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您所在的位置: 首頁  |  AMC  |  AMC10每日一題(2000年真題#16)

AMC10每日一題(2000年真題#16)

宋老師
6年從業經驗

免費留學咨詢表 (趣考不是留學中介,能給你客觀建議?。?/font>

聯系老師

2000年AMC 10競賽試題/問題16

The problem

The diagram shows

lattice points, each one unit from its nearest neighbors. Segment

meets segment

in

. Find the length of segment

。

The solution

Solution 1

let

be the line containing

and

And let

be the line containing

and

. If we set the bottom left point at

And then

,

,

,

。

This line

is given by the equation

. the

-intercept is

, so

. We are given two points on

, hence we can compute the slope,

is

, so

Is the line

In the same way,

Is made up of

. The slope in this case is

, so

. Plugging in the point

Give us the

, so

Is the line

。

in

, the intersection point, both of the equations must be true, so

We have the coordinates of

and

, so we can use the distance formula here:

which is answer choice

Solution 2

Draw the perpendiculars from

and

to

, respectively. As it turns out,

. let

be the point on

the

。

,

, so by AA similarity,

By the Pythagorean Theorem, we have

,

,

. let

, so

And then

This is answer choice

Also, you could extend CD to the end of the box and create two similar triangles. Then use ratios and find that the distance is 5/9 of the diagonal AB. Thus, the answer is B.

Solution 3

Drawing line

and parallel line

We see

by AA similarity. Thus

. Reciprocating, we know that

so

. Reciprocating again, we have

. We know that

, so by the pythagorean theorem,

. so

. Applying the pythagorean theorem again, we have

. We finally have

AMC真題AMC歷年真題AMCAMC10

2019-07-30 15:37:07

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