Given a triangle $ABC$, let $D$ and $E$ be points on the side $BC$ such that $\angle BAD = \angle CAE$. If $M$ and $N$ are, respectively, the points of tangency of the incircles of the triangles $ABD$ and $ACE$ with the line $BC$, then show that \[\frac{1}{MB}+\frac{1}{MD}= \frac{1}{NC}+\frac{1}{NE}. \]
Problem
Source: IMO Shortlist 1993, Spain 2; India TST 1994
Tags: geometry, inradius, incenter, trigonometry, Law of Sines, IMO Shortlist
Gabriel(fr)
28.03.2006 06:30
Considering that $\angle CAE=\angle BAD=2\theta$, $\angle ABC=\alpha$, $\angle ACB=\beta$ denoting $R$ as the circunradium of triangle $ABC$, $r_1$, $r_2$, as the inradius of the triangles $ABD$ and $ACE$ respectively and assuming that the lines that pass thought the angles of $ABD$ and $ACE$ and it incenters are the angle bissectors of the same, we have,
$tg(\theta)=\frac{r_1}{AB-MB}$ but by the Law of Sines we know that $AB=2R.sin(\beta)$ consequently,
$r_1=tg(\theta)(2R.sin(\beta)-MB)$ therefore,
$tg\left(\frac{\alpha}{2}\right)=\frac{r_1}{MB}=\frac{tg(\theta)(2R.sin(\beta)-MB)}{MB} \implies MB=\frac{2R.sin(\beta)tg(\theta)}{tg\left(\frac{\alpha}{2}\right)+tg(\theta)}$, hence
$r_1=\frac{2R.sin(\beta)tg(\theta)tg\left(\frac{\alpha}{2}\right)}{tg\left(\frac{\alpha}{2}\right)+tg(\theta)}$ considering that $\angle ADB=180^o-2\theta-\alpha$ we get,
$tg\left(\frac{\angle ADB}{2}\right)=\frac{1}{tg\left(\theta+\frac{\alpha}{2}\right)}=\frac{r_1}{MD}$ yields $MD=\frac{2R.sin(\beta)tg(\theta)tg\left(\frac{\alpha}{2}\right)}{1-tg\left(\frac{\alpha}{2}\right).tg(\theta)}$ consequently,
$\frac{1}{MB} + \frac{1}{MD}=\frac{1}{{2R.sin(\beta)tg(\theta)}}\left(tg\left(\frac{\alpha}{2}\right)+tg(\theta)+\frac{1-tg\left(\frac{\alpha}{2}\right).tg(\theta)}{tg\left(\frac{\alpha}{2}\right)}\right)$
$=\frac{1}{{2R.sin(\beta)tg(\theta)}}\left(tg\left(\frac{\alpha}{2}\right)+\frac{1}{tg\left(\frac{\alpha}{2}\right)}\right)$
$=\frac{1}{{R.tg(\theta)sin(\beta)2.sin\left(\frac{\alpha}{2}\right)cos\left(\frac{\alpha}{2}\right)}}$
$=\frac{1}{{R.tg(\theta)sin(\beta)sin(\alpha)}}$
analogously we have,
$r_2=\frac{2R.sin(\alpha)tg(\theta)tg\left(\frac{\beta}{2}\right)}{tg\left(\frac{\beta}{2}\right)+tg(\theta)}$ considering that $\angle AEC=180^o-2\theta-\beta$ we get,
$NC=\frac{2R.sin(\alpha)tg(\theta)}{tg\left(\frac{\beta}{2}\right)+tg(\theta)}$
$NE=\frac{2R.sin(\alpha)tg(\theta)tg\left(\frac{\beta}{2}\right)}{1-tg\left(\frac{\beta}{2}\right).tg(\theta)}$ and,
$\frac{1}{NC} + \frac{1}{NE}=\frac{1}{{R.tg(\theta)sin(\alpha)2.sin\left(\frac{\beta}{2}\right)cos\left(\frac{\beta}{2}\right)}}$
$=\frac{1}{{R.tg(\theta)sin(\beta)sin(\alpha)}}$
in conclusion, $\frac{1}{MB} + \frac{1}{MD} = \frac{1}{NC} + \frac{1}{NE}$
fagot
28.03.2006 18:56
Let $\angle BAD=\alpha$ and the altitude $AH=h$. Then \[ {1\over MB}+{1\over MD}={BD\over MB\cdot MD}={4BD\over (BD-AD+AB)(BD+AD-AB)}={4BD\over BD^2-(AD-AB)^2}={2BD\over AB\cdot AD(1-\cos\alpha)}={BD \sin \alpha\over S(ABD)(1-\cos \alpha)}={2\sin \alpha\over h(1-\cos\alpha)}={1\over NC}+{1\over NE} \]
M4RI0
11.03.2007 18:27
I was looking for a solution without trigonometry.
M4RI0
19.12.2007 19:37
Lemma: In a triangle $ ABC$, the points of tangency of the incircle with $ AB$, $ BC$, $ CA$ are $ X$,$ Y$,$ Z$, $ \measuredangle BAC = \alpha$, and let $ S$ denothe the area, then: $ \boxed{S = (BY)(YC)cot\frac {\alpha}{2}}$ Proof: $ r$:inradius $ S = \frac {(m + rcot\frac {\alpha}{2})(n + rcot\frac {\alpha}{2})}{2}sin\alpha$ $ S = \frac {mn}{2} sin\frac {\alpha}{2} + \frac {1}{2}[rctg\frac {\alpha}{2}(m + n + rctg\frac {\alpha}{2})]sin\alpha$ $ S = \frac {mn}{2} sin\frac {\alpha}{2} + Scos^2\frac {\alpha}{2}$ $ \boxed{S = (BY)(YC)cot\frac {\alpha}{2}}$ By the lemma, it follows that: $ \frac {(AB)(AD)}{2}sin\alpha = (MB)(MD)ctg\frac {\alpha}{2}$ ; $ \alpha = \measuredangle BAD=\measuredangle CAE$ $ \frac {(AC)(AE)}{2}sin\alpha = (NC)(NE)ctg\frac {\alpha}{2}$ $ \frac {(AB)(AD)}{(MB)(MD)} = \frac {(AC)(AE)}{(NC)(NE)}$ Also we have: $ \frac {[ABD]}{BD} = \frac {[ACE]}{CE}$ $ \frac {(AB)(AD)}{BD} = \frac {(AC)(AE)}{CE}$ $ \frac {BD}{(MB)(MD)} = \frac {CE}{(NC)(NE)}$ $ \boxed{\frac {1}{MB} + \frac {1}{MD} = \frac {1}{NC} + \frac {1}{NE}}$ $ \mathbb{QED}$
djmathman
20.05.2020 00:07
$12\tfrac12$ year revive, but hey, this solution is different! Observe that, using standard triangle notation, \[ \frac{1}{s-b}+\frac{1}{s-c} = \frac{a}{(s-b)(s-c)} = \frac{as(s-a)}{K^2} = \frac{as(s-a)}{\frac12ah_a\cdot \frac12rs} = \frac{4(s-a)}{h_ar} = \frac{4\cot\frac12 A}{h_a}, \]where $h_a$ is the height of the altitude from $A$. Since $h_a$ and $A$ are the same for both triangle $ABD$ and triangle $ACE$, the result follows.