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Geometry Difficulty 4.9 AIME Find the answer

7. In a right triangular prism, it is known that the base area is s m2s \mathrm{~m}^{2}, and the areas of the three lateral faces are m m2,n m2,p m2m \mathrm{~m}^{2}, n \mathrm{~m}^{2}, p \mathrm{~m}^{2}. Then its volume is \qquad m3\mathrm{m}^{3}.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Let the height of a right triangular prism be h mh \mathrm{~m}, and the lengths of the three sides of the base triangle be a ma \mathrm{~m}, b mb \mathrm{~m}, and c mc \mathrm{~m}. Thus,
ah=m,bh=n,ch=p. Therefore, S=14(a+b+c)(a+b−c)(c+a−b)(b+c−a)=14m+n+ph⋅m+n−ph⋅p+m−nh⋅n+p−mh=14h2(m+n+p)(m+n−p)(p+m−n)(n+p−m).  \begin{array}{l} a h=m, b h=n, c h=p . \\ \text { Therefore, } S=\frac{1}{4} \sqrt{(a+b+c)(a+b-c)(c+a-b)(b+c-a)} \\ =\frac{1}{4} \sqrt{\frac{m+n+p}{h} \cdot \frac{m+n-p}{h} \cdot \frac{p+m-n}{h} \cdot \frac{n+p-m}{h}} \\ =\frac{1}{4 h^{2}} \sqrt{(m+n+p)(m+n-p)(p+m-n)(n+p-m)} \text {. } \\ \end{array}

Then h2=14S(m+n+p)(m+n−p)(p+m−n)(n+p−m)h^{2}=\frac{1}{4 S} \sqrt{(m+n+p)(m+n-p)(p+m-n)(n+p-m)},
V2=S2h2=S4(m+n+p)(m+n−p)(p+m−n)(n+p−m),V=S2(m+n+p)(m+n−p)(p+m−n)(n+p−m)4. \begin{array}{l} V^{2}=S^{2} h^{2} \\ =\frac{S}{4} \sqrt{(m+n+p)(m+n-p)(p+m-n)(n+p-m)}, \\ V=\frac{\sqrt{S}}{2} \sqrt[4]{(m+n+p)(m+n-p)(p+m-n)(n+p-m)} . \end{array}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.