Simple question on ADC sample code

The bold part of code below states that if A1 > 0.5*AVcc, P1.0 set, else reset.

But I don't quite understand how 0x1FF represents 0.5*AVcc.

Does anybody have idea?

//******************************************************************************

// MSP430G2x31 Demo - ADC10, Sample A1, AVcc Ref, Set P1.0 if > 0.5*AVcc
//
// Description: A single sample is made on A1 with reference to AVcc.
// Software sets ADC10SC to start sample and conversion - ADC10SC
// automatically cleared at EOC. ADC10 internal oscillator times sample (16x)
// and conversion. In Mainloop MSP430 waits in LPM0 to save power until ADC10
// conversion complete, ADC10_ISR will force exit from LPM0 in Mainloop on
// reti. If A1 > 0.5*AVcc, P1.0 set, else reset.
//
// MSP430G2x31
// -----------------
// /|\| XIN|-
// | | |
// --|RST XOUT|-
// | |
// >---|P1.1/A1 P1.0|-->LED
//
// D. Dang
// Texas Instruments Inc.
// October 2010
// Built with CCS Version 4.2.0 and IAR Embedded Workbench Version: 5.10
//******************************************************************************
#include "msp430g2231.h"

void main(void)
{
 WDTCTL = WDTPW + WDTHOLD; // Stop WDT
 ADC10CTL0 = ADC10SHT_2 + ADC10ON + ADC10IE; // ADC10ON, interrupt enabled
 ADC10CTL1 = INCH_1; // input A1
 ADC10AE0 |= 0x02; // PA.1 ADC option select
 P1DIR |= 0x01; // Set P1.0 to output direction

 for (;;)
 {
 ADC10CTL0 |= ENC + ADC10SC; // Sampling and conversion start
 __bis_SR_register(CPUOFF + GIE); // LPM0, ADC10_ISR will force exit
 if (ADC10MEM < 0x1FF)
 P1OUT &= ~0x01; // Clear P1.0 LED off
 else
 P1OUT |= 0x01; // Set P1.0 LED on
 }
}

// ADC10 interrupt service routine
#pragma vector=ADC10_VECTOR
__interrupt void ADC10_ISR(void)
{
 __bic_SR_register_on_exit(CPUOFF); // Clear CPUOFF bit from 0(SR)
}