In a right angled triangle, a semicircle is drawn such that its diameter lies on the hypotenuse and its center divides the hypotenuse into two segments of lengths 15 and 20.Find the length of the arc of the semicircle between the points at which the legs touch the semicircle.
[Math] A Part of a semicircle between the two legs of a right angle triangle
circlesgeometrytriangles
Related Solutions
Writing $a := |BC|$, $b := |CA|$, $c := |AB| = \sqrt{a^2+b^2}$, and $r = |PE| = |PF|$ (so that $|PD| = 3r$), we have $$\begin{align} |\triangle ABC| &= |\triangle ABP| + |\triangle BCP| + |\triangle CAP| \\[4pt] \implies \qquad \frac{1}{2} |BC||CA| &= \frac{1}{2} \left(\; |AB| |PF| + |BC||PD| + |CA||PE| \;\right) \\[4pt] \implies \qquad a b &= c r + 3 a r + b r = r ( 3 a + b + c )\\[6pt] \implies \qquad r &= \frac{ab}{3 a + b + c} = \frac{ab}{3 a + b + \sqrt{a^2+b^2}} \end{align}$$
To address @DanielV's suggestion of generalizing to higher dimensions, consider a right-corner tetrahedron $OABC$, with right corner at $O$ and edge lengths $a := |OA|$, $b := |OB|$, $c := |OC|$. (Note that I'm changing notation slightly from the above.) Let a sphere with center $P$ and radius $r$ be tangent to the faces around vertex $A$, and let a congruent sphere (tangent to the first) be tangent to the faces around vertex $O$. Then $P$ has distance $r$ from faces $\triangle OAB$, $\triangle OCA$, $\triangle ABC$ (the ones touching $A$), and distance $3r$ from face $\triangle OBC$ (the one opposite $A$).
Here's a poor attempt at a diagram:
(In this case, the altitudes from $P$ are color-coded to match their parallel counterparts through $O$. The black altitude is to face $\triangle ABC$.)
Thus,
$$\begin{align} |OABC| &= |OABP| + |OBCP| + |OCAP| + |ABCP| \\[4pt] \implies \qquad \frac{1}{6}a b c &= \frac{1}{3}\left(\; r\;|\triangle OAB| + r \;|\triangle OCA| + r\;|\triangle ABC| + 3r\;|\triangle OBC| \;\right) \\[4pt] &= \frac{1}{3}r \cdot \frac{1}{2} \left(\; a b + c a + 3 b c + 2\;|\triangle ABC| \;\right) \\[6pt] \implies \qquad r &= \frac{abc}{3bc + ab + ca + 2\;|\triangle ABC|} \qquad (\star) \end{align}$$
Fun fact: The Pythagorean Theorem for Right-Corner Tetrahedra says that $$|\triangle ABC|^2 = |\triangle OBC|^2 + |\triangle OCA|^2 + |\triangle OAB|^2$$ so that we have $$|\triangle ABC| = \frac{1}{2} \sqrt{\; b^2 c^2 + c^2 a^2 + a^2 b^2 \;}$$ and $(\star)$ becomes $$r = \frac{abc}{3bc + ab + ca + \sqrt{\; b^2 c^2 + c^2 a^2 + a^2 b^2 \;}}$$
In $4$-dimensional space (where there's an analogous Pythagorean Theorem, as there is in any-dimensional space), we have $$r = \frac{abcd}{3bcd + acd + abd + abc + \sqrt{\;b^2 c^2 d^2 + a^2 c^2 d^2 + a^2 b^2 d^2 + a^2 b^2 c^2\;}}$$ and so forth.
Incidentally, the matching-notation version of the initial answer is $$r = \frac{ab}{3b + a + \sqrt{\;b^2 + a^2\;}}$$
Best Answer
Let the tangent length shown in sketch be T. The power of circle
$$ T^2 = (15-R) (15+R) \tag{1}$$
From similar triangles, (radius/hypotenuse) of right side right angled triangle:
$$ \frac{T}{R}= \frac{15}{20}= \frac{3}{4} \tag{2}$$
Solving
$$ R= 12, \; T = 9 \tag{3} $$
Arc Length is quarter circle $$ =\pi \, R/ 2 = \frac{ \pi \cdot 12}{2} = 6 \pi \tag{4}$$