Laurent Schwartz on learning mathematics

Laurent Schwartz, as quoted from  A Mathematician Grappling with His Century, Birkhäuser Basel, 2001, pp. 30-31. [With thanks to Jonathan Crabtree]

I was always deeply uncertain about my own intellectual capacity; I thought I was unintelligent. And it is true that I was, and still am, rather slow. I need time to seize things because I always need to understand them fully. Even when I was the first to answer the teacher’s questions, I knew it was because they happened to be questions to which I already knew the answer. But if a new question arose, usually students who weren’t as good as I was answered before me. Towards the end of eleventh grade, I secretly thought of myself as stupid. I worried about this for a long time. Not only did I believe I was stupid, but I couldn’t understand the contradiction between this stupidity and my good grades. I never talked about this to anyone, but I always felt convinced that my imposture would someday be revealed: the whole world and myself would finally see that what looked like intelligence was really just an illusion. If this ever happened, apparently no one noticed it, and I’m still just as slow. When a teacher dictated something to us, I had real trouble taking notes; it’s still difficult for me to follow a seminar.

At the end of eleventh grade, I took the measure of the situation, and came to the conclusion that rapidity doesn’t have a precise relation to intelligence. What is important is to deeply understand things and their relations to each other. This is where intelligence lies. The fact of being quick or slow isn’t really relevant. Naturally, it’s help to be quick, like it is to have a good memory. But it’s neither necessary or sufficient for intellectual success.The laurels I won in the Concourse General liberated me definitely from my anguish. I won first prize in Latin theme and first access it in Latin version; I was no longer merely a brilliant high school student, I acquired national fame. The Concourse General counted a lot in in my life, by helping me to get rid of a terrible complex. Of course, I was not instantly metamorphosed, and I’ve always had to confront the same problems; it’s just that since that day I know that these obstacles are not unsurmountable and that in spite of delicate and even painful moments,they will not block my way to accomplishment, which research represent for me. Fortunately, I had an excellent memory. For instance, in twelfth grade, in math, I believe that at the end of the year I remembered every single thing I had learned, without ever have written anything down. At that point, I knew my limits but I had a solid feeing of confidence in my possibility of success.

This type of competition is an excellent thing. Many young people feel self-doubt, for one reason or another. The refusal of any kind of comparison which reigns in our classrooms as a concession to egalitarianism, is all too often quite destructive; it prevents the young people who doubt their own capacities, and particularly those from modest backgrounds, from acquiring real confidence in themselves. But self-confidence is a condition of success. Of course, one must be modest, and every intellectual needs to recall this. I’m perfectly conscious of the immensity of my ignorance compared with what I know. It’s enough to meet other intellectuals to see that my knowledge is just a drop of water in an ocean. Every intellectual needs to be capable of considering himself relatively, and measuring the immensity of his ignorance. But he must also have confidence in himself and in his possibilities of succeeding, through the constant and tenacious search for truth.

Source. 

XXXIII Olimpiada Iberoamericana de Matemáticas - Problema 2

Sea $ABC$ un triángulo tal que $\angle{BAC}=90^\circ$ y $BA=CA$. Sea $M$ el punto medio de $BC$. Un punto $D\neq{A}$ es elegido en la semicircunferencia de diámetro $BC$ que contiene a $A$. La circunferencia circunscrita al triangulo $DAM$ intersecta a las rectas $DB$ y $DC$ en los puntos $E$ y $F$, respectivamente. Demuestre que $BE=CF$. 

