of one ratio to the numerator of the other ratio where it becomes a multiplication; e.g.: 3 = (9)(5)⁄15; (3)(15)⁄5 = 9 etc. If this is done with both denominators and both numerators, it is called "cross multiplication;" e.g. (3)(15) = (5)(9)
You can invert both ratios and the proportion is still true; e.g.: 5/3 = 15/9.
These manipulations permit us to solve some problems with similar triangles . . . so let’s summarize and get to it!!
4. We Now Know!
(a) Similar triangles have the same sized angles, regardless of the triangle size.
(b) The sides opposite the equal angles (corresponding sides) in similar triangles are proportional.
(c) Proportions can be manipulated to easily find a missing value.
(d) Neat sketches using the information sensibly (if one line is given longer than another, then draw it that way but no need for scale diagrams) can help the solving of a problem.
5. An Example
Triangles MNO and PQR are similar triangles (implying that the pairs of equal angles are by the order of the vertices in the names) and MN:PQ = 5:11 (Remember that 5 and 11 aren’t necessarily actual lengths - just the ratio - so the sides could be 10 and 22, etc.)
(a) If ∡NMO = 98°, then ∡ QPR = _____.
(b) If, in addition, ∡MNO = 44°, then ∡PRQ = _____.
(c) If side MO = 6 cm, then side PR = _____.
(d) If side QR = 15.4 cm, then side NO =_____.
Answers: 98°, 38°, 13.2 cm, 7 cm.
The Puzzles
Please use the capital letters, A, B, C, etc. to the right to list your answers, one digit per letter.
1. △POR and △QYZ are similar triangles such that PO corresponds to QY and so on.
(a)If ∡POR = 37°, then the size of ∡QYZ = ____; so AB = __ __.
(b) If ∡YZQ = 61°, then
(i) the size of ∡ORP = ____ ; so CD = __ __
(ii) and the size of ∡OPR = ____; so EF = __ __.
2. △STU and △VWX are similar triangles such that side ST corresponds to side VW etc. and ST:VW = 2:3.
(a) If side SU were 14 cm long, how long would side VX be? GH = __ __
(b)If side ST were 9 cm long, how long would side VW be? IJK = __ __ . __.
(c)If side WX were 25.5 cm long, how long would side TU be? LM = __ __.
The Co-ordinates
The co-ordinates of the cache are: N 44°ab.cde’ and W 078° fg.hij where there is no relationship between the upper case letters above and the lower case letters used here.You have established that: A = ___, B = ___, C = ___, D = ___, E = ___, F = ___, G = ___, H = ___, I = ___, J = ___, K = ___, L = ___, M = ___. N is the thousandths digit of the minutes of the west co-ordinate of the correct location of Bernoulli which, it is assumed, you have done already so N = ___.
a = B + F - E = ___; b = C + J - D = ___;
c = K + M - A = ___; d = G - L + I + (2)(H) = ___;
e = K + M - N = ___; f = E + (2)(H) - K = ___;
g = E - (C + L) = ___; h = B + I - (F + G) = ___;
i = J + H + D = ___; j = N - A - (2)(L) = ___.
So you are looking for:
N 44° __ __ . __ __ __ ’ and W 078° __ __ . __ __ __’.
Other Notes
- Please provide your own pen or pencil.
- Parking to the west of the cache location is preferable.
- Be sure to record your co-ordinates for this cache for use later in the series.
- The cache holds only a log.
- It is difficult to assess the difficulty of a cache like this when people come with such a range of backgrounds; some will find it less difficult than 2.0 - others more difficult. So be it!
Check Your Answer
You can check your answers for this puzzle on GeoChecker.com.