Track / Stage 4 / 97 of 340 #357 of 2000

Problem 357

AMC 12 late, AIME early

Geometry Difficulty 4.7 Find the answer

4. In $\triangle A B C$ , $\angle B A C=60^{\circ}$ , the angle bisector $A D$ of $\angle B A C$ intersects $B C$ at $D$ , and $\overrightarrow{A D}=\frac{1}{4} \overrightarrow{A C}+t \overrightarrow{A B}$ . If $A B=8$ , then $A D=$ . $\qquad$

Your answer

Official solution

Answer: $6 \sqrt{3}$ .
[Analysis] It is easy to know that $t=\frac{3}{4}$ , thus: $A C=24, A D^{2}=\frac{1}{16} \times 24^{2}+\frac{9}{16} \times 8^{2}+\frac{3}{16} \times 8 \times 24=108$ , hence: $A D=6 \sqrt{3}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.