### Hallway Borders

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### Not a Polite Question

When asked how old she was, the teacher replied: My age in years is not prime but odd and when reversed and added to my age you have a perfect square...

Four rods, two of length a and two of length b, are linked to form a kite. The linkage is moveable so that the angles change. What is the maximum area of the kite?

# Adding in Pairs

##### Age 11 to 14 ShortChallenge Level

Answer: $30$

Using algebra and adding the sums together
Let $39=a+b$, $48=a+c$ and $51=b+c$.

$c$ is the largest number because $a+b=39$ is the smallest sum.

Adding together the sums containing $c$, \begin{align}48+51&=a+c+b+c\\ \Rightarrow 99&=\underbrace{a+b}_{39}+2c\\ 99&=39+2c\\ \Rightarrow60&=2c\\ \Rightarrow30&=c\end{align}

Using algebra and elimination
Let $39=a+b$, $48=a+c$ and $51=b+c$.

$c$ is the largest number because $a+b=39$ is the smallest sum.

Subtracting $39=a+b$ from $48=a+c$ we can eliminate $a$:
\begin{align}48-39&=a+c-(a+b)\\ \Rightarrow 9&=c-b\end{align}

Adding $9=c-b$ and $51=b+c$ we can eliminate $b$:
\begin{align}9+51&=c-b+c+b\\ \Rightarrow60&=2c\\ \Rightarrow30&=c\end{align}

Trying out numbers
51 = largest number added to middle number.
51 = 25 + 26, so the largest number must be at least 26.
largest number middle number (=51-largest number) smallest number (=39- middle number) smallest number + largest number (should be 48)
26 25 14 40 - much too small
32 19 20 - too large 52 - too large
29 22 17  46 - too small
30 21 18 48
So the largest number is 30.

Finding the differences between the numbers
39 is the smallest number so it must be the sum of the smallest and middle numbers.
51 is the largest number so it must be the the sum of the largest and middle numbers.
48 must be the sum of the largest and smallest numbers.

So when the smallest number is added to the middle and largest numbers, the sums are 39 and 48 respectively. So the difference between 39 and 48 must be the same as the difference between the middle and largest numbers. So the largest number must be 9 more than the middle number.

So 51 is the sum of two numbers whose difference is 9.
Ways to make 51
25 + 26 = 51     difference: 1
23 + 28 = 51     difference: 5
21 + 30 = 51     difference: 9

So the largest number is 30.

Adding the sums together
39 is the smallest number so it must be the sum of the smallest and middle numbers.
51 is the largest number so it must be the the sum of the largest and middle numbers.
48 must be the sum of the largest and smallest numbers.

We are interested in the largest number. Adding together the sums containing the largest number,
the largest number + the middle number + the largest number + the smallest number = 51 + 48 = 99.

But that is the largest number twice, plus the sum of the smallest and middle numbers. But we know that the sum of the smallest and middle numbers is 39! So the largest number twice is 99 $-$ 39 = 60.

So the largest number is 30.

You can find more short problems, arranged by curriculum topic, in our short problems collection.