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AMC数学竞赛真题2016年10A 24

2018-11-06 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 24

A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?

$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$

Solution 1 (Algebra)

To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by $200$ for now, then multiply it back at the end of our solution.

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Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$ . Let the intersection of $BD$ and $OC$ be point $E$ . Notice that $BD$ and $OC$ are perpendicular because $BCDO$ is a kite.

We set lengths $BE=ED$ equal to $x$ . By the Pythagorean Theorem,

$\sqrt{1^2-x^2}+\sqrt{(\sqrt{2})^2-x^2}=\sqrt{2}$

We solve for $x$ :

$1-x^2+2-x^2+2\sqrt{(1-x^2)(2-x^2)}=2$

$2\sqrt{(1-x^2)(2-x^2)}=2x^2-1$

$4(1-x^2)(2-x^2)=(2x^2-1)^2$

$8-12x^2+4x^4=4x^4-4x^2+1$

$8x^2=7$

$x=\frac{\sqrt{14}}{4}$

By Ptolemy's Theorem,

$AB \cdot CD + BC \cdot AD = AC \cdot BD = BD^2 = (2 \cdot BE)^2$

Substituting values,

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$1+AD=\frac{7}{2}$

$AD=\frac{5}{2}$

Finally, we multiply back the $200$ that we divided by at the beginning of the problem to get $AD=\boxed{500 (E)}$ .

Solution 2 (HARD Algebra)

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Let quadrilateral $ABCD$ be inscribed in circle $O$ , where $AD$ is the side of unknown length. Draw the radii from center $O$ to all four vertices of the quadrilateral, and draw the altitude of $\triangle BOC$ such that it passes through side $AD$ at the point $G$ and meets side $BC$ at the point $H$ .

By the Pythagorean Theorem, the length of $OH$ is

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Note that $[ABCDO] = [AOB] + [BOC] + [COD] = [AOD] + [ABCD].$ Let the length of $OG$ be $h$ and the length of $AD$ be $x$ ; then we have that

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Furthermore,

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Substituting this value of $h$ into the previous equation and evaluating for $x$ , we get:

$\frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2}$

$\frac{3 \times 200 \times 100\sqrt{7}}{2} = \frac{x\sqrt{80000 - \frac{x^2}{4}}}{2} + \frac{\left(100\sqrt{7} - \sqrt{80000 - \frac{x^2}{4}}\right)(200 + x)}{2}$

$60000\sqrt{7} = \left(x\sqrt{80000 - \frac{x^2}{4}}\right) + \left(20000\sqrt{7}\right) + \left(100x\sqrt{7}\right) - \left(200\sqrt{80000 - \frac{x^2}{4}}\right) - \left(x\sqrt{80000 - \frac{x^2}{4}}\right)$

$40000\sqrt{7} = 100x\sqrt{7} - 200\sqrt{80000 - \frac{x^2}{4}}$

$400\sqrt{7} = x\sqrt{7} - 2\sqrt{80000 - \frac{x^2}{4}}$

$(x - 400)\sqrt{7} = 2\sqrt{80000 - \frac{x^2}{4}}$

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$7x^2 - 5600x + 1120000 = 320000 - x^2$

$8x^2 - 5600x + 800000 = 0$

$x^2 - 700x + 100000 = 0$

The roots of this quadratic are found by using the quadratic formula:

$\begin{align*} x &= \frac{-(-700) \pm \sqrt{(-700)^2 - 4 \times 1 \times 100000}}{2 \times 1} \\ &= \frac{700 \pm \sqrt{490000 - 400000}}{2} \\ &= \frac{700}{2} \pm \frac{\sqrt{90000}}{2} \\ &= 350 \pm \frac{300}{2} \\ &= 200, 500 \end{align*}$

If the length of $AD$ is $200$ , then quadrilateral $ABCD$ would be a square and thus, the radius of the circle would be

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Which is a contradiction. Therefore, our answer is $\boxed{500}.$

Solution 3 (Trigonometry Bash)

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Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$ . Apply law of cosines on $\Delta BOC$ ; let $\theta = \angle BOC$ . We get the following equation:

$(BC)^{2}=(OB)^{2}+(OC)^{2}-2\cdot OB \cdot OC\cdot \cos\theta$

Substituting the values in, we get

$(200)^{2}=2\cdot (200)^{2}+ 2\cdot (200)^{2}- 2\cdot 2\cdot (200)^{2}\cdot \cos\theta$

