Difference between revisions of "2020 AIME II Problems/Problem 13"
(→Solutions) |
(→Solutions (Misplaced problem?)) |
||
Line 3: | Line 3: | ||
==Solutions (Misplaced problem?)== | ==Solutions (Misplaced problem?)== | ||
− | Assume the incircle touches <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math>, <math>EA</math> at <math>P,Q,R,S,T</math> respectively. Then let <math>PB=x=BQ=RD=SD</math>, <math>ET=y=ES=CR=CQ</math>, <math>AP=AT=z</math>. So we have <math>x+y=6</math>, <math>x+z=5</math> and <math>y+z</math>=7, solve it we have <math>x=2</math>, <math>z=3</math>, <math>y=4</math>. Let the center of the incircle be <math>I</math>, by SAS we can proof triangle <math>BIQ</math> is congruent to triangle <math>DIS</math>, and triangle <math>CIR</math> is congruent to triangle <math>SIE</math>. Then we have <math>\angle AED=\angle BCD</math>, <math>\angle ABC=\angle CDE</math>. Extend <math>CD</math>, cross ray <math>AB</math> at <math>M</math>, ray <math>AE</math> at <math>N</math>, then by AAS we have triangle <math>END</math> is congruent to triangle <math>BMC</math>. Thus <math>\angle M=\angle N</math>. Let <math>EN=MC=a</math>, then <math>BM=DN=a+2</math>. So by low of cosine in triangle <math>END</math> and triangle <math>ANM</math> we can obtain < | + | Assume the incircle touches <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math>, <math>EA</math> at <math>P,Q,R,S,T</math> respectively. Then let <math>PB=x=BQ=RD=SD</math>, <math>ET=y=ES=CR=CQ</math>, <math>AP=AT=z</math>. So we have <math>x+y=6</math>, <math>x+z=5</math> and <math>y+z</math>=7, solve it we have <math>x=2</math>, <math>z=3</math>, <math>y=4</math>. Let the center of the incircle be <math>I</math>, by SAS we can proof triangle <math>BIQ</math> is congruent to triangle <math>DIS</math>, and triangle <math>CIR</math> is congruent to triangle <math>SIE</math>. Then we have <math>\angle AED=\angle BCD</math>, <math>\angle ABC=\angle CDE</math>. Extend <math>CD</math>, cross ray <math>AB</math> at <math>M</math>, ray <math>AE</math> at <math>N</math>, then by AAS we have triangle <math>END</math> is congruent to triangle <math>BMC</math>. Thus <math>\angle M=\angle N</math>. Let <math>EN=MC=a</math>, then <math>BM=DN=a+2</math>. So by low of cosine in triangle <math>END</math> and triangle <math>ANM</math> we can obtain <cmath>\frac{2a+8}{2(a+7)}=\cos N=\frac{a^2+(a+2)^2-36}{2a(a+2)}</cmath>, solved it gives us <math>a=8</math>, which yield triangle <math>ANM</math> to be a triangle with side length 15, 15, 24, draw a height from <math>A</math> to <math>NM</math> divides it into two triangles with side lengths 9, 12, 15, so the area of triangle <math>ANM</math> is 108. Triangle <math>END</math> is a triangle with side lengths 6, 8, 10, so the area of two of them is 48, so the area of pentagon is <math>108-48=\boxed{60}</math>. |
-Fanyuchen20020715 | -Fanyuchen20020715 |
Revision as of 10:34, 8 June 2020
Problem
Convex pentagon has side lengths
,
, and
. Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of
.
Solutions (Misplaced problem?)
Assume the incircle touches ,
,
,
,
at
respectively. Then let
,
,
. So we have
,
and
=7, solve it we have
,
,
. Let the center of the incircle be
, by SAS we can proof triangle
is congruent to triangle
, and triangle
is congruent to triangle
. Then we have
,
. Extend
, cross ray
at
, ray
at
, then by AAS we have triangle
is congruent to triangle
. Thus
. Let
, then
. So by low of cosine in triangle
and triangle
we can obtain
, solved it gives us
, which yield triangle
to be a triangle with side length 15, 15, 24, draw a height from
to
divides it into two triangles with side lengths 9, 12, 15, so the area of triangle
is 108. Triangle
is a triangle with side lengths 6, 8, 10, so the area of two of them is 48, so the area of pentagon is
.
-Fanyuchen20020715
Video Solution
https://youtu.be/bz5N-jI2e0U?t=327
2020 AIME II (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.