Difference between revisions of "2014 AIME I Problems/Problem 13"
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dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); dot("$I$",I,dir(90)); dot("$J$",J,dir(270)); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); dot("$O$",O,dir(-30)); dot("$K$",K,dir(-180)); pair P = extension(F,H,E,G); dot("$P$",P,dir(180+60)); | dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); dot("$I$",I,dir(90)); dot("$J$",J,dir(270)); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); dot("$O$",O,dir(-30)); dot("$K$",K,dir(-180)); pair P = extension(F,H,E,G); dot("$P$",P,dir(180+60)); | ||
</asy> | </asy> | ||
− | Let the area of the square be <math>1360a</math>. Then | + | Let the area of the square be <math>1360a</math>. Then <math>[HPOI]=71a</math> and <math>[FPOJ]=65a</math>. This is because <math>\overline{HF}</math> is perpendicular to <math>\overline{EG}</math> (given in the problem), so <math>\overline{IJ}</math> is also perpendicular to <math>\overline{EG}</math>. These two orthogonal lines also pass through the center of the square, so they split it into 4 congruent quadrilaterals. |
− | |||
Let the side length of the square be <math>d=\sqrt{1360a}</math>. | Let the side length of the square be <math>d=\sqrt{1360a}</math>. | ||
− | Draw <math>\overline{OK}\parallel \overline{HI}</math> and intersects <math>\overline{HF}</math> at <math>K</math>. < | + | Draw <math>\overline{OK}\parallel \overline{HI}</math> and intersects <math>\overline{HF}</math> at <math>K</math>. Then <cmath>OK=d\cdot\tfrac{[HFJI]}{[ABCD]}=\frac{1}{10}d.</cmath> |
− | + | Then <math>[HKOI]=\tfrac12\cdot [HFJI]=68a</math>, so <math>[POK]=3a</math>. | |
− | |||
− | |||
Let <math>\overline{PO}=h</math>. Then <math>KP=\frac{6a}{h}</math> | Let <math>\overline{PO}=h</math>. Then <math>KP=\frac{6a}{h}</math> | ||
Revision as of 14:03, 31 December 2021
Problem 13
On square , points
, and
lie on sides
and
respectively, so that
and
. Segments
and
intersect at a point
, and the areas of the quadrilaterals
and
are in the ratio
Find the area of square
.
Solution
Notice that . This means
passes through the center of the square.
Draw with
on
,
on
such that
and
intersects at the center of the square which I'll label as
.
Let the area of the square be
. Then
and
. This is because
is perpendicular to
(given in the problem), so
is also perpendicular to
. These two orthogonal lines also pass through the center of the square, so they split it into 4 congruent quadrilaterals.
Let the side length of the square be
.
Draw and intersects
at
. Then
Then
, so
.
Let
. Then
Consider the area of .
Thus, . Now we solve
to get
or
.
The former leads to a square with diagonal less than , which can't be, since
; therefore
and the area of
Lazy Solution
, a multiple of
. In addition,
, which is
.
Therefore, we suspect the square of the "hypotenuse" of a right triangle, corresponding to
and
must be a multiple of
. All of these triples are primitive:
The sides of the square can only equal the longer leg, or else the lines would have to extend outside of the square. Substituting :
Thus, is the only valid answer.
Solution 3
Continue in the same way as solution 1 to get that has area
, and
. You can then find
has length
.
Then, if we drop a perpendicular from to
at
, We get
.
Thus, , and we know
, and
. Thus, we can set up an equation in terms of
using the Pythagorean theorem.
is extraneous, so
. Since the area is
, we have it is equal to
-Alexlikemath
See also
2014 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.