ABC is a triangle with ∠BAC=60∘. It has an incircle Γ, which is tangential to BC at D. It is given that BD=3 and DC=4. What is the value of [ABC]^2?
432
How???
To find the value of \([ABC]^2\), we need to determine the area of triangle ABC. Let's break down the problem step by step.
1. First, draw triangle ABC and label the points and lengths as given in the problem:
B
/ \
/ \
/ \
/ \
D /_________\ C
----3-----4----
2. Since ∠BAC = 60° and triangle ABC is a triangle with an incircle Γ, it means that AD is an angle bisector. Let's label AD as x.
B
/ \
/ \
/ \
/ \
D /_____x_____ C
----3-----4----
3. Now, we can use the angle bisector theorem to find the lengths of AB and AC. According to the theorem, the lengths of AB and AC are proportional to BD and CD, respectively. Therefore:
AB/BD = AC/CD
Substituting the given values, we have:
AB/3 = AC/4
Cross-multiplying, we get:
4AB = 3AC
4. Now, let's find the length of BC. We can use the Law of Cosines for triangle ABC. The formula is:
BC² = AB² + AC² - 2 * AB * AC * cos(∠BAC)
Since ∠BAC is 60°, we have:
BC² = AB² + AC² - 2 * AB * AC * cos(60°)
5. Since AB = 3AC/4 (from step 3), we can substitute this into the equation from step 4:
BC² = (3AC/4)² + AC² - 2 * (3AC/4) * AC * cos(60°)
Simplifying, we have:
BC² = 9AC²/16 + AC² - 3AC²/2
Combining like terms, we get:
BC² = 25AC²/16
6. Now, we need to find the area of triangle ABC. We can use Heron's Formula, which states:
[ABC] = √(s * (s - AB) * (s - AC) * (s - BC))
Where s is the semi-perimeter of the triangle:
s = (AB + AC + BC)/2
Substituting the given values, we have:
s = (3AC/4 + AC + √(25AC²/16))/2
Simplifying, we get:
s = (13AC/8 + √(25AC²/16))/2
7. Now, let's substitute the value of s into Heron's Formula:
[ABC]² = (s * (s - AB) * (s - AC) * (s - BC))²
= ((13AC/8 + √(25AC²/16))/2 * ((13AC/8 + √(25AC²/16))/2 - 3AC/4) * ((13AC/8 + √(25AC²/16))/2 - AC) * ((13AC/8 + √(25AC²/16))/2 - √(25AC²/16)))²
Simplifying this equation will yield the final value of [ABC]².