Hello, I am newbie at Spring. I have an application that uses tiles that I would like to implement an interceptor. I tried to implement it but it's not working

Here is my spring-context.xml

Code:
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
	xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
	xmlns:p="http://www.springframework.org/schema/p"
	xmlns:context="http://www.springframework.org/schema/context"
	xmlns:mvc="http://www.springframework.org/schema/mvc"
	xsi:schemaLocation="http://www.springframework.org/schema/mvc 
						http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd
	http://www.springframework.org/schema/beans 
	http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
	http://www.springframework.org/schema/context 
	http://www.springframework.org/schema/context/spring-context-3.0.xsd">

	<context:component-scan base-package="br.com.project.livros" />
	<mvc:annotation-driven />

	<mvc:resources location="/resources/" mapping="/resources/**"/>
	
	<mvc:default-servlet-handler/>

	<bean class="org.springframework.web.servlet.view.UrlBasedViewResolver"
      id="tilesViewResolver">
	     <property name="viewClass" value="org.springframework.web.servlet.view.tiles2.TilesView" />
	</bean>

	<mvc:interceptors>
		<bean class="br.com.coldsoft.livros.interceptor.AutorizadorInterceptor" />
	</mvc:interceptors>


	<bean id="tilesConfigurer" class="org.springframework.web.servlet.view.tiles2.TilesConfigurer">
	  <property name="definitions">
	    <list>
	      <value>/WEB-INF/tiles-defs.xml</value>
	    </list>
	  </property>
	</bean>
</beans>
Interceptor class:

Code:
public class AutorizadorInterceptor extends HandlerInterceptorAdapter {

	@Override
	public boolean preHandle(HttpServletRequest request,
			HttpServletResponse response, Object handler) throws Exception {

		String uri = request.getRequestURI();
		if (uri.endsWith("loginForm") || uri.endsWith("efetuaLogin")){
			return true;
		}
		
		if(request.getSession().getAttribute("usuarioLogado") != null){
			return true;
		}
		response.sendRedirect("loginForm");
		return false;
			
	}
}
LoginController class

Code:
@Controller
public class LoginController {
	
	@RequestMapping("logout")
	public String logout(HttpSession session){
		session.invalidate();
		return "redirect:loginForm";
	}
	
	@RequestMapping("loginForm")
	public String loginForm(){
		return "formulario-login";
	}
	
	@RequestMapping("efetuaLogin")
	public String efetuaLogin(Usuario usuario, HttpSession session){
		if(new UsuarioDao().userExists(usuario)){
			session.setAttribute("usuarioLogado", usuario);
			return "redirect:main";
		}
		return "redirect:loginForm";
	}
	
}
The error is:
Code:
javax.servlet.ServletException: Could not resolve view with name 'formulario-login' in servlet with name 'springmvc
My file formulario-login.jsp is at WebContent folder

How can I use a Spring Interceptor with Tiles 2.2.2 ? Anyone can help me ?