Maths Olympiad Prep

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, 2020

Geometry Difficulty 4.8 AIME Find the answer Canada

In the diagram, rectangle PQRSPQRS has PS=2PS=2 and PQ=4PQ=4. Points T,U,V,WT,U,V,W are positioned so that RT=RU=PW=PV=aRT=RU=PW=PV=a.

If VUVU and WTWT pass through the centre of the rectangle, for what value of aa is the shaded region 18\frac{1}{8} the area of PQRSPQRS?

Pick one

Solution

If the first positive integer in the list is aa and the second is bb, then the third integer is a+ba+b, the fourth is b+(a+b)b+(a+b) or a+2ba+2b, and the fifth is (a+b)+(a+2b)(a+b)+(a+2b) or 2a+3b2a+3b.

Thus, we are asked to find the number of pairs of positive integers aa and bb, where aa is less than bb (since the list is increasing), and for which 2a+3b=1242a+3b=124.

What is the largest possible value for bb?

If b=42b=42, then 3b=3×42=1263b=3\times42=126 which is too large since 2a+3b=1242a+3b=124. (Note that a larger value of bb makes 3b3b even larger.)

If b=41b=41, then 3b=3×41=1233b=3\times41=123.

However in this case, we get that 2a=124123=12a=124-123=1, which is not possible since aa is a positive integer.

If b=40b=40, then 3b=3×40=1203b=3\times40=120 and so 2a=42a=4 or a=2a=2.

Thus, the largest possible value for bb is 40.

What is the smallest value for bb?

If b=26b=26, then 3b=3×26=783b=3\times26=78 and so 2a=12478=462a=124-78=46 or a=23a=23.

If b=25b=25, then 3b=3×25=753b=3\times25=75.

However in this case, we get that 2a=12475=492a=124-75=49, which is not possible since aa is a positive integer.

If b=24b=24, then 3b=3×24=723b=3\times24=72 and so 2a=12472=522a=124-72=52 or a=26a=26.

However, if the first integer in the list is 26, then the second integer can not equal 24 since the list is increasing.

Smaller values of bb will give larger values of aa, and so the smallest possible value of bb is 26.

From the values of bb attempted thus far, we notice that when bb is an odd integer, 3b3b is also odd (since the product of two odd integers is odd), and 1243b124-3b is odd (since the difference between an even integer and an odd integer is odd).

So when bb is odd, 1243b124-3b is odd, and so 2a2a is odd (since 2a=1243b2a=124-3b).

However, 2a2a is even for every choice of the integer aa and so bb cannot be odd.

Conversely, when bb is even, 1243b124-3b is even (as required), and so all even integer values of bb from 26 to 40 inclusive will satisfy the requirements.
These values of bb are 26,28,30,32,34,36,38,4026,28,30,32,34,36,38,40, and so there are 8 such lists of five integers that have 124 as the fifth integer.

Here are the 8 lists: 2,40,42,82,1242,40, 42, 82, 124; 5,38,43,81,1245,38,43,81,124; 8,36,44,80,1248,36,44,80,124; 11,34,45,79,12411,34,45,79,124; 14,32,46,78,12414,32,46,78,124; 17,30,47,77,12417,30,47,77,124; 20,28,48,76,12420,28,48,76,124; 23,26,49,75,12423,26,49,75,124.

Want a route through all this instead of an archive? The track puts 2,444 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.