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Difference between revisions of "2021 Fall AMC 10B Problems/Problem 25"

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(→‎Solution)
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Since the larger square by definition has all equal sides, we can set the sum of the lengths of the sides equal to each other.
 
Since the larger square by definition has all equal sides, we can set the sum of the lengths of the sides equal to each other.
−
<math>3a+b+b+a = 3b+a \implies 3a = b</math>
+
<math>3a+b+b+a = 3b+a \implies 3a = b</math>. Now let's draw some more perpendiculars and rename the side lengths.
 +
 
 +
<asy>
 +
size(8cm);
 +
draw((0,0)--(10,0));
 +
draw((0,0)--(0,10));
 +
draw((10,0)--(10,10));
 +
draw((0,10)--(10,10));
 +
draw((1,6)--(0,9));
 +
draw((1,6)--(0,6));
 +
draw((1,6)--(4,6));
 +
draw((4,6)--(4,10));
 +
draw((0,9)--(3,10));
 +
draw((3,10)--(4,7));
 +
draw((4,7)--(1,6));
 +
draw((0,3)--(1,6));
 +
draw((1,6)--(10,3));
 +
draw((10,3)--(9,0));
 +
draw((9,0)--(0,3));
 +
draw((6,13/3)--(10,22/3));
 +
draw((10,22/3)--(8,10));
 +
draw((8,10)--(4,7));
 +
draw((4,7)--(6,13/3));
 +
label("$3$",(9/2,3/2),N);
 +
label("$3$",(11/2,9/2),S);
 +
label("$1$",(1/2,9/2),E);
 +
label("$1$",(19/2,3/2),W);
 +
label("$1$",(1/2,15/2),E);
 +
label("$1$",(3/2,19/2),S);
 +
label("$1$",(5/2,13/2),N);
 +
label("$1$",(7/2,17/2),W);
 +
label("$R$",(7,43/6),W);
 +
label("$3a$",(-1/8,9/2),W);
 +
label("$3a$",(-1/8,7),W);
 +
label("$a$",(1/3,6),N);
 +
label("$a$",(-1/8,28/3),W);
 +
label("$3a$",(1,81/8),N);
 +
label("$3a$",(3,6),N);
 +
label("$a$",(401/100,13/2),W);
 +
label("$3a$",(401/100,17/2),E);
 +
label("$a$",(7/2,81/8),N);
 +
label("$3a$",(-1/8,1),W);
 +
label("$9a$",(4,-1/8),S);
 +
label("$a$",(19/2,-1/8),S);
 +
label("$3a$",(81/8,1),E);
 +
</asy>
  
 
Solution in progress
 
Solution in progress

Revision as of 13:25, 23 November 2021

Problem

A rectangle with side lengths $1{ }$ and $3,$ a square with side length $1,$ and a rectangle $R$ are inscribed inside a larger square as shown. The sum of all possible values for the area of $R$ can be written in the form $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. What is $m+n?$ [asy] size(8cm); draw((0,0)--(10,0)); draw((0,0)--(0,10)); draw((10,0)--(10,10)); draw((0,10)--(10,10)); draw((1,6)--(0,9)); draw((0,9)--(3,10)); draw((3,10)--(4,7)); draw((4,7)--(1,6)); draw((0,3)--(1,6)); draw((1,6)--(10,3)); draw((10,3)--(9,0)); draw((9,0)--(0,3)); draw((6,13/3)--(10,22/3)); draw((10,22/3)--(8,10)); draw((8,10)--(4,7)); draw((4,7)--(6,13/3)); label("$3$",(9/2,3/2),N); label("$3$",(11/2,9/2),S); label("$1$",(1/2,9/2),E); label("$1$",(19/2,3/2),W); label("$1$",(1/2,15/2),E); label("$1$",(3/2,19/2),S); label("$1$",(5/2,13/2),N); label("$1$",(7/2,17/2),W); label("$R$",(7,43/6),W); [/asy] $(\textbf{A})\: 14\qquad(\textbf{B}) \: 23\qquad(\textbf{C}) \: 46\qquad(\textbf{D}) \: 59\qquad(\textbf{E}) \: 67$


Solution

We see that the polygon bounded by the small square, large square, and rectangle of known lengths is an isosceles triangle. Let’s draw a perpendicular from the vertex of this triangle to its opposing side;

