Much about your question is unclear, but I can talk about the median in the gamma distribution and you may be able to resolve the problem from that.
The gamma distribution is typically written either in the rate form or the scale form (I'll avoid using $\beta$ here, since your intention isn't clear):
Rate form:
$$f(x;\alpha,\phi) = \frac{\phi^\alpha}{\Gamma(\alpha)} x^{\alpha-1}e^{-\phi x} \quad \text{ for } x > 0 \text{ and } \alpha, \phi > 0$$
Scale form:
$$f(x;\alpha,\theta) = \frac{x^{\alpha-1}e^{-\frac{x}{\theta}}}{\theta^\alpha\Gamma(\alpha)} \quad \text{ for } x > 0 \text{ and } \alpha, \theta > 0$$
It's sometimes also written in the mean form (especially for GLMs):
$$f(x;\alpha,\mu) = \frac{\alpha^\alpha}{\mu^\alpha\Gamma(\alpha)}x^{\alpha-1}e^{-\frac{x\alpha}{\mu}} \quad \text{ for } x > 0 \text{ and } \alpha, \mu > 0$$
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The mean is $\alpha\theta = \alpha/\phi=\mu$.
The median doesn't have a simple closed form, but for larger $\alpha$, it can be approximated.
The median is approximately $\frac{3 \alpha - 0.8}{3 \alpha + 0.2}$ times the mean (as long as $\alpha$ is not too small).
Another way to approach the median is via the Wilson-Hilferty transformation; since the cube root of a gamma is approximately normal, one can approximate the relationship between the mean and the median via that relationship (the median directly transforms both directions, but the mean is more complicated - one can use Taylor expansions to approximate the mean).
For more accuracy, if one specifies the median and one specifies the scale or rate parameter, one can use the gamma cdf or the inverse gamma cdf (the gamma quantile function) to iteratively solve for the required alpha.
If $F$ is the gamma cdf and $F^{-1}$ is its inverse, you use root finding to solve $F_{\alpha,\beta}(m) - \frac{1}{2}=0$ for $\alpha$, or to solve $m - F^{-1}_{\alpha,\beta}(\frac{1}{2})=0$
A good starting point would be from one of the previously mentioned approximations.
I think I know what an appropriate β parameter is (though I could be wrong on this point, feel free to make comments if I'm thinking about that incorrectly)
It's a bit hard to comment about whether you have a misconception here if you aren't explicit about what you actually think and why.