Difference between revisions of "2001 AMC 12 Problems/Problem 22"
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</math> | </math> | ||
− | + | == Solution 1 == | |
− | |||
− | == Solution == | ||
− | |||
<asy> | <asy> | ||
unitsize(0.5cm); | unitsize(0.5cm); | ||
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draw(H--D, dashed); | draw(H--D, dashed); | ||
</asy> | </asy> | ||
− | |||
− | |||
− | |||
− | |||
Note that the triangles <math>AFH</math> and <math>CEH</math> are similar, as they have the same angles. Hence <math>\frac {AH}{HC} = \frac{AF}{EC} = \frac 23</math>. | Note that the triangles <math>AFH</math> and <math>CEH</math> are similar, as they have the same angles. Hence <math>\frac {AH}{HC} = \frac{AF}{EC} = \frac 23</math>. | ||
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Therefore <math>[EHJ]=[ACD]-[AHD]-[DEH]-[EJC]=35-14-\frac {21}2-\frac{15}2 = \boxed{3}</math>. | Therefore <math>[EHJ]=[ACD]-[AHD]-[DEH]-[EJC]=35-14-\frac {21}2-\frac{15}2 = \boxed{3}</math>. | ||
− | + | == Solution 2 == | |
As in the previous solution, we note the similar triangles and prove that <math>H</math> is in <math>2/5</math> and <math>J</math> in <math>4/7</math> of <math>AC</math>. | As in the previous solution, we note the similar triangles and prove that <math>H</math> is in <math>2/5</math> and <math>J</math> in <math>4/7</math> of <math>AC</math>. | ||
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As <math>E</math> is the midpoint of <math>CD</math>, the height from <math>E</math> onto <math>AC</math> is <math>1/2</math> of the height from <math>D</math> onto <math>AC</math>. Therefore we have <math>[EHJ] = \frac{6}{35} \cdot \frac 12 \cdot [ACD] = \frac 3{35} \cdot 35 = \boxed{3}</math>. | As <math>E</math> is the midpoint of <math>CD</math>, the height from <math>E</math> onto <math>AC</math> is <math>1/2</math> of the height from <math>D</math> onto <math>AC</math>. Therefore we have <math>[EHJ] = \frac{6}{35} \cdot \frac 12 \cdot [ACD] = \frac 3{35} \cdot 35 = \boxed{3}</math>. | ||
− | + | == Solution 3 == | |
Because we see that there are only lines and there is a rectangle, we can coordbash (place this figure on coordinates). Because this is a general figure, we can assume the sides are <math>7</math> and <math>10</math> (or any other two positive real numbers that multiply to 70). We can find <math>H</math> and <math>J</math> by intersecting lines, and then we calculate the area of <math>EHJ</math> using shoelace formula. This yields <math>\boxed{3}</math>. | Because we see that there are only lines and there is a rectangle, we can coordbash (place this figure on coordinates). Because this is a general figure, we can assume the sides are <math>7</math> and <math>10</math> (or any other two positive real numbers that multiply to 70). We can find <math>H</math> and <math>J</math> by intersecting lines, and then we calculate the area of <math>EHJ</math> using shoelace formula. This yields <math>\boxed{3}</math>. | ||
− | + | == Solution 4 == | |