[asy] size(8cm); draw((0,0)--(10,0)); draw((0,0)--(0,10)); draw((10,0)--(10,10)); draw((0,10)--(10,10)); draw((1,6)--(0,9)); draw((1,6)--(0,6)); draw((1,6)--(4,6)); draw((4,6)--(4,10)); draw((0,9)--(3,10)); draw((3,10)--(4,7)); draw((4,7)--(1,6)); draw((0,3)--(1,6)); draw((1,6)--(10,3)); draw((10,3)--(9,0)); draw((9,0)--(0,3)); draw((6,13/3)--(10,22/3)); draw((10,22/3)--(8,10)); draw((8,10)--(4,7)); draw((4,7)--(6,13/3)); label("$3$",(9/2,3/2),N); label("$3$",(11/2,9/2),S); label("$1$",(1/2,9/2),E); label("$1$",(19/2,3/2),W); label("$1$",(1/2,15/2),E); label("$1$",(3/2,19/2),S); label("$1$",(5/2,13/2),N); label("$1$",(7/2,17/2),W); label("$R$",(7,43/6),W); label("$b$",(-1/8,9/2),W); label("$b$",(-1/8,7),W); label("$a$",(1/3,6),N); label("$a$",(-1/8,28/3),W); label("$b$",(1,81/8),N); label("$b$",(3,6),N); label("$a$",(401/100,13/2),W); label("$b$",(401/100,17/2),E); label("$a$",(7/2,81/8),N); [/asy]

We see that this creates two congruent triangles. Let the smaller side of the triangle have length $a$ and let the larger side of the triangle have length $b$. Now we see by AAS congruency that if we draw perpendiculars that surround the smaller square, each outer triangle will be congruent to these two triangles.

Now notice that these small triangles are also similar to the large triangle bounded by the bigger square and the rectangle by AA, and the ratio of the sides are 1:3, so we can fill in the lengths of that triangle. Similarly, the small triangle on the right bounded by the rectangle and the square is also congruent to the other small triangles by AAS, so we can fill in those sides;

[asy] size(8cm); draw((0,0)--(10,0)); draw((0,0)--(0,10)); draw((10,0)--(10,10)); draw((0,10)--(10,10)); draw((1,6)--(0,9)); draw((1,6)--(0,6)); draw((1,6)--(4,6)); draw((4,6)--(4,10)); draw((0,9)--(3,10)); draw((3,10)--(4,7)); draw((4,7)--(1,6)); draw((0,3)--(1,6)); draw((1,6)--(10,3)); draw((10,3)--(9,0)); draw((9,0)--(0,3)); draw((6,13/3)--(10,22/3)); draw((10,22/3)--(8,10)); draw((8,10)--(4,7)); draw((4,7)--(6,13/3)); label("$3$",(9/2,3/2),N); label("$3$",(11/2,9/2),S); label("$1$",(1/2,9/2),E); label("$1$",(19/2,3/2),W); label("$1$",(1/2,15/2),E); label("$1$",(3/2,19/2),S); label("$1$",(5/2,13/2),N); label("$1$",(7/2,17/2),W); label("$R$",(7,43/6),W); label("$b$",(-1/8,9/2),W); label("$b$",(-1/8,7),W); label("$a$",(1/3,6),N); label("$a$",(-1/8,28/3),W); label("$b$",(1,81/8),N); label("$b$",(3,6),N); label("$a$",(401/100,13/2),W); label("$b$",(401/100,17/2),E); label("$a$",(7/2,81/8),N); label("$3a$",(-1/8,1),W); label("$3b$",(4,-1/8),S); label("$a$",(19/2,-1/8),S); label("$b$",(81/8,1),E); [/asy]

Since the larger square by definition has all equal sides, we can set the sum of the lengths of the sides equal to each other. $3a+b+b+a = 3b+a \implies 3a = b$. Now let's draw some more perpendiculars and rename the side lengths.

[asy] size(8cm); draw((0,0)--(10,0)); draw((0,0)--(0,10)); draw((10,0)--(10,10)); draw((0,10)--(10,10)); draw((1,6)--(0,9)); draw((1,6)--(0,6)); draw((1,6)--(4,6)); draw((4,6)--(4,10)); draw((0,9)--(3,10)); draw((3,10)--(4,7)); draw((4,7)--(1,6)); draw((0,3)--(1,6)); draw((1,6)--(10,3)); draw((10,3)--(9,0)); draw((9,0)--(0,3)); draw((6,13/3)--(10,22/3)); draw((10,22/3)--(8,10)); draw((8,10)--(4,7)); draw((4,7)--(6,13/3)); label("$3$",(9/2,3/2),N); label("$3$",(11/2,9/2),S); label("$1$",(1/2,9/2),E); label("$1$",(19/2,3/2),W); label("$1$",(1/2,15/2),E); label("$1$",(3/2,19/2),S); label("$1$",(5/2,13/2),N); label("$1$",(7/2,17/2),W); label("$R$",(7,43/6),W); label("$3a$",(-1/8,9/2),W); label("$3a$",(-1/8,7),W); label("$a$",(1/3,6),N); label("$a$",(-1/8,28/3),W); label("$3a$",(1,81/8),N); label("$3a$",(3,6),N); label("$a$",(401/100,13/2),W); label("$3a$",(401/100,17/2),E); label("$a$",(7/2,81/8),N); label("$3a$",(-1/8,1),W); label("$9a$",(4,-1/8),S); label("$a$",(19/2,-1/8),S); label("$3a$",(81/8,1),E); [/asy]

Solution in progress ~KingRavi

Video Solution

https://www.youtube.com/watch?v=5mPvkipCvhE


See Also

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Problem 1
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 10 Problems and Solutions

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