This is from a competition math problem I had recently that I just couldn't figure out.

If $ (x+y):(y+z):(x+z) = 1:2:4$ and $x+y+z=35$ compute the value of x.

I can tell that $7*(x+y)=2x+2y+2z$ which can be simplified as $5x+5y=2z$

Also $7/2(y+z)=2x+2y+2z$ which can be written as $1.5z+1.5y=2x$

And finally $7/4(x+z)=2x+2y+2z$ or $-0.25x-0.25z=2y$

Then by simple replacement algebra you can find that $4.375y-0.625z=2z$ or $4.375y=2.625z$ or $y=0.6z$ this makes the 2nd formula into $2.4y=2x$ or $1.2y=x$

With all of these in mind, $x+y+z=(6/5)y+y+(5/3)y= (58/15)y=35$ so $y=525/58$ since $1.2y=x$ we can tell that x is $2625/348$

This answer may or may not be right (I never got the answers) but in itself the method is too complicated. I was supposed to do this in less than 10 minutes. Is there a faster or simpler way to do this that I just missed?

  • $\begingroup$ Have you tried $4(x+y):2(y+z):(x+z) = 1:1:1 \to 4(x+y)=2(y+z)=(x+z)$ $\endgroup$ – user137794 Apr 13 '14 at 19:15
  • $\begingroup$ I guess that simplifies it a bit. Honestly i stumbled my way through this one. I was more wondering if there was a way to look at the 1:2:4 ratio and a formula(which may or may not exist) from those 3 numbers to the final answer, just as a general case. $\endgroup$ – Asimov Apr 13 '14 at 19:19

If you are attuned to the cyclic symmetry of things, you might notice it can be rewritten

$$(x+y):(y+z):(z+x)=1:2:4\qquad\text{and}\qquad (x+y)+(y+z)+(z+x)=70$$

which gives $(x+y)+2(x+y)+4(x+y)=70$, or $x+y=10$. From this it follows that $y+z=2(x+y)=20$, and we can finish by invoking $x+y+z=35$, which gives $x=15$.

  • $\begingroup$ This answer makes much more sense than the horrifying fraction i created. Thanks $\endgroup$ – Asimov Apr 13 '14 at 19:30
  • $\begingroup$ @JohnJPershing, you're very welcome. It's often a sign that something's gone wrong when the algebra and/or arithmetic gets messy, especially if it's a contest problem (as one of the tags indicates). $\endgroup$ – Barry Cipra Apr 13 '14 at 21:23

Let the sum $S$ denote $2x+2y+2z$. Then, from the ratios, you have:

$$x+y = \frac{1}{7}S$$ $$y+z = \frac{2}{7}S$$ $$y+z = \frac{4}{7}S$$

From $x+y+z=35$ you also get: $$\frac{1}{2}S = 35$$

This is a linear system of four equations with four unknowns, which can be solved by a method of your choice, for example Gaussian elimination.

  • 1
    $\begingroup$ Thank you for making this more concise than my own convoluted methods. Clarity and concision are the name of the game in contest math. $\endgroup$ – Asimov Apr 13 '14 at 19:26
  • $\begingroup$ @JohnJPershing Clarity is gold when it comes to math in general. :) $\endgroup$ – naslundx Apr 13 '14 at 19:26

A more straightforward algebraic approach, as I assume this is a HS competition...

So if (x+y):(y+z):(x+z) = 1:2:4, why not multiply the expressions to make an equality?

4(x+y) = 2(y+z) = (x+z)

Next, use the fact that, since x + y + z = 35,

(x+y) = 35-z, (y+z) = 35-x, (x+z) = 35-y,

Substitute these expressions into your equality to get

4(35-z) = 2(35-x) = 35-y

Use these equalities to express y and z in terms of x, getting

z = x/2 + 70/4
y = 2x - 35

Substitute these values into x + y + z = 35 and solve for x.

I got x = 15.


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