Iran MO 1997

For discussing Olympiad level Geometry Problems
User avatar
Thamim Zahin
Posts:98
Joined:Wed Aug 03, 2016 5:42 pm
Iran MO 1997

Unread post by Thamim Zahin » Mon Oct 10, 2016 9:06 pm

In triangle $ABC$, angles $B,C$ are acute. Point $D$ is on the side $BC$ such that $AD\perp{BC}$. Let the interior bisectors of $\angle B,\angle C$ meet $AD$ at $E,F$, respectively. If $BE=CF$, prove that $ABC$ is isosceles.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.

tanmoy
Posts:312
Joined:Fri Oct 18, 2013 11:56 pm
Location:Rangpur,Bangladesh

Re: Iran MO 1997

Unread post by tanmoy » Wed Oct 12, 2016 1:36 pm

A trigonometric solution:
For the sake of contradiction,let's assume that $AB \neq AC$.WLOG,suppose $AB > AC$.We will prove that $BE > CF$.

Now,$BE^{2}=BD^{2}+DE^{2}$ and $CF^{2}=CD^{2}+DF^{2}$...............(1)

Claim 1: $BD>CD$.
Proof: $\text{tan}$ $\angle ACD$=$\dfrac {AD} {DC}$ and $\text{tan}$ $\angle ABD$=$\dfrac {AD} {DB}$.
So,$\dfrac {BD} {CD}=\dfrac {tan \angle ACD} {tan \angle ABD} >1$.(Since $\angle ACD>\angle ABD$ and $f(x)= tan x$ is an increasing function on $[0,\dfrac {\pi} {2})$).
Therefore,$BD>CD$

Claim 2: $DE>DF$.
Proof: $\dfrac {AD} {DF}=\dfrac {DC+CA} {DC}$ and $\dfrac {AD} {DE}=\dfrac {DB+BA} {BD}$.So,$DE<DF$ $\Rightarrow$ $\dfrac {DB+BA} {BD}>\dfrac {DC+CA} {DC}$ $\Rightarrow$ $\dfrac {BA} {BD}>\dfrac {CA} {CD}$ $\Rightarrow$ $\text{Sec}$ $\angle ABD$ $>$ $\text{Sec}$ $\angle ACD$.But $\angle ACD>\angle ABD$,so contradiction.(Since,$f(x)=secx$ is increasing on $[0,\dfrac {\pi} {2}]$).Therefore $ DE>DF $

From Claim 1,Claim 2 and (1),we get that $BE>CF$.But this contradicts with the problem's condition.So,our first assumption was wrong and $AB=AC$.
"Questions we can't answer are far better than answers we can't question"

tanmoy
Posts:312
Joined:Fri Oct 18, 2013 11:56 pm
Location:Rangpur,Bangladesh

Re: Iran MO 1997

Unread post by tanmoy » Wed Oct 12, 2016 3:14 pm

Synthetic and easier proofs of claim 1 and claim 2:
Proof of claim 1: $BD^{2}=AB^{2}-AD^{2}$ and $CD^{2}=AC^{2}-CD^{2}$.As $AB>AC$(our assumption),so,$BD>CD$.

Proof of claim 2: $AB^{2}-BD^{2}=AC^{2}-CD^{2}$.From claim 1,$BD>CD$,so,dividing the LHS by $BD$ and dividing the RHS by $CD$,we get that

$\dfrac {AB^{2}} {BD^{2}}-1<\dfrac {AC^{2}} {CD^{2}}-1$

$\Rightarrow$ $\dfrac {AB} {BD}<\dfrac {AC} {CD}$

$\Rightarrow$ $\dfrac {AE} {ED}<\dfrac {AF} {FD}$

$\Rightarrow$ $\dfrac {AD} {ED}<\dfrac {AD} {FD}$ $\Rightarrow$ $ED>FD$.
"Questions we can't answer are far better than answers we can't question"

Post Reply