# UKMT Special (Problem $2$)

The largest of $4$ different real numbers is $d$.

When the numbers are summed in pairs, the four largest sums are

$9, 10, 12, 13$

What are the possible values of $d$?

[UKMT Hamilton Olympiad $2017$, H$4$] Note by Yajat Shamji
8 months, 2 weeks ago

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You have until next Sunday, $3:00$pm!

- 8 months, 2 weeks ago

Let the other $3$ numbers be $a, b, c$ in increasing order. Then we know that $c+d$ is the largest sum, and $b+d$ is the next largest. Whether $a+d$ or $b+c$ is greater cannot be guaranteed. Let us do this in two cases -

Case $1$ - $a+d$ is larger than $b+c$

Then, we know that $b+c$ must be fourth largest. Thus, we have the following equations,

$c + d = 13 \ldots \text{eq(1)} \\ b + d = 12 \ldots \text{eq(2)} \\ a+d = 10 \ldots \text{eq(3)} \\ b + c = 9 \ldots \text{eq(4)}$

From equations $1$ and $4$, we can simplify to get $b = d - 4 \ldots \text{eq(5)}$

From equations $2$ and $5$ now, we can simplify to get $\to d = 8$ (and $b=4$, which can help us solve for $a$ and $c$, but that is not required)

First Case : $d = 8$

Case $2$ - $b+c$ is larger than $a+d$

Then, we know that $b+c$ must be fourth largest. Thus, we have the following equations,

$c + d = 13 \ldots \text{eq(1)} \\ b + d = 12 \ldots \text{eq(2)} \\ b+c = 10 \ldots \text{eq(3)} \\ a+d = 9 \ldots \text{eq(4)}$

From equations $2$ and $3$, we can simplify to get $c = d - 2 \ldots \text{eq(5)}$

From equations $1$ and $5$ now, we can simplify to get $\to d = 7.5$ (and $c=5.5$, which can help us solve for $a$ and $b$, but again, that is not required)

Second Case : $d = 7.5$

Thus,

$d = 8 \text{ or } 7.5$

- 8 months, 2 weeks ago

@Yajat Shamji - Where did you find these problems? They are interestingly easy lol

- 8 months, 2 weeks ago

Go to this note:

- 8 months, 2 weeks ago

Nice :)

- 8 months, 2 weeks ago

But... slightly different method.

- 8 months, 2 weeks ago

Ok (some text)

- 8 months, 2 weeks ago