Canceling out, we get

$\cos\theta=\frac{3}{4}$

Because $\angle AOB$ , $\angle BOC$ , and $\angle COD$ are congruent, $\angle AOD = 3\theta$ . To find the remaining side ( $AD$ ), we simply have to apply the law of cosines to $\Delta AOD$ . Now, to find $\cos 3\theta$ , we can derive a formula that only uses $\cos\theta$ :

$\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- (2\sin\theta \cos\theta) \cdot \sin \theta$

$\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-1+2\cos^{2}\theta)$

$\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta$

Plugging in $\cos\theta=\frac{3}{4}$ , we get $\cos 3\theta= -\frac{9}{16}$ . Now, applying law of cosines on triangle $OAD$ , we get

$(AD)^{2}= 2\cdot (200)^{2}+ 2\cdot (200)^{2}+2\cdot 200\sqrt2 \cdot 200\sqrt2 \cdot \frac{9}{16}$

$\Rightarrow 2\cdot (200)^{2} \cdot (1+1+ \frac{9}{8})=(200)^{2}\cdot \frac{25}{4}$

$AD=200 \cdot \frac{5}{2}=\boxed{500}$

Solution 4 (Easier Trigonometry)

$[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=foot(A,B,C); F=foot(D,B,C); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label($

Construct quadrilateral $ABCD$ on the circle $O$ with $AD$ being the desired side. Then, drop perpendiculars from $A$ and $D$ to the extended line of $\overline{BC}$ and let these points be $E$ and $F$ , respectively. Also, let $\theta = \angle BOC$ . From the Law of Cosines on $\triangle BOC$ , we have $\cos \theta = \frac{3}{4}$ .

Now, since $\triangle BOC$ is isosceles with $\overline{OB} \cong \overline{OC}$ , we have that $\angle BCO = \angle CBO = 90 - \frac{\theta}{2}$ . In addition, we know that $\overline{BC} \cong \overline{CD}$ as they are both equal to $200$ and $\overline{OB} \cong \overline{OC} \cong \overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\triangle OBC \cong \triangle OCD$ , so we have that $\angle OCD = \angle BCO = 90 - \frac{\theta}{2}$ , so $\angle DCF = \theta$ .

Thus, we have $\frac{FC}{DC} = \cos \theta = \frac{3}{4}$ , so $FC = 150$ . Similarly, $BE = 150$ , and $AD = 150 + 200 + 150 = \boxed{500}$ .

Solution 5 (Just Geometry)

$[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label($

Label AD intercept OB at E and OC at F.

$\overarc{AB}= \overarc{BC}= \overarc{CD}=\theta$

$\angle{BAD}=\frac{1}{2} \cdot \overarc{BCD}=\theta=\angle{AOB}$

From there, $\triangle{OAB} \sim \triangle{ABE}$ , thus:

$\frac{OA}{AB} = \frac{AB}{BE} = \frac{OB}{AE}$

$OA = OB$ because they are both radii of $\odot{O}$ . Since $\frac{OA}{AB} = \frac{OB}{AE}$ , we have that $AB = AE$ . Similarly, $CD = DF$ .

$OE = 100\sqrt{2} = \frac{OB}{2}$ and $EF=\frac{BC}{2}=100$ , so $AD=AE + EF + FD = 200 + 100 + 200 = \boxed{\textbf{(E) } 500}$

Solution 6 (Ptolemy's Theorem)

$[asy] pathpen = black; pointpen = black; size(6cm); draw(unitcircle); pair A = D($

Let $s = 200$ . Let $O$ be the center of the circle. Then $AC$ is twice the altitude of $\triangle OBC$ . Since $\triangle OBC$ is isosceles we can compute its area to be $s^2 \sqrt7/4$ , hence $CA = 2 \tfrac{2 \cdot s^2\sqrt7/4}{s\sqrt2} = s\sqrt{7/2}$ .

Now by Ptolemy's Theorem we have $CA^2 = s^2 + AD \cdot s \implies AD = (7/2-1)s.$ This gives us:

$\boxed{\textbf{(E) } 500.}$

Solution 7 (Trigonometry)

Since all three sides equal $200$ , they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths $100,100\sqrt{7},200\sqrt{2}$ by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is $\frac{100}{200\sqrt{2}}=\frac{\sqrt{2}}{4}$ . Similarly, the cosine is $\frac{100\sqrt{7}}{200\sqrt{2}}=\frac{\sqrt{14}}{4}$ . Since there are three sides, and since $\sin\theta=\sin\left(180-\theta\right)$ ,we seek to find $2r\sin 3\theta$ . First, $\sin 2\theta=2\sin\theta\cos\theta=2\cdot\left(\frac{\sqrt{2}}{4}\right)\left(\frac{\sqrt{14}}{4}\right)=\frac{2\sqrt{2}\sqrt{14}}{16}=\frac{\sqrt{7}}{4}$ and $\cos 2\theta=\frac{3}{4}$ by Pythagorean.

