AMC10 2013 A

AMC10 2013 A · Q23

AMC10 2013 A · Q23. It mainly tests Circle theorems, Primes & prime factorization.

In $\triangle ABC$, $AB = 86$, and $AC = 97$. A circle with center $A$ and radius $AB$ intersects $BC$ at points $B$ and $X$. Moreover $BX$ and $CX$ have integer lengths. What is $BC$ ?

在$\triangle ABC$中，$AB=86$，$AC=97$。以$A$为圆心、$AB$为半径的圆与$BC$相交于点$B$和$X$。而且$BX$和$CX$为整数长度。$BC$是多少？

(A) 11 11

(B) 28 28

(D) 61 61

(E) 72 72

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): By the Power of a Point Theorem, $BC\cdot CX=AC^2-r^2$ where $r=AB$ is the radius of the circle. Thus $BC\cdot CX=97^2-86^2=2013$. Since $BC=BX+CX$ and $CX$ are both integers, they are complementary factors of 2013. Note that $2013=3\cdot 11\cdot 61$, and $CX<BC<AB+AC=183$. Thus the only possibility is $CX=33$ and $BC=61$.

答案（D）：根据点的幂定理，$BC\cdot CX=AC^2-r^2$，其中$r=AB$为圆的半径。因此$BC\cdot CX=97^2-86^2=2013$。由于$BC=BX+CX$且$CX$都是整数，它们是2013的一对互补因子。注意$2013=3\cdot 11\cdot 61$，并且$CX<BC<AB+AC=183$。因此唯一可能是$CX=33$且$BC=61$。

Topics