circleseuclidean-geometrygeometrytrianglestrigonometry
As title suggests, the question is to find the largest length's measure (the blue length) in the figure below. Note that the blue length is NOT passing through the center of the circle and the orange length is tangent to the circle. I tried several different approaches to this problem, both purely geometric and trigonometric, but none of them provided a simple solution until I recalled an important theorem. Please post your own solutions as well, I will have mine as an answer below!
The large square's area is always the sum of the other two squares' areas.
Say $\triangle ABC$ is our isosceles right triangle, with right angle at $A$. $D$ and $E$ are on hypotenuse $BC$, and $\angle DAE$ measures $45^\circ$. Reflect $C$ across line $AD$ to find point $F$ — that is, $\angle CAD = \angle DAF$ and $AC=AF$. Then
$$ 90^\circ = \angle CAB = \angle CAD + \angle DAE + \angle EAB = \angle CAD + \angle EAB + 45^\circ $$ $$ 45^\circ = \angle CAD + \angle EAB $$ $$ 45^\circ = \angle DAE = \angle DAF + \angle FAE $$
Since by construction $\angle CAD = \angle DAF$, this gives $\angle EAB = \angle FAE$. We then have (SAS) two congruent triangle pairs, $\triangle ACD \cong \triangle AFD$ and $\triangle ABE \cong \triangle AFE$.
The congruences give $\angle DFA = \angle DCA = 45^\circ$ and $\angle AFE = \angle ABE = 45^\circ$, so $\angle DFE = \angle DFA + \angle AFE = 90^\circ$, and $\triangle DEF$ is a right triangle. By the Pythagorean theorem, $DF^2+EF^2=DE^2$. By the congruent triangles again, $CD=DF$ and $BE=EF$. Finally,
$$ CD^2+BE^2=DE^2 $$
This is going to be my approach to the problem:
Here's my approach for the problem:
1.) Extend $CD$ to meet $E$ that lies on the circumference. Join the center of the circle $O$ with $E$ via segment $OE$. Since $OE$ is radius, we know that $OE=9$. Notice that $\angle ECB$ is the inscribed angle of $\angle EOC$, therefore we can conclude that $\angle EOB=2\alpha$
2.) Notice that $\angle EOA=\angle EDB=180-2\alpha$, therefore $OE=ED=9$. Now we can use the intersecting chords theorem (which can easily be proven via similarity of triangles in any general cyclic quadrilateral) and conclude that:
$$9\cdot CD=6\cdot12=72$$
$$\Rightarrow CD=8$$
Best Answer
Triangle cosine Rule applied twice, alternate segment angle, outside tangent squared is product segment theorem are used
I quickly got two results by above trig numerical method. It is only slightly different from the first one by @Goku. Others complex, discarded. Can you spot my mistake in the second solution on
Mathematica? Or, is a second solution also admissible?Two values of long side $ BC= x+y ~$ are
$$(1.5+4.5, 1.50735+4.46341=6, 5.97076)$$
$ (x,y)$ are segments in the blue line outside and inside respectively.
$$ ( x^2=3^2+2^2-2\cdot 3 \cdot 2~\cos \beta,~y^2+4^2-2 \cdot 4 y ~\cos \beta = 2^2,~ x*(x+y)=9 ~)$$