AMC12 2024 A

AMC12 2024 A · Q19

AMC12 2024 A · Q19. It mainly tests Triangles (properties), Circle theorems.

Cyclic quadrilateral $ABCD$ has lengths $BC=CD=3$ and $DA=5$ with $\angle CDA=120^\circ$. What is the length of the shorter diagonal of $ABCD$?

循环四边形 $ABCD$ 有边长 $BC=CD=3$ 和 $DA=5$，且 $\angle CDA=120^\circ$。求 $ABCD$ 的较短对角线的长度？

(A) \frac{31}7 \frac{31}7

(B) \frac{33}7 \frac{33}7

(D) \frac{39}7 \frac{39}7

(E) \frac{41}7 \qquad \frac{41}7 \qquad

Answer

Correct choice: (D)

正确答案：（D）

Solution

Since $ABCD$ is a cyclic quadrilateral, opposite angles sum to $180^\circ$, so \[ \angle CBA = 60^\circ. \] Let $AC = u$. Apply the Law of Cosines in $\triangle ACD$ (where $\angle ACD = 120^\circ$): \begin{align*} u^2 &= 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cdot \cos 120^\circ \\ u^2 &= 9 + 25 - 30 \cdot \left(-\dfrac{1}{2}\right) \\ u^2 &= 34 + 15 = 49 \\ u &= 7. \end{align*} Let $AB = v$. Apply the Law of Cosines in $\triangle ABC$ (where $\angle BAC = 60^\circ$): \begin{align*} 7^2 &= 3^2 + v^2 - 2 \cdot 3 \cdot v \cdot \cos 60^\circ \\ 49 &= 9 + v^2 - 6v \cdot \dfrac{1}{2} \\ 49 &= 9 + v^2 - 3v \\ v^2 - 3v - 40 &= 0. \end{align*} Solve the quadratic equation: \[ v = \dfrac{3 \pm \sqrt{9 + 160}}{2} = \dfrac{3 \pm \sqrt{169}}{2} = \dfrac{3 \pm 13}{2}. \] The positive root is \[ v = \dfrac{3 + 13}{2} = 8. \] (We discard $v = -5$ since length is positive.) Now apply **Ptolemy's theorem** for cyclic quadrilateral $ABCD$: \[ AB \cdot CD + AD \cdot BC = AC \cdot BD \] \[ 8 \cdot 3 + 5 \cdot 3 = 7 \cdot BD \] \[ 24 + 15 = 7 \cdot BD \] \[ 39 = 7 \cdot BD \implies BD = \dfrac{39}{7}. \] Since \[ \dfrac{39}{7} \approx 5.571 < 7, \] the answer is $\boxed{\dfrac{39}{7}}$.

由于 $ABCD$ 是循环四边形，对角相加为 $180^\circ$，所以 \[\angle CBA = 60^\circ.\] 设 $AC = u$。在 $\triangle ACD$ 中应用余弦定律（其中 $\angle ACD = 120^\circ$）： \begin{align*}u^2 &= 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cdot \cos 120^\circ \\u^2 &= 9 + 25 - 30 \cdot \left(-\dfrac{1}{2}\right) \\u^2 &= 34 + 15 = 49 \\u &= 7.\end{align*} 设 $AB = v$。在 $\triangle ABC$ 中应用余弦定律（其中 $\angle BAC = 60^\circ$）： \begin{align*}7^2 &= 3^2 + v^2 - 2 \cdot 3 \cdot v \cdot \cos 60^\circ \\49 &= 9 + v^2 - 6v \cdot \dfrac{1}{2} \\49 &= 9 + v^2 - 3v \\v^2 - 3v - 40 &= 0.\end{align*} 解二次方程： \[v = \dfrac{3 \pm \sqrt{9 + 160}}{2} = \dfrac{3 \pm \sqrt{169}}{2} = \dfrac{3 \pm 13}{2}.\] 正根为 \[v = \dfrac{3 + 13}{2} = 8.\] （丢弃 $v = -5$ 因为长度为正。）现在对循环四边形 $ABCD$ 应用**托勒密定理**： \[AB \cdot CD + AD \cdot BC = AC \cdot BD\] \[8 \cdot 3 + 5 \cdot 3 = 7 \cdot BD\] \[24 + 15 = 7 \cdot BD\] \[39 = 7 \cdot BD \implies BD = \dfrac{39}{7}.\] 由于 \[\dfrac{39}{7} \approx 5.571 < 7,\] 答案为 $\boxed{\dfrac{39}{7}}$。

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