$\sin 3\theta=\sin(2\theta+\theta)=\sin 2\theta\cos\theta+\sin \theta\cos 2\theta=\frac{\sqrt{7}}{4}\left(\frac{\sqrt{14}}{4}\right)+\frac{\sqrt{2}}{4}\left(\frac{3}{4}\right)=\frac{7\sqrt{2}+3\sqrt{2}}{16}=\frac{5\sqrt{2}}{8}$

$2r\sin 3\theta=2\left(200\sqrt{2}\right)\left(\frac{5\sqrt{2}}{8}\right)=400\sqrt{2}\left(\frac{5\sqrt{2}}{8}\right)=\frac{800\cdot 5}{8}=\boxed{\textbf{(E)}\text{ 500}}$

Solution 8 (Area By Brahmagupta's Formula)

For simplicity, scale everything down by a factor of 100. Let the inscribed trapezoid be $ABCD$ , where $AB=BC=CD=2$ and $DA$ is the missing side length. Let $DA=2x$ . If $M$ and $N$ are the midpoints of $BC$ and $AD$ , respectively, the height of the trapezoid is $OM-ON$ . By the pythagorean theorem, $OM=\sqrt{OB^2-BM^2}=\sqrt7$ and $ON=\sqrt{OA^2-AN^2}=\sqrt{8-x^2}$ . Thus the height of the trapezoid is $\sqrt7-\sqrt{8-x^2}$ , so the area is $\frac{(2+2x)(\sqrt7-\sqrt{8-x^2})}{2}=(x+1)(\sqrt7-\sqrt{8-x^2})$ . By Brahmagupta's formula, the area is $\sqrt{(x+1)(x+1)(x+1)(3-x)}$ . Setting these two equal, we get $(x+1)(\sqrt7-\sqrt{8-x^2})=\sqrt{(x+1)(x+1)(x+1)(3-x)}$ . Dividing both sides by $x+1$ and then squaring, we get $7-2(\sqrt7)(\sqrt{8-x^2})+8-x^2=(x+1)(3-x)$ . Expanding the right hand side and canceling the $x^2$ terms gives us $15-2(\sqrt7)(\sqrt{8-x^2})=2x+3$ . Rearranging and dividing by two, we get $(\sqrt7)(\sqrt{8-x^2})=6-x$ . Squaring both sides, we get $56-7x^2=x^2-12x+36$ . Rearranging, we get $8x^2-12x-20=0$ . Dividing by 4 we get $2x^2-3x-5=0$ . Factoring we get, $(2x-5)(x+1)=0$ , and since $x$ cannot be negative, we get $x=2.5$ . Since $DA=2x$ , $DA=5$ . Scaling up by 100, we get $\boxed{\textbf{(E)}\text{ 500}}$ .

Solution 9 (Cheap Solution - For when you are running out of time.)

WLOG, let $AB=BC=CD=200$ , and let ABCD be inscribed in a clrcle with radius $200\sqrt2$ . We draw perpendiculars from $B$ and $C$ to $AD$ , and label the intersections $E$ and $F$ , respectively. We can see that $EF=200$ (because BCFE is a rectangle), and since $AD$ is clearly greater than 200, and and since $EF$ , which is part of segment $AD$ , is an integer, than we conclude that $AD$ is also an integer or of the form $200+2*AE$ . There is no reason for $AE$ to be of the form $a\sqrt{b} - 100$ because it seems too arbitrary. The only other integer choice is $\boxed{\textbf{(E)}\text{ 500}}$ .

Solution 10 (Measuring)

A quick aside -- this problem takes about 90 seconds to solve with an accurate compass and a good ruler. Draw the diagram, to scale, and measure the last side. It becomes clear that the answer is $\boxed{\textbf{(E)}\text{ 500}}$ .

以上就是小编对AMC10数学竞赛真题以及解析的介绍，希望对你有所帮助，更多学习资料请持续关注AMC数学竞赛网。