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If have two circles intersecting along their diameter, let $A,B$ with radius $r_1, r_2$ in the ratio of $1:n/2$ (for $n$ being a positive real number), and overlapping area is the same as radius of smaller circle (let, $r_1$); then how to prove geometrically that the lune formed will be having the intersection point (shown as point $H$, on the line joining the diameters) with length from the center of the smaller circle as : $\frac{r_1}{n}$. enter image description here

I can prove for different ratios of two circles individually and trying algebraically the two equations of different circles with diameters lying only on the $x$-axis.
Say, for ratio $n=5$, the two equations are given below with $b =r_1+2\cdot r_2$ being the total length of the two circles with intersecting length equal to $r_1$, assuming $r_1= \frac{b}{2^3}, r_2 = \frac{5r_1}{2}$ (or, diameter of circle $B=5r_1$) :
$$circle \,\,A :> (x-\frac{b}{8})^2 + y^2 = (\frac{b}{8})^2$$ $$circle \,\,B :> (x-\frac{7b}{16})^2 + y^2 = (\frac{5b}{16})^2$$

Have drawn at desmos too, the graph for the above at : https://www.desmos.com/calculator/e6dhp99eiu.
The graph also shows that the intersection point (shown by dotted line) is having length = $\frac{r_1}{5}$.

But a geometrical proof is needed.

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  • $\begingroup$ Please point out any error in problem construction. $\endgroup$
    – jiten
    Apr 17 '18 at 6:22
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Using Andrei's notation:

enter image description here

Since $C_AK$ is a diameter, $$\angle C_AMK=90^\circ.$$ Hence $$\angle C_AMK = \angle C_AHM.$$

Also $\angle MC_AK = \angle HC_AM.$

Hence $\triangle C_AMK$ is similar to $\triangle C_AHM$, hence we have

$$\frac{C_AH}{C_AM}=\frac{C_AM}{C_AK}=\frac{r_1}{2r_2}=\frac{1}{n}$$

$$\frac{C_AH}{C_AM}=\frac{1}{n}$$

$$\frac{C_AH}{r_1}=\frac{1}{n}$$

$$C_AH=\frac{r_1}{n}$$

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  • $\begingroup$ Want to draw in desmos the graph of 'Treasure hunt' problem, with varying position of gallows; but am unable to use the intersecting circles approach that has pine and oak as centers of two different circles, or alternate configuration that places pine and oak on x-axis, for different reasons. Also, have put a post at : math.stackexchange.com/q/2744464/424260 $\endgroup$
    – jiten
    Apr 19 '18 at 11:36
  • $\begingroup$ Please help with the above comment, as the general cases are represented by the $2$nd & $3$rd diagrams, for which getting no idea. $\endgroup$
    – jiten
    Apr 20 '18 at 1:19
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    $\begingroup$ will look at it when i m freer. I thought someone answered that... $\endgroup$ Apr 20 '18 at 1:23
  • $\begingroup$ Please see the post and my comments. Sorry, for not waiting. But am just lost and in total mess. So, will attempt the problem after your response. $\endgroup$
    – jiten
    Apr 20 '18 at 8:57
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    $\begingroup$ i will take at least $24$ hours and even up to one week... but yup, let's see how things go.. haven't read what is needed exactly yet. $\endgroup$ Apr 20 '18 at 9:12
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Let's call the upper intersection point $M$, and the two centers $C_A$ and $C_B$. The triangles $C_AMH$ and $C_BMH$ are right angle triangles, so we can write $MH$ from Pythagoras' theorem in each triangle: $$MH^2=C_AM^2-C_AH^2=C_BM^2-C_BH^2=C_BM^2-(C_AC_B-C_AH)^2$$ We notice that $C_BM=C_AC_B$, so from the second and fourth terms above $$C_AM^2-C_AH^2=C_AC_B^2-C_AC_B^2+2C_AC_B\cdot C_AH-C_AH^2\\ C_AM^2=2C_AC_B\cdot C_AH\\C_AH=\frac{C_AM^2}{2C_AC_B}=C_AM\frac{1}{2n/2}=\frac{r_1}{n}$$

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  • $\begingroup$ Thanks a lot. It is the correct framing of the problem that mattered, particularly the ratio ($r_1: r_2 = 1: n/2$), apart from the appln. of Pythagoras' theorem. $\endgroup$
    – jiten
    Apr 17 '18 at 6:44

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