AMC12 2001 A

AMC12 2001 A · Q24

AMC12 2001 A · Q24. It mainly tests Angle chasing, Triangles (properties).

In $\triangle ABC$, $\angle ABC=45^\circ$. Point $D$ is on $\overline{BC}$ so that $2\cdot BD=CD$ and $\angle DAB=15^\circ$. Find $\angle ACB.$

在 $\triangle ABC$ 中，$\angle ABC=45^\circ$。点 $D$ 在 $\overline{BC}$ 上，使得 $2\cdot BD=CD$，且 $\angle DAB=15^\circ$。求 $\angle ACB$。

(A) 54° 54°

(B) 60° 60°

(D) 75° 75°

(E) 90° 90°

Answer

Correct choice: (D)

正确答案：（D）

Solution

Draw a good diagram! Now, let's call $BD=t$, so $DC=2t$. Given the rather nice angles of $\angle ABD = 45^\circ$ and $\angle ADC = 60^\circ$ as you can see, let's do trig. Drop an altitude from $A$ to $BC$; call this point $H$. We realize that there is no specific factor of $t$ we can call this just yet, so let $AH=kt$. Notice that in $\triangle{ABH}$ we get $BH=kt$. Using the 60-degree angle in $\triangle{ADH}$, we obtain $DH=\frac{\sqrt{3}}{3}kt$. The comparable ratio is that $BH-DH=t$. If we involve our $k$, we get: $kt(\frac{3}{3}-\frac{\sqrt{3}}{3})=t$. Eliminating $t$ and removing radicals from the denominator, we get $k=\frac{3+\sqrt{3}}{2}$. From there, one can easily obtain $HC=3t-kt=\frac{3-\sqrt{3}}{2}t$. Now we finally have a desired ratio. Since $\tan\angle ACH = 2+\sqrt{3}$ upon calculation, we know that $\angle ACH$ can be simplified. Indeed, if you know that $\tan(75)=2+\sqrt{3}$ or even take a minute or two to work out the sine and cosine using $\sin(x)^2+\cos(x)^2=1$, and perhaps the half- or double-angle formulas, you get $\boxed{75^\circ}$.

画出清晰的图！令 $BD=t$，则 $DC=2t$。由图可见 $\angle ABD = 45^\circ$ 且 $\angle ADC = 60^\circ$，因此用三角函数。过 $A$ 向 $BC$ 作高，垂足为 $H$。此时还无法给出 $t$ 的具体倍数，设 $AH=kt$。注意在 $\triangle{ABH}$ 中有 $BH=kt$。利用 $\triangle{ADH}$ 中的 $60$ 度角，得到 $DH=\frac{\sqrt{3}}{3}kt$。又因为 $BH-DH=t$，代入 $k$ 得： $kt(\frac{3}{3}-\frac{\sqrt{3}}{3})=t$。约去 $t$ 并将分母有理化，得 $k=\frac{3+\sqrt{3}}{2}$。于是 $HC=3t-kt=\frac{3-\sqrt{3}}{2}t$。现在得到所需比值。计算可得 $\tan\angle ACH = 2+\sqrt{3}$，因此 $\angle ACH$ 可化简。确实，若知道 $\tan(75)=2+\sqrt{3}$，或用 $\sin(x)^2+\cos(x)^2=1$ 以及半角/倍角公式计算正弦余弦，也可得到 $\boxed{75^\circ}$。

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