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Question Number 130339 by mnjuly1970 last updated on 24/Jan/21

Answered by MJS_new last updated on 25/Jan/21

$$A=\left(\begin{array}{c}0\\ a\end{array}\right)B=\left(\begin{array}{c}0\\ 0\end{array}\right)C=\left(\begin{array}{c}c\\ 0\end{array}\right)$$$$AC:y=-\frac{a}{c}x+a$$$$0⩽d\in \mathbb{R}$$$$BD:y=dx$$$$D=\left(\begin{array}{c}\frac{ac}{a+cd}\\ \frac{acd}{a+cd}\end{array}\right)$$$$\mathrm{radius}\mathrm{of}\mathrm{incircle}ABD$$$$\frac{ac}{a+cd+\sqrt{a^2+c^2}+c\sqrt{d^2+1}}=r_{ABD}$$$$\mathrm{radius}\mathrm{of}\mathrm{incircle}BCD$$$$\frac{acd}{a+cd+d\sqrt{a^2+c^2}+a\sqrt{d^2+1}}=r_{BCD}$$$$r_{ABD}=r_{BCD}\mathrm{leads}\mathrm{to}$$$$(a-cd)\sqrt{d^2+1}=(a+cd)(d-1)$$$$\Rightarrow d=\frac{(a\sqrt{2}+\sqrt{ac}-c\sqrt{2})\sqrt{a}}{(2a-c)\sqrt{c}}$$$$\mathrm{this}\mathrm{means},\mathrm{for}\mathrm{any}\mathrm{rectangular}\mathrm{triangle}\mathrm{we}$$$$\mathrm{can}\mathrm{find}\mathrm{a}d\mathrm{in}\mathrm{order}\mathrm{to}\mathrm{get}2\mathrm{part}\mathrm{triangles}$$$$\mathrm{with}\mathrm{equal}\mathrm{incircles}(\underset{c\to 2a}{\lim }d=\frac{3}{4})$$$$$$$$\mathrm{now}\mathrm{we}\mathrm{know}\mid BD\mid =13$$$$\frac{ac\sqrt{d^2+1}}{a+cd}=13$$$$\mathrm{the}\mathrm{question}\mathrm{is}\mathrm{if}\mathrm{the}\mathrm{area}\mathrm{is}\mathrm{unique}\mathrm{under}$$$$\mathrm{these}\mathrm{conditions}$$$$\mathrm{indeed}\mathrm{because}\mathrm{of}a>0\wedge c>0\mathrm{inserting}\mathrm{for}d$$$$\mathrm{we}\mathrm{get}$$$$ac=338$$$$\Rightarrow \mathrm{area}\mathrm{is}\frac{ac}{2}=169$$

Commented by mnjuly1970 last updated on 25/Jan/21

$$gratefulmrmjs.thanksalot...$$

Answered by Olaf last updated on 24/Jan/21

$$\mathrm{By}\mathrm{symmetry}\mathrm{AB}=\mathrm{BC}$$$$\mathrm{D}\mathrm{middle}\mathrm{of}\left[\mathrm{AC}\right]$$$$\mathrm{BD}=\mathrm{AD}=\mathrm{DC}$$$$\mathrm{AC}=26$$$$\mathrm{Area}\left[\mathrm{ABC}\right]=\frac{1}{2}\mathrm{AC}\times \mathrm{BD}=169$$

Commented by MJS_new last updated on 25/Jan/21

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{necessary}\mathrm{to}\mathrm{have}\mid AB\mid =\mid BC\mid .{\mathrm{it}}^{\prime }\mathrm{s}$$$$\mathrm{always}\mathrm{true}\mathrm{for}\mid AB\mid \times \mid BC\mid =338$$

Commented by mnjuly1970 last updated on 25/Jan/21

$$perfect.thankyoumrolaf...$$

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