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Difference between revisions of "2001 IMO Shortlist Problems/G1"
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<cmath>\frac{\sin\angle CBB_2}{\sin\angle ABB_2}=\frac{\sin{(\gamma+45^{\circ})}}{\sin{(\alpha+45^{\circ})}}</cmath>
 
<cmath>\frac{\sin\angle CBB_2}{\sin\angle ABB_2}=\frac{\sin{(\gamma+45^{\circ})}}{\sin{(\alpha+45^{\circ})}}</cmath>
  
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An application of the trigonometric version of Ceva's Theorem shows that <math>AA_2</math>, <math>BB_2</math>, and <math>CC_2</math> are concurrent, which shoes that <math>AA_1</math>, <math>BB_1</math>, and <math>CC_1</math> are concurrent.
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An application of the trigonometric version of Ceva's Theorem shows that <math>AA_2</math>, <math>BB_2</math>, and <math>CC_2</math> are concurrent, which shows that <math>AA_1</math>, <math>BB_1</math>, and <math>CC_1</math> are concurrent.
  
 
== Resources ==
 
== Resources ==

Latest revision as of 14:07, 12 December 2011

Problem

Let $A_1$ be the center of the square inscribed in acute triangle $ABC$ with two vertices of the square on side $BC$. Thus one of the two remaining vertices of the square is on side $AB$ and the other is on $AC$. Points $B_1,\ C_1$ are defined in a similar way for inscribed squares with two vertices on sides $AC$ and $AB$, respectively. Prove that lines $AA_1,\ BB_1,\ CC_1$ are concurrent.

Solution

Let $BC=a$, $CA=b$, $AB=c$, $\angle BAC=\alpha$, $\angle CBA=\beta$, and $\angle ACB=\gamma$. Let $A_2$ be the point on the other side of $BC$ than $A$ such that $BA_2C$ is an isosceles right triangle. Define $B_2$ and $C_2$ similarly. Let $E$ and $F$ be the points on $AB$ and $CA$ that are the vertices of the square centered at $A_1$. We then have that $EA_1F$ is also an isosceles right triangle. It's clear that $EF$ is parallel to $BC$, so $\triangle AEF\sim \triangle ABC$ and $\triangle A_1EF\sim \triangle A_2BC$. The ratio of similarity of both relations is $\frac{EF}{BC}$, which implies that quadrilaterals $AEA_1F$ and $ABA_2C$ are similar. Therefore $\angle BAA_2=\angle EAA_1$ and $\angle CAA_2=EAA_1$. It then follows that $A$, $A_1$, and $A_2$ are collinear. Similarly $B$, $B_1$, and $B_2$ are collinear, as are $C$, $C_1$, and $C_2$. It therefore suffices to show that $AA_2$, $BB_2$, and $CC_2$ are concurrent.

Let $AC_2=s$ and $CC_2=d$, for positive reals $s$ and $d$. Also let $\angle ACC_2=\theta$ and $\angle BCC_2=\phi$. Note that $\angle C_2AC=\angle C_2AB+\angle BAC=45^{\circ}+\alpha$, and likewise $\angle C_2BC = 45^{\circ}+\beta$. It then follows from the Law of Sines on triangles $ACC_2$ and $BCC_2$ that

\[\frac{d}{\sin{(\alpha+45^{\circ})}}=\frac{s}{\sin{\theta}}\]

and

\[\frac{d}{\sin{(\beta+45^{\circ})}}=\frac{s}{\sin{\phi}}\]

Solving for $\sin{\theta}$ and $\sin{\phi}$ gives that

\[\sin{\theta}=\frac{s\sin{(\alpha+45^{\circ})}}{d}\]

and

\[\sin{\phi}=\frac{s\sin{(\beta+45^{\circ})}}{d}\]

Therefore

\[\frac{\sin{\theta}}{\sin{\phi}}=\frac{\sin\angle ACC_2}{\sin\angle BCC_2}=\frac{\sin{(\alpha+45^{\circ})}}{\sin{(\beta+45^{\circ})}}\]

Similar lines of reasoning show that

\[\frac{\sin\angle BAA_2}{\sin\angle CAA_2}=\frac{\sin{(\beta+45^{\circ})}}{\sin{(\gamma+45^{\circ})}}\]

and

\[\frac{\sin\angle CBB_2}{\sin\angle ABB_2}=\frac{\sin{(\gamma+45^{\circ})}}{\sin{(\alpha+45^{\circ})}}\]

An application of the trigonometric version of Ceva's Theorem shows that $AA_2$, $BB_2$, and $CC_2$ are concurrent, which shows that $AA_1$, $BB_1$, and $CC_1$ are concurrent.

Resources

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