Difference between revisions of "2007 IMO Problems/Problem 4"
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<math>(mathdummy)</math> | <math>(mathdummy)</math> | ||
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+ | ==Solution 3== | ||
+ | |||
+ | [img]https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYi85LzQ5OTJlZGNhYTQ0YjJjODcxMTBmZGNmMTdiZDdkMGRjZGUyOWQ5LnBuZw==&rn=U2NyZWVuIFNob3QgMjAxOS0wOC0wOCBhdCAxMi4yMC4zOCBQTS5wbmc=[/img] | ||
+ | |||
+ | WLOG, let the diameter of <math>(ACBD)</math> be <math>1.</math> | ||
+ | |||
+ | We see that <math>PK = \dfrac{1}{2}a \tan \dfrac{1}{2}C</math> and <math>QL = \dfrac{1}{2}b \tan \dfrac{1}{2}C</math> from right triangles <math>\triangle PKC</math> and <math>\triangle QLC.</math> | ||
+ | |||
+ | We now look at <math>AR.</math> By the Extended Law of Sines on <math>\triangle ACR,</math> we get that <math>AR = \sin\frac{1}{2}C.</math> Similarly, <math>BR = \sin \frac{1}{2}C.</math> | ||
+ | |||
+ | We now look at <math>CR.</math> By Ptolemy's Theorem, we have <cmath>AR \cdot BC + BR \cdot AC = AB \cdot CR,</cmath> which gives us <cmath>\sin \frac{1}{2}C (a + b) = c(CR).</cmath> This means that <cmath>CR = \dfrac{\sin \frac{1}{2}C (a + b)}{c}.</cmath> We now seek to relate the lengths computed with the areas. | ||
+ | |||
+ | To do this, we consider the altitude from <math>R</math> to <math>PK.</math> This is to find the area of <math>RPK.</math> Finding the area of <math>\triangle RQL</math> is similar. | ||
+ | |||
+ | We claim that <math>RF = \dfrac{1}{2}b.</math> In order to prove this, we will prove that <math>\triangle RFP \cong \triangle QLC.</math> In other words, we wish to prove that <math>PR = QC.</math> This is equivalent to proving that <math>PC + QC = CR.</math> | ||
+ | |||
+ | Note that <math>PC = \dfrac{PK}{\sin \frac{1}{2}C}</math> and <math>QC = \dfrac{QL}{\sin \frac{1}{2}C}.</math> Therefore, we get that \begin{align*} | ||
+ | &PC + QC = \dfrac{PK}{\sin \frac{1}{2}C} + \dfrac{QL}{\sin\frac{1}{2}C}\\ | ||
+ | & = \dfrac{PK + QL}{\sin\frac{1}{2}C}\\ | ||
+ | & = \dfrac{PK(1 + \frac{b}{a})}{\sin\frac{1}{2}C} \\ | ||
+ | & = \dfrac{PK(\frac{a + b}{a})}{\sin\frac{1}{2}C} \\ | ||
+ | & = \dfrac{\frac{1}{2}a\tan\frac{1}{2}C(a + b)}{a\sin\frac{1}{2}C} \\ | ||
+ | & = \dfrac{\frac{1}{2}a\sin\frac{1}{2}C(a + b)}{a\sin\frac{1}{2}C\cos\frac{1}{2}C} \\ | ||
+ | & = \dfrac{\frac{1}{2}C(a + b)}{2\sin{1}{2}C\cos\frac{1}{2}C} \\ | ||
+ | & = \dfrac{\frac{1}{2}C(a + b)}{\sin C} \\ | ||
+ | & = \dfrac{\frac{1}{2}C(a + b)}{c} \\ | ||
+ | &= CR. | ||
+ | \end{align*} Thus, <math>RF = \dfrac{1}{2}b.</math> In this way, we get that the altidude from <math>R</math> to <math>QL</math> has length <math>\dfrac{1}{2}a.</math> Therefore, we see that <math>[RPK] = \dfrac{1}{8}ab \tan \frac{1}{2}C</math> and <math>[RQL] = \dfrac{1}{8}ab \tan \frac{1}{2}C,</math> so the two areas are equal. | ||
+ | |||
+ | Solution by Ilikeapos | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 18:32, 8 August 2019
Problem
In the bisector of
intersects the circumcircle again at
, the perpendicular bisector of
at
, and the perpendicular bisector of
at
. The midpoint of
is
and the midpoint of
is
. Prove that the triangles
and
have the same area.
Solution
The area of is given by
and the area of
is
. Let
,
, and
. Now
and
, thus
.
, so
, or
. The ratio of the areas is
. The two areas are only equal when the ratio is 1, therefore it suffices to show
. Let
be the center of the circle. Then
, and
. Using law of sines on
we have:
so
.
by law of sines, and
, thus 1)
. Similarly, law of sines on
results in
or
. Cross multiplying we have
or 2)
. Dividing 1) by 2) we have
Solution 2 (Power of a point)
, and similarly
, we have
. Using triangle area formula
, the problem is equivalent to proving
, or
. Draw line
perpendicular to BC and intersects BC at
, then
, and
. Now the problem is equivalent to proving
, or
. Since
, we have
. Let the radius of the circumcircle be
, then the diameter through
is divided by point
into lengths of
and
. By power of point,
. Similarly,
. Therefore
.
Solution 3
WLOG, let the diameter of be
We see that and
from right triangles
and
We now look at By the Extended Law of Sines on
we get that
Similarly,
We now look at By Ptolemy's Theorem, we have
which gives us
This means that
We now seek to relate the lengths computed with the areas.
To do this, we consider the altitude from to
This is to find the area of
Finding the area of
is similar.
We claim that In order to prove this, we will prove that
In other words, we wish to prove that
This is equivalent to proving that
Note that and
Therefore, we get that \begin{align*}
&PC + QC = \dfrac{PK}{\sin \frac{1}{2}C} + \dfrac{QL}{\sin\frac{1}{2}C}\\
& = \dfrac{PK + QL}{\sin\frac{1}{2}C}\\
& = \dfrac{PK(1 + \frac{b}{a})}{\sin\frac{1}{2}C} \\
& = \dfrac{PK(\frac{a + b}{a})}{\sin\frac{1}{2}C} \\
& = \dfrac{\frac{1}{2}a\tan\frac{1}{2}C(a + b)}{a\sin\frac{1}{2}C} \\
& = \dfrac{\frac{1}{2}a\sin\frac{1}{2}C(a + b)}{a\sin\frac{1}{2}C\cos\frac{1}{2}C} \\
& = \dfrac{\frac{1}{2}C(a + b)}{2\sin{1}{2}C\cos\frac{1}{2}C} \\
& = \dfrac{\frac{1}{2}C(a + b)}{\sin C} \\
& = \dfrac{\frac{1}{2}C(a + b)}{c} \\
&= CR.
\end{align*} Thus,
In this way, we get that the altidude from
to
has length
Therefore, we see that
and
so the two areas are equal.
Solution by Ilikeapos
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
2007 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |