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There was this problem given in a math textbook:

A right triangle has a hypotenuse equal to 10 and an altitude to the hypotenuse equal to 6. What is the area of the triangle?

Well, as any good geometry math student can see, there is an error with the way the question was worded – the person who made this question just pulled the numbers out of thin air and created a seemingly simple problem. For most students the answer would be $A=\frac12bh=30$, right?

Well, if we were to place the hypotenuse between $(-5, 0)$ and $(5, 0)$ and make it a diameter of a radius-5 circle, the 90° angle of the third vertex would need to be on the circumference on the circle, as any right-angled triangle needs to fit inside a half circle. But as can be seen, the largest height to a circle of this kind is the radius – 5 in this case, not 6 as stated in the question!

So, since the formula for the area still holds true, is there any way I can determine the coordinates of the triangle, given that they must contain complex numbers? Many have suggested complex analysis. One suggested the alternate formula $A=\frac12ab\sin\theta$, which in this case yielded $25\sin\theta$. Then using Euler's relationship $$\sin\theta=\sin\left(\frac\pi2-i\ln \frac{6+\sqrt{11}}5\right)=\frac65$$ The third vertex would then be $(i\sqrt{11},6)$. OK, this makes sense, but… I have no idea of how he gets here!! Can someone give me a simple step-by-step on how this is arrived at?

Parcly Taxel
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Craigers
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3 Answers3

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This is a classic attributed to V. I. Arnold (see more at V.I. Arnold says Russian students can't solve this problem, but American students can — why?).

Short answer is that there is no such right triangle, as you noted already in the real case.

Looking in $\mathbb{C}^2$, instead of $\mathbb{R}^2$, is not going to find any such right triangle, either. Reason is that $\mathbb{C}^2$ is isomorphic to $\mathbb{R}^4$ (as real vector spaces). Any $3$ non-collinear points in $\mathbb{R}^4$ define a unique 2D plane, so any geometric result you stated about a triangle in $\mathbb{R}^4$ could be restated in the $(x,y)$ plane in $\mathbb{R}^2$ by a suitable rotation of the axes. But the construction is impossible in $\mathbb{R}^2$, as noted above.

Community
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dxiv
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  • Yes, I realize that it would be impossible to construct in R^2, as it would be unable to also visualize it, but... that doesn't mean that one can't get the coordinates of each of the vertices, and some of them will be complex numbers, right? – Craigers Oct 25 '16 at 02:56
  • @Craigers The point is precisely that if you could solve it in complex numbers, then you could find a corresponding solution in real numbers as well. In this case, however, there is no solution either way. – dxiv Oct 25 '16 at 02:58
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There is nothing wrong with the wording of the question.   The hypotenuse is just not usually considered the base of a right triangle; and the phrase "altitude to the hypotenuse" should clue that this is the case.

In preschool* you should have learned that a triangle with sides in the ratio $3{:}4{:}5$ has a right angle opposite the hypotenuse (the longest side).   Which makes the numbers used convenient, but adequate for a primary school problem.

PS: The answer is $24$.

$\tiny\text{(* hyperbole)}$

Graham Kemp
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The altitude is perpendicular to the hypotenuse, it's not one of the bases; we're not looking at a 6-8-10.

This kind of problem is usually solved using the altitude-on-hypoteneuse theorem, i.e., the altitude divides the triangle into similar subtriangles. You can figure out the lengths (and coordinates) using similar triangles.

Dean C Wills
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