Difference between revisions of "2023 AIME II Problems/Problem 3"
MRENTHUSIASM (talk | contribs) (→Solution 1) |
MRENTHUSIASM (talk | contribs) (→Solution 1) |
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Moreover, we have <math>\angle PBA = \angle PCB = 45^\circ-\theta,</math> as shown below: | Moreover, we have <math>\angle PBA = \angle PCB = 45^\circ-\theta,</math> as shown below: | ||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(250); | ||
+ | pair A, B, C, P; | ||
+ | |||
+ | A = origin; | ||
+ | B = (0,10*sqrt(5)); | ||
+ | C = (10*sqrt(5),0); | ||
+ | P = intersectionpoints(Circle(A,10),Circle(C,20))[0]; | ||
+ | |||
+ | dot("$A$",A,1.5*SW,linewidth(4)); | ||
+ | dot("$B$",B,1.5*NW,linewidth(4)); | ||
+ | dot("$C$",C,1.5*SE,linewidth(4)); | ||
+ | dot("$P$",P,1.5*NE,linewidth(4)); | ||
+ | |||
+ | markscalefactor=0.125; | ||
+ | draw(rightanglemark(B,A,C,10),red); | ||
+ | draw(anglemark(P,A,B,25),red); | ||
+ | draw(anglemark(P,B,C,25),red); | ||
+ | draw(anglemark(P,C,A,25),red); | ||
+ | draw(anglemark(A,B,P,25),green); | ||
+ | draw(anglemark(B,C,P,25),green); | ||
+ | add(pathticks(anglemark(P,A,B,25), n = 1, r = 0.1, s = 10, red)); | ||
+ | add(pathticks(anglemark(P,B,C,25), n = 1, r = 0.1, s = 10, red)); | ||
+ | add(pathticks(anglemark(P,C,A,25), n = 1, r = 0.1, s = 10, red)); | ||
+ | |||
+ | draw(A--B--C--cycle^^P--A^^P--B^^P--C); | ||
+ | label("$10$",midpoint(A--P),dir(-30),blue); | ||
+ | label("$x$",midpoint(A--B),1.5*W,blue); | ||
+ | label("$x$",midpoint(A--C),1.5*S,blue); | ||
+ | label("$x\sqrt2$",midpoint(B--C),1.5*NE,blue); | ||
+ | label("$\sqrt{x^2-100}$",midpoint(P--C),dir(225),blue); | ||
+ | label("$\theta$",A,9*dir(76),red); | ||
+ | label("$\theta$",C,9*dir(168),red); | ||
+ | label("$\theta$",B,9*dir(305),red); | ||
+ | label("$45^\circ-\theta$",B,6*dir(235),green); | ||
+ | label("$45^\circ-\theta$",C,6*dir(85),green); | ||
+ | </asy> | ||
Note that <math>\triangle PAB \sim \triangle PBC</math> by the AA Similarity. The ratio of similitude is <math>\frac{PA}{PB} = \frac{PB}{PC} = \frac{AB}{BC},</math> or <cmath>\frac{10}{PB} = \frac{PB}{\sqrt{x^2-100}} = \frac{x}{x\sqrt2} = \frac{1}{\sqrt2}.</cmath> | Note that <math>\triangle PAB \sim \triangle PBC</math> by the AA Similarity. The ratio of similitude is <math>\frac{PA}{PB} = \frac{PB}{PC} = \frac{AB}{BC},</math> or <cmath>\frac{10}{PB} = \frac{PB}{\sqrt{x^2-100}} = \frac{x}{x\sqrt2} = \frac{1}{\sqrt2}.</cmath> | ||
From <math>\frac{10}{PB} = \frac{1}{\sqrt2},</math> we get <math>PB=10\sqrt2.</math> It follows that from <math>\frac{10}{PB} = \frac{PB}{\sqrt{x^2-100}},</math> we get <math>x^2=500.</math> | From <math>\frac{10}{PB} = \frac{1}{\sqrt2},</math> we get <math>PB=10\sqrt2.</math> It follows that from <math>\frac{10}{PB} = \frac{PB}{\sqrt{x^2-100}},</math> we get <math>x^2=500.</math> |
Revision as of 15:27, 17 February 2023
Problem
Let be an isosceles triangle with
There exists a point
inside
such that
and
Find the area of
Diagram
~MRENTHUSIASM
Solution 1
This solution refers to the Diagram section.
Let and
from which
and
By the Pythagorean Theorem on right
we have
Moreover, we have as shown below:
Note that
by the AA Similarity. The ratio of similitude is
or
From
we get
It follows that from
we get
Finally, the area of is
~s214425
~MRENTHUSIASM
Solution 2
Since the triangle is a right isosceles triangle, .
Let the common angle be . Note that
, thus
. From there, we know that
.
Note that , so from law of sines we have
Dividing by
and multiplying across yields
From here use the sine subtraction formula, and solve for
:
Substitute this to find that
, thus the area is
.
~SAHANWIJETUNGA
Solution 3
Since the triangle is a right isosceles triangle, .
Do some angle chasing yielding:
We have since
is a right triangle. Since
is a
-
-
triangle,
, and
.
Note that by a factor of
. Thus,
, and
.
From Pythagorean theorem, so the area of
is
.
~SAHANWIJETUNGA
Solution 4
Since the triangle is a right isosceles triangle, .
Notice that in triangle ,
, so
. Similar logic shows
.
Now, we see that with ratio
(as
is a
-
-
triangle). Hence,
. We use the Law of Cosines to find
.
Since
is a right triangle, the area is
.
~Kiran
See also
2023 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.