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AMC数学竞赛真题2017年10B 21-22

2018-08-30 重点归纳

AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10数学竞赛真题以及官方解答吧：

Problem 21

In $\triangle ABC$ , $AB=6$ , $AC=8$ , $BC=10$ , and $D$ is the midpoint of $\overline{BC}$ . What is the sum of the radii of the circles inscribed in $\triangle ADB$ and $\triangle ADC$ ?

$\textbf{(A)}\ \sqrt{5}\qquad\textbf{(B)}\ \frac{11}{4}\qquad\textbf{(C)}\ 2\sqrt{2}\qquad\textbf{(D)}\ \frac{17}{6}\qquad\textbf{(E)}\ 3$

Solution

We note that by the converse of the Pythagorean Theorem, $\triangle ABC$ is a right triangle with a right angle at $A$ . Therefore, $AD = BD = CD = 5$ , and $[ADB] = [ADC] = 12$ . Since $A = rs$ , the inradius of $\triangle ADB$ is $amc真题$ , and the inradius of $\triangle ADC$ is $\frac{12}{(5+5+8)/2} = \frac 43$ . Adding the two together, we have $\boxed{\textbf{(D) } \frac{17}6}$ .

Problem 22

The diameter $\overline{AB}$ of a circle of radius $2$ is extended to a point $D$ outside the circle so that $BD=3$ . Point $E$ is chosen so that $ED=5$ and line $ED$ is perpendicular to line $AD$ . Segment $\overline{AE}$ intersects the circle at a point $C$ between $A$ and $E$ . What is the area of $\triangle ABC$ ?

$美国数学竞赛$

Solutions

$amc数学竞赛$

Solution 1

Notice that $ADE$ and $ABC$ are right triangles. Then $AE = \sqrt{7^2+5^2} = \sqrt{74}$ . $\sin{DAE} = \frac{5}{\sqrt{74}} = \sin{BAE} = \sin{BAC} = \frac{BC}{4}$ , so $BC = \frac{20}{\sqrt{74}}$ . We also find that $AC = \frac{28}{\sqrt{74}}$ , and thus the area of $ABC$ is $\frac{\frac{20}{\sqrt{74}}\cdot\frac{28}{\sqrt{74}}}{2} = \frac{\frac{560}{74}}{2} = \boxed{\textbf{(D) } \frac{140}{37}}$ .

Solution 2

We note that $\triangle ACB ~ \triangle ADE$ by $AA$ similarity. Also, since the area of $\triangle ADE = \frac{7 \cdot 5}2 = \frac{35}2$ and $AE = \sqrt{74}$ , $amc竞赛$ , so the area of $\triangle ABC = \boxed{\textbf{(D) } \frac{140}{37}}$ .

Solution 3

As stated before, note that $\triangle ACB ~ \triangle ADE$ . By similarity, we note that $\frac{\overline{AC}}{\overline{BC}}$ is equivalent to $\frac{7}{5}$ . We set $\overline{AC}$ to $7x$ and $\overline{BC}$ to $5x$ . By the Pythagorean Theorem, $(7x)^2+(5x)^2 = 4^2$ . Combining, $49x^2+25x^2=16$ . We can add and divide to get $x^2=\frac{8}{37}$ . We square root and rearrange to get $x=\frac{2\sqrt{74}}{37}$ . We know that the legs of the triangle are $7x$ and $5x$ . Mulitplying $x$ by $7$ and $5$ eventually gives us $\frac {14\sqrt{74}}{37}$ $\frac {10\sqrt{74}}{37}$ . We divide this by 2, since $\frac{1}{2}bh$ is the formula for a triangle. This gives us $\boxed{\textbf{(D) } \frac{140}{37}}$ .

Solution 4

Let's call the center of the circle that segment $AB$ is the diameter of, $O$ . Note that $\triangle ODE$ is an isosceles right triangle. Solving for side $OE$ , using the Pythagorean theorem, we find it to be $5\sqrt{2}$ . Calling the point where segment $OE$ intersects circle $O$ , the point $I$ , segment $IE$ would be $5\sqrt{2}-2$ . Also, noting that $\triangle ADE$ is a right triangle, we solve for side $AE$ , using the Pythagorean Theorem, and get $\sqrt{74}$ . Using Power of Point on point $E$ , we can solve for $CE$ . We can subtract $CE$ from $AE$ to find $AC$ and then solve for $CB$ using Pythagorean theorem once more.

$(AE)(CE)$ = (Diameter of circle $O$ + $IE$ ) $(IE)$ $\rightarrow$ ${\sqrt{74}}(CE)$ = $(5\sqrt{2}+2)(5\sqrt{2}-2)$ $\Rightarrow$ $CE$ = $\frac{23\sqrt{74}}{37}$

$AC = AE - CE$ $\rightarrow$ $AC$ = ${\sqrt74}$ - $\frac{23\sqrt{74}}{37}$ $\Rightarrow$ $AC$ = $\frac{14\sqrt{74}}{37}$

Now to solve for $CB$ :

$AB^2$ - $AC^2$ = $CB^2$ $\rightarrow$ $4^2$ + $\frac{14\sqrt{74}}{37}^2$ = $CB^2$ $\Rightarrow$ $CB$ = $amc数学竞赛真题英语$

Note that $\triangle ABC$ is a right triangle because the hypotenuse is the diameter of the circle. Solving for area using the bases $AC$ and $BC$ , we get the area of triangle $ABC$ to be $\boxed{\textbf{(D) } \frac{140}{37}}$ .