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Algebra Difficulty 1.0 Junior Prove it Canada

What is the integer tt for which 2t3+3t2=26\dfrac{2t}{3} + \dfrac{3t}{2} = 26?
What is the integer xx for which 3+x4=6+x8\dfrac{3+x}{4} = \dfrac{6+x}{8} ?
Suppose that y gt; 0\text{y gt; 0} and 32+42+122=32+42+y2\sqrt{3^2+4^2+12^2} = \sqrt{3^2+4^2} + \sqrt{y^2}. Determine the value of yy.

Solution

Multiplying both sides of the equation by 23=62 \cdot 3 = 6, we obtain 62t3+63t2=6266 \cdot \dfrac{2t}{3} + 6\cdot\dfrac{3t}{2} = 6 \cdot 26 or 4t+9t=1564t + 9t = 156.

Simplifying, we obtain 13t=15613t = 156
and so t=12t = 12.

Alternatively, using a common denominator of 23=62 \cdot 3 = 6, we obtain 22t23+33t32=26\dfrac{2 \cdot 2t}{2 \cdot 3} + \dfrac{3 \cdot 3t}{3 \cdot 2} = 26 or 4t6+9t6=26\dfrac{4t}{6} + \dfrac{9t}{6} = 26.

Simplifying, we obtain 13t6=26\dfrac{13t}{6} = 26 and so 13t=62613t = 6 \cdot 26
or t=62=12t = 6 \cdot 2 = 12.
Multiplying both sides of the equation by 88, we obtain 8(3+x)4=6+x\dfrac{8(3+x)}{4} = 6 + x which
simplifies to 2(3+x)=6+x2(3+x) = 6 + x.

Simplifying further, we obtain 6+2x=6+x6 + 2x = 6 + x and so x=0x = 0.

Alternatively, splitting each fraction into two pieces, we obtain
34+x4=68+x8\dfrac{3}{4} + \dfrac{x}{4} = \dfrac{6}{8} + \dfrac{x}{8}.

Since 34=68\dfrac{3}{4} = \dfrac{6}{8},
we obtain x4=x8\dfrac{x}{4} = \dfrac{x}{8} and so 8x=4x8x = 4x or x=0x = 0.
Since y gt; 0\text{y gt; 0}, then y2=y\sqrt{y^2} = y.

From the given equation, we obtain 9+16+144=9+16+y\sqrt{9 + 16 + 144} = \sqrt{9 + 16} + y.

Thus, 169=25+y\sqrt{169} = \sqrt{25} + y or
13=5+y13 = 5 + y and so y=8y = 8.

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