Demostración. Considera el cuadrilátero cíclico $ABCD$. Como $\angle{ACB}=45^\circ$, entonces, $\angle{BDA}=180^\circ-45^\circ=135^\circ$. Ahora considera el cuadrilátero cíclico $AMED$. Como $\angle{BDA}=135^\circ$, entonces, $\angle{AME}=180^\circ-135^\circ=45^\circ$. Al ser $ABC$ un triángulo rectángulo isósceles, $\angle{MAC}=\angle{AME}=45^\circ$, significando que $AC\parallel{EM}$. Por propiedad del triángulo medial sabemos que una línea que pase por $M$ y sea paralela al lado $AC$ debe contener el punto medio de $AB$, de modo que $EM$ es la mediatriz del lado $AB$. Análogamente, $FM$ es la mediatriz del lado $AC$. Llamemos $N$ y $O$ a los puntos medios de los lados $AB$ y $AC$, respectivamente. Note que por propiedad de ángulos inscritos en una circunferencia, $\angle{ABD}=\angle{ACD}$. También $\angle{BNE}=\angle{COF}=90^\circ$ y $BN=CO$. Por el criterio de congruencia de triángulos $ALA$, $\triangle{BNE}\cong{\triangle{COF}}$, y por lo tanto, $BE=CF$. 

$\square$

Tres Círculos Congruentes y un Triángulo Equilátero

En la figura adjunta los tres círculos son iguales de radio 4 y tangentes externos dos a dos, siendo $P_1$, $P_2$ y $P_3$ puntos de tangencia y los segmentos $P_1P_2=P_2P_3=P_3P_1$. Hallar el área del triángulo $P_1P_2P_3$. Este problema fue propuesto por Walter Sivoli en el grupo de Facebook Matemáticas Puras y Aplicadas.

Solución.

En general, denotaremos el radio de los tres círculos con $r$ y el lado del triángulo con $l$. En la figura, note que al ser $P_2P_3$ tangente al círculo $(O_2, O_2P_2)$ en $P_2$ el triángulo $O_2P_2P_3$ es rectángulo, por lo que, aplicando el Teorema de Pitágoras tenemos

$$O_2P_3^2=P_2P_3^2+O_2P_2^2$$

$$O_2P_3^2=l^2+r^2$$

$$O_2P_3=\sqrt{l^2+r^2}$$

Note que al ser $P_1P_3$ tangente al círculo $(O_3, O_3P_3)$ en $P_3$, el ángulo $\angle{O_3P_3P_1}=90^\circ$. Como $P_1P_2P_3$ es un triángulo equilátero, tenemos que 

$$\angle{O_3P_3P_2}=\angle{O_3P_3P_1}-\angle{P_1P_2P_3}=90^\circ-60^\circ=30^\circ$$

El seno y el coseno del ángulo $\angle{O_2P_3P_2}$ están dados por $\frac{r}{\sqrt{l^2+r^2}}$ y $\frac{l}{\sqrt{l^2+r^2}}$, respectivamente. Ahora, fíjese que por la Ley de Cosenos tenemos la siguiente ecuación:

$$O_2P_3^2+O_3P_3^2-2\cdot{O_2P_3}\cdot{O_3P_3}\cdot{\cos{(\angle{O_3P_3P_2}+\angle{P_2P_3O_2})}}=O_2O_3^2$$

Reescribiendo en términos de $l$ y $r$, y usando la identidad para el coseno de la suma de dos ángulos, tenemos

$$l^2+r^2+r^2-2\cdot{\sqrt{l^2+r^2}}\cdot{r}[(\frac{\sqrt{3}}{2})(\frac{l}{\sqrt{l^2+r^2}})-(\frac{1}{2})(\frac{r}{\sqrt{l^2+r^2}})]=4r^2$$

Simplificando,

$$l^2+3r^2-r\cdot{l}\sqrt{3}=4r^2$$

Resolviendo la ecuación para $l$:

$$l=\frac{r(\sqrt{7}+\sqrt{3})}{2}$$

Así, para el caso del problema donde $r=4$, 

$$l=2(\sqrt{7}+\sqrt{3})$$

El área para un triángulo equilátero es $\frac{l^2\sqrt{3}}{4}$. Así el área buscada es

$$10\sqrt{3}+6\sqrt{7}$$