Note that triangle <math>AFH</math> is similar to triangle <math>CEH</math> with ratio <math>\frac{2}{3}</math>. Similarly, triangle <math>AGJ</math> is similar to triangle <math>ECJ</math> with ratio <math>\frac{4}{3}</math>. Thus, if <math>AC = a</math> then we know that <math>AH = \frac{2}{5}a</math> and <math>JC = \frac{3}{7}a</math> meaning <math>HJ = \frac{6}{35}a</math> and thus the ratio of <math>HJ</math> to <math>JC</math> is <math>\frac{\frac{6}{35}}{\frac{3}{7}} = \frac{2}{5}</math> which equals the ratio of the areas of <math>HJE</math> to <math>JEC</math>. If <math>y = AD, x = DC</math>, then we know that <math>JEC = \text{(altitude from J to EC)} \cdot EC = \frac{3}{7}y \cdot \frac{1}{2}x \cdot \frac{1}{2}</math> and since <math>xy = 70</math> and we want to find <math>\frac{2}{5}</math> of this, we get our answer is <math>\frac{2}{5} \cdot \frac{3}{7} \cdot \frac{1}{2} \cdot 70 \cdot \frac{1}{2} = \boxed{3}</math>. -SuperJJ | Note that triangle <math>AFH</math> is similar to triangle <math>CEH</math> with ratio <math>\frac{2}{3}</math>. Similarly, triangle <math>AGJ</math> is similar to triangle <math>ECJ</math> with ratio <math>\frac{4}{3}</math>. Thus, if <math>AC = a</math> then we know that <math>AH = \frac{2}{5}a</math> and <math>JC = \frac{3}{7}a</math> meaning <math>HJ = \frac{6}{35}a</math> and thus the ratio of <math>HJ</math> to <math>JC</math> is <math>\frac{\frac{6}{35}}{\frac{3}{7}} = \frac{2}{5}</math> which equals the ratio of the areas of <math>HJE</math> to <math>JEC</math>. If <math>y = AD, x = DC</math>, then we know that <math>JEC = \text{(altitude from J to EC)} \cdot EC = \frac{3}{7}y \cdot \frac{1}{2}x \cdot \frac{1}{2}</math> and since <math>xy = 70</math> and we want to find <math>\frac{2}{5}</math> of this, we get our answer is <math>\frac{2}{5} \cdot \frac{3}{7} \cdot \frac{1}{2} \cdot 70 \cdot \frac{1}{2} = \boxed{3}</math>. -SuperJJ | ||
− | == See | + | == Solution 5 == |
+ | |||
+ | <math>[CEF] = \frac{[ABCD]}{4} = \frac{35}{2}</math> | ||
+ | |||
+ | <math>\triangle CEH \sim \triangle AFH</math>, <math>\frac{HE}{HF} = \frac{CE}{AF} = \frac{3}{2}</math>, <math>\frac{HE}{EF} = \frac{3}{5}</math> | ||
+ | |||
+ | <math>[CEH] = \frac{HE}{EF} \cdot [CEF] = \frac{3}{5} \cdot \frac{35}{2} = \frac{21}{2}</math> | ||
+ | |||
+ | <math>\triangle CEH \sim \triangle AFH</math>, <math>\frac{AH}{HC} = \frac{AF}{CE} = \frac{2}{3}</math>, <math>\frac{AH}{AC} = \frac{2}{5}</math>, <math>\frac{CH}{AC} = \frac{3}{5}</math> | ||
+ | |||
+ | <math>\triangle CEJ \sim \triangle AGJ</math>, <math>\frac{AJ}{JC} = \frac{AG}{CE} = \frac{4}{3}</math>, <math>\frac{AJ}{AC} = \frac{4}{7}</math> | ||
+ | |||
+ | <math>\frac{HJ}{AC} = \frac{AJ}{AC} - \frac{AH}{AC} = \frac{4}{7} - \frac{2}{5} = \frac{6}{35}</math> | ||
+ | |||
+ | <math>\frac{HJ}{CH} = \frac{HJ}{AC} \cdot \frac{AC}{CH} = \frac{6}{35} \cdot \frac{5}{3} = \frac{2}{7}</math> | ||
+ | |||
+ | <math>[EHJ] = \frac{HJ}{CH} \cdot [CEH] = \frac{2}{7} \cdot \frac{21}{2} = \boxed{\textbf{(C) }3}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | == Solution 6 (Mass Points) == | ||
+ | |||
+ | We need one more pair of ratios to fully define our mass point system. Let's use <math>\triangle AFH \sim \triangle CEH\implies EH:HF = 3:2</math> and now do mass points on <math>\triangle AEG</math>: | ||
+ | |||
+ | <asy> | ||
+ | size(250); | ||
+ | pair A = (0,0), B = (10,0), C = (10,7), D = (0,7); | ||
+ | pair F = (10/3,0), G = (20/3,0), E = (5,7); | ||
+ | pair H = intersectionpoint(A--C, E--F); | ||
+ | pair J = intersectionpoint(A--C, E--G); | ||
+ | filldraw(A--E--G--cycle, rgb(1,1,1)+opacity(0.3), red+2bp); | ||
+ | draw(A--B--C--D--cycle); | ||
+ | draw(A--C); | ||
+ | draw(E--F); | ||
+ | draw(E--G); | ||
+ | draw(A--E, dashed); | ||
+ | draw(E--B, dashed); | ||
+ | dot("$A$", A, SW); | ||
+ | dot("$B$", B, SE); | ||
+ | dot("$C$", C, NE); | ||
+ | dot("$D$", D, NW); | ||
+ | dot("$E$", E, N); | ||
+ | dot("$F$", F, S); | ||
+ | dot("$G$", G, S); | ||
+ | dot("$H$", H, ESE); | ||
+ | dot("$J$", J, W); | ||
+ | |||
+ | // mass point labels | ||
+ | pair mass = A + SW; | ||
+ | label(scale(0.8)*"$3$", mass, UnFill); | ||
+ | draw(circle(mass, .4), linewidth(1)); | ||
+ | pair mass = F + S; | ||
+ | label(scale(0.8)*"$6$", mass, UnFill); | ||
+ | draw(circle(mass, .4), linewidth(1)); | ||
+ | pair mass = G + SE; | ||
+ | label(scale(0.8)*"$3$", mass, UnFill); | ||
+ | draw(circle(mass, .4), linewidth(1)); | ||
+ | pair mass = J + .7*ESE; | ||
+ | label(scale(0.8)*"$7$", mass, UnFill); | ||
+ | draw(circle(mass, .4), linewidth(1)); | ||
+ | pair mass = E + N; | ||
+ | label(scale(0.8)*"$4$", mass, UnFill); | ||
+ | draw(circle(mass, .4), linewidth(1)); | ||
+ | pair mass = H + .7*WNW; | ||
+ | label(scale(0.8)*"$10$", mass, UnFill); | ||
+ | draw(circle(mass, .4), linewidth(1)); | ||
+ | </asy> | ||
+ | |||
+ | Now it's just a standard "Area Reduction by Ratios"™ problem going from: | ||
+ | |||
+ | <cmath>[ABCD]\xrightarrow[]{\frac{1}{2}}[AEB]\xrightarrow[]{\frac{2}{3}}[AEG]\xrightarrow[]{\frac{3}{7}}[AEJ]\xrightarrow[]{\frac{3}{10}}[HEJ]</cmath> | ||
+ | |||
+ | or, | ||
+ | |||
+ | <cmath>70 \cdot \frac{1}{2}\cdot \frac{2}{3}\cdot \frac{3}{7}\cdot \frac{3}{10} = \boxed{\textbf{(C) }3}</cmath> | ||
+ | |||
+ | ~ proloto | ||
+ | |||
+ | ==See also== | ||
{{AMC12 box|year=2001|num-b=21|num-a=23}} | {{AMC12 box|year=2001|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 02:27, 29 August 2023
Contents
Problem
In rectangle , points
and
lie on
so that
and
is the midpoint of
. Also,
intersects
at
and
at
. The area of the rectangle
is
. Find the area of triangle
.
Solution 1
Note that the triangles
and
are similar, as they have the same angles. Hence
.
Also, triangles and
are similar, hence
.
We can now compute as
. We have:
.
is
of
, as these two triangles have the same base
, and
is
of
, therefore also the height from
onto
is
of the height from
. Hence
.
is
of
, as the base
is
of the base
, and the height from
is
of the height from
. Hence
.
is
of
for similar reasons, hence
.
Therefore .
Solution 2
As in the previous solution, we note the similar triangles and prove that is in
and
in
of
.
We can then compute that .
As is the midpoint of
, the height from
onto
is
of the height from
onto
. Therefore we have
.
Solution 3
Because we see that there are only lines and there is a rectangle, we can coordbash (place this figure on coordinates). Because this is a general figure, we can assume the sides are and
(or any other two positive real numbers that multiply to 70). We can find
and
by intersecting lines, and then we calculate the area of
using shoelace formula. This yields
.
Solution 4
Note that triangle is similar to triangle
with ratio
. Similarly, triangle
is similar to triangle
with ratio
. Thus, if
then we know that
and
meaning
and thus the ratio of
to
is
which equals the ratio of the areas of
to
. If
, then we know that
and since
and we want to find
of this, we get our answer is
. -SuperJJ
Solution 5
,
,
,
,
,
,
,
Solution 6 (Mass Points)
We need one more pair of ratios to fully define our mass point system. Let's use and now do mass points on
:
Now it's just a standard "Area Reduction by Ratios"™ problem going from:
or,
~ proloto
